OXFORD IB STUDY GUIDES
Ti Kik
Physics
o r T h e I B d I p lo m a
2014 edition
2
3
Great Clarendon Street, Oxford, OX2 6DP, United Kingdom
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Introduction and acknowledgements
Many people seem to think that you have to be really
clever to understand Physics and this puts some people
o studying it in the rst place. So do you really need a
brain the size o a planet in order to cope with IB Higher
Level Physics? The answer, you will be pleased to hear, is
No. In act, it is one o the worlds best kept secrets that
Physics is easy! There is very little to learn by heart and
even ideas that seem really difcult when you rst meet
them can end up being obvious by the end o a course o
study. But i this is the case why do so many people seem
to think that Physics is really hard?
I think the main reason is that there are no saety nets
or short cuts to understanding Physics principles. You
wont get ar i you just learn laws by memorising them
and try to plug numbers into equations in the hope o
getting the right answer. To really make progress you need
to be amiliar with a concept and be completely happy
that you understand it. This will mean that you are able
to apply your understanding in unamiliar situations. The
hardest thing, however, is oten not the learning or the
understanding o new ideas but the getting rid o wrong
and conused every day explanations.
This book should prove useul to anyone ollowing a preuniversity Physics course but its structure sticks very
closely to the recently revised International Baccalaureate
syllabus. It aims to provide an explanation (albeit very
brie) o all o the core ideas that are needed throughout
the whole IB Physics course. To this end each o the
sections is clearly marked as either being appropriate
or everybody or only being needed by those studying at
Higher level. The same is true o the questions that can be
ound at the end o the chapters.
I would like to take the opportunity to thank the many
people that have helped and encouraged me during the
writing o this book. In particular I need to mention David
Jones and Paul Ruth who provided many useul and
detailed suggestions or improvement unortunately
there was not enough space to include everything. The
biggest thanks, however, need to go to Betsan or her
support, patience and encouragement throughout the
whole project.
Tim Kirk
October 2002
Third edition
Since the IB Study Guide's rst publication in 2002, there
have been two signicant IB Diploma syllabus changes.
The aim, to try and explain all the core ideas essential or
the IB Physics course in as concise a way as possible,
has remained the same. I continue to be grateul to all the
teachers and students who have taken time to comment
and I would welcome urther eedback. In addition to
the team at OUP, I would particularly like to thank my
exceptional colleagues and all the outstanding students
at my current school, St. Dunstan's College, London. It
goes without saying that this third edition could not have
been achieved without Betsan's continued support and
encouragement.
This book is dedicated to the memory o my ather,
Francis Kirk.
Tim Kirk
August 2014
I n tr o d u c tI o n an d ac kn o wle d g e m e n ts
iii
Contents
(Italics denote topics which are exclusively Higher Level)
1 measurement and uncertaIntIes
The realm o physics range o magnitudes o
quantities in our universe
The SI system o undamental and derived units
Estimation
Uncertainties and error in experimental measurement
Uncertainties in calculated results
Uncertainties in graphs
Vectors and scalars
IB Questions measurement and uncertainties
1
2
3
4
5
6
7
8
2 mechanIcs
Motion
Graphical representation o motion
Uniormly accelerated motion
Projectile motion
Fluid resistance and ree-all
Forces and ree-body diagrams
Newtons rst law
Equilibrium
Newtons second law
Newtons third law
Mass and weight
Solid riction
Work
Energy and power
Momentum and impulse
IB Questions mechanics
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
3 thermal PhYsIcs
Thermal concepts
Heat and internal energy
Specic heat capacity
Phases (states) o matter and latent heat
The gas laws 1
The gas laws 2
Molecular model o an ideal gas
IB Questions thermal physics
25
26
27
28
29
30
31
32
4 waVes
Oscillations
Graphs o simple harmonic motion
Travelling waves
Wave characteristics
Electromagnetic spectrum
Investigating speed o sound experimentally
iv
co n te n ts
33
34
35
36
37
38
Intensity
Superposition
Polarization
Uses o polarization
Wave behaviour Refection
Snells law and reractive index
Reraction and critical angle
Diraction
Two-source intererence o waves
Nature and production o standing (stationary)
waves
Boundary conditions
IB Questions waves
39
40
41
42
43
44
45
46
47
48
49
50
5 electrIcItY and magnetIsm
Electric charge and Coulomb's law
Electric elds
Electric potential energy and electric potential
dierence
Electric current
Electric circuits
Resistors in series and parallel
Potential divider circuits and sensors
Resistivity
Example o use o Kircho's laws
Internal resistance and cells
Magnetic orce and elds
Magnetic orces
Examples o the magnetic eld due to currents
IB Questions electricity and magnetism
51
52
53
54
55
56
57
58
59
60
61
62
63
64
6 cIrcular motIon and graVItatIon
Uniorm circular motion
Angular velocity and vertical circular motion
Newtons law o gravitation
IB Questions circular motion and gravitation
65
66
67
68
7 atomIc, nuclear and
PartIcle PhYsIcs
Emission and absorption spectra
Nuclear stability
Fundamental orces
Radioactivity 1
Radioactivity 2
Hal-lie
Nuclear reactions
69
70
71
72
73
74
75
Fission and usion
Structure o matter
Description and classifcation o particles
Quarks
Feynman diagrams
IB Questions atomic, nuclear and particle physics
76
77
78
79
80
81
8 energY ProductIon
Energy and power generation Sankey diagram
Primary energy sources
Fossil uel power production
Nuclear power process
Nuclear power saety and risks
Solar power and hydroelectric power
Wind power and other technologies
Thermal energy transer
Radiation: Wiens law and the SteanBoltzmann law
Solar power
The greenhouse eect
Global warming
IB Questions energy production
82
83
84
85
86
87
88
89
90
91
92
93
94
9 WavE phEnomEna
Sie ri ti
Eergy ges drig sie ri ti
Dirti
Tw-sre itereree wes:
Ygs dbe-sit exeriet
mtie-sit dirti
Ti re fs
Resti
Te Der eet
xes d itis te Der eet
IB Qestis we ee
95
96
97
98
99
100
101
102
103
104
10 IElDS
pteti (gritti d eetri)
Eqitetis
Gritti teti eergy d teti
orbit ti
Eetri teti eergy d teti
Eetri d Gritti ieds red
IB Qestis feds
105
106
107
108
109
110
111
11 ElEcTRomaGnETIc InDucTIon
Induced electromotive force (emf)
lez's w d rdy's w
atertig rret (1 )
atertig rret (2)
Retifti d stig irits
cite
112
113
114
115
116
117
118
119
120
citr disrge
citr rge
IB Qestis eetrgeti idti
12 QuanTum anD nuclEaR phYSIcS
121
122
123
124
125
pteetri eet
mtter wes
ati setr d ti eergy sttes
Br de te t
Te Srdiger de te t
Te heiseberg ertity riie d
te ss deteriis
Teig, teti brrier d trs etig
teig rbbiity
Te es
ner eergy ees d rditie dey
IB Qestis qt d er ysis
126
127
128
129
130
13 oPtIon a relatIVItY
Reerence rames
Maxwells equations
Special relativity
Lorentz transormations
Velocity addition
Invariant quantities
Time dilation
Length contraction and evidence to support
special relativity
Spacetime diagrams ( Minkowski diagrams) 1
Spacetime diagrams 2
The twin paradox 1
Twin paradox 2
Spacetime diagrams 3
mss d eergy
Retiisti et d eergy
Retiisti eis exes
Geer retiity te eqiee riie
Gritti red sit
Srtig eidee
crtre setie
Bk es
IB Questions option A relativity
131
132
133
134
135
136
137
138
139
140
140
141
142
143
144
145
146
147
148
149
150
151
14 oPtIon B engIneerIng PhYsIcs
Translational and rotational motion
Translational and rotational relationships
Translational and rotational equilibrium
Equilibrium examples
Newtons second law moment o inertia
Rotational dynamics
c o n te n ts
152
153
154
155
156
157
v
Solving rotational problems
Thermodynamic systems and concepts
Work done by an ideal gas
The rst law o thermodynamics
Second law o thermodynamics and entropy
Heat engines and heat pumps
lids t rest
lids i motio Berolli efect
Berolli exmples
viscosity
orced oscilltios d resoce (1 )
Resoce (2)
IB Questions option B engineering physics
158
159
160
161
162
163
164
165
166
167
168
169
170
15 oPtIon c ImagIng
Image ormation
Converging lenses
Image ormation in convex lenses
Thin lens equation
Diverging lenses
Converging and diverging mirrors
The simple magniying glass
Aberrations
The compound microscope and
astronomical telescope
Astronomical refecting telescopes
Radio telescopes
Fibre optics
Dispersion, attenuation and noise in optical bres
Channels o communication
X-rys
X-ry imgig techiqes
ultrsoic imgig
Imgig cotied
IB Questions option C imaging
vi
co n te n ts
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
16 oPtIon d astroPhYsIcs
Objects in the universe (1)
Objects in the universe (2)
The nature o stars
Stellar parallax
Luminosity
Stellar spectra
Nucleosynthesis
The HertzsprungRussell diagram
Cepheid variables
Red giant stars
Stellar evolution
The Big Bang model
Galactic motion
Hubbles law and cosmic scale actor
The accelerating universe
ncler sio the Jes criterio
ncleosythesis of the mi seqece
Types o speroe
The cosmologicl priciple d mthemticl models
Rottio cres d drk mtter
The history o the uierse
The tre o the uierse
Drk eergy
astrophysics reserch
IB Questions astrophysics
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
17 aPPendIX
Graphs
215
Graphical analysis and determination o relationships 216
Grphicl lysis logrithmic ctios
217
ANSWERS
ORIGIN OF INDIVIDUAL QUESTIONS
INDEX
218
218
219
1 M E AS U R E M E N T AN D U N CE R TAI N TI E S
Te ream o pysics rane o manitudes o
quantities in our universe
ORDERS Of MAgNITUDE
INClUDINg ThEIR RATIOS
Physics seeks to explain nothing less than the
Universe itself. In attempting to do this, the
range of the magnitudes of various quantities
will be huge.
If the numbers involved are going to mean
anything, it is important to get some feel
for their relative sizes. To avoid getting lost
among the numbers it is helpful to state them
to the nearest order of magnitude or power
of ten. The numbers are just rounded up or
down as appropriate.
Comparisons can then be easily made because
working out the ratio between two powers of
ten is just a matter of adding or subtracting
whole numbers. The diameter of an atom,
1 0 - 1 0 m, does not sound that much larger
than the diameter of a proton in its nucleus,
1 0 - 1 5 m, but the ratio between them is 1 0 5 or
1 00,000 times bigger. This is the same ratio as
between the size of a railway station (order
of magnitude 1 0 2 m) and the diameter of the
Earth (order of magnitude 1 0 7 m) .
electrons
RANgE Of MASSES
10 52
10 48
10 44
10 40
10 36
10 32
10 28
10 24
10 20
10 16
10 12
10 8
10 4
10 0
10 -4
10 -8
10 - 12
10 - 16
10 - 20
10 - 24
10 - 28
10 - 32
Mass / kg
total mass of observable
Universe
mass of local galaxy
(Milky Way)
mass of Sun
mass of Earth
total mass of oceans
total mass of atmosphere
laden oil supertanker
elephant
human
mouse
grain of sand
blood corpuscle
bacterium
haemoglobin molecule
proton
electron
RANgE Of lENgThS
10 26
10 24
10 22
10 20
10 18
10 16
10 14
10 12
10 10
10 8
10 6
10 4
10 2
10 0
10 - 2
10 - 4
10 - 6
10 - 8
10 - 10
10 - 12
10 - 14
10 - 16
Size / m
radius of observable Universe
radius of local galaxy (Milky Way)
distance to nearest star
distance from Earth to Sun
distance from Earth to Moon
radius of the Earth
deepest part of the
ocean / highest mountain
tallest building
length of ngernail
thickness of piece of paper
human blood corpuscle
wavelength of light
diameter of hydrogen atom
wavelength of gamma ray
diameter of proton
protons
RANgE Of TIMES
Carbon atom
railway
station
Earth
For example, you would probably feel very
pleased with yourself if you designed a new,
environmentally friendly source of energy
that could produce 2.03 1 0 3 J from 0.72 kg
of natural produce. But the meaning of these
numbers is not clear is this a lot or is it a
little? In terms of orders of magnitudes, this
new source produces 1 0 3 joules per kilogram
of produce. This does not compare terribly
well with the 1 0 5 joules provided by a slice of
bread or the 1 0 8 joules released per kilogram
of petrol.
You do NOT need to memorize all of the
values shown in the tables, but you should
try and develop a familiarity with them.
10 20
10 18
10 16
10 14
10 12
10 10
10 8
10 6
10 4
10 2
10 0
10 - 2
10 - 4
10 - 6
10 - 8
10 - 10
10 - 12
10 - 14
10 - 16
10 - 18
10 - 20
10 - 22
10 - 24
Time / s
age of the Universe
RANgE Of ENERgIES
10 44
age of the Earth
10 34
age of species Homo
sapiens
10 30
typical human lifespan
1 year
1 day
10 26
10 22
10 18
10 14
heartbeat
10 10
period of high-frequency
sound
10 6
10 2
passage of light across
a room
10 - 2
10 - 6
vibration of an ion in a solid
period of visible light
Energy / J
energy released in a supernova
energy radiated by Sun in 1 second
energy released in an earthquake
energy released by annihilation of
1 kg of matter
energy in a lightning discharge
energy needed to charge a car
battery
kinetic energy of a tennis ball
during game
energy in the beat of a ys wing
10 - 10
10 - 14
passage of light across
an atom
passage of light across
a nucleus
10 - 18
10 - 22
energy needed to remove electron
from the surface of a metal
10 - 26
M E A S U R E M E N T A N D U N C E R TA I N T I E S
1
The SI system o undamenta and deried units
fUNDAMENTAl UNITS
Any measurement and every quantity can be thought o as
being made up o two important parts:
1.
the number and
2.
the units.
Without both parts, the measurement does not make sense.
For example a persons age might be quoted as seventeen
but without the years the situation is not clear. Are they
1 7 minutes, 1 7 months or 1 7 years old? In this case you would
know i you saw them, but a statement like
length = 4.2
actually says nothing. Having said this, it is really surprising to
see the number o candidates who orget to include the units in
their answers to examination questions.
In order or the units to be understood, they need to be defned.
There are many possible systems o measurement that have
DERIvED UNITS
Having fxed the undamental units, all other measurements
can be expressed as dierent combinations o the undamental
units. In other words, all the other units are derived units. For
example, the undamental list o units does not contain a unit
or the measurement o speed. The defnition o speed can be
used to work out the derived unit.
been developed. In science we use the International System o
units (SI) . In SI, the fundamental or base units are as ollows
Quantity
SI unit
SI symbol
Mass
kilogram
kg
Length
metre
m
Time
second
s
Electric current
ampere
A
Amount o substance
mole
mol
Temperature
kelvin
K
(Luminous intensity
candela
cd)
You do not need to know the precise defnitions o any o these
units in order to use them properly.
are so large that the SI unit (the metre) always involves large
orders o magnitudes. In these cases, the use o a dierent
(but non SI) unit is very common. Astronomers can use the
astronomical unit (AU) , the light-year (ly) or the parsec (pc)
as appropriate. Whatever the unit, the conversion to SI units is
simple arithmetic.
1 AU = 1 .5 1 0 1 1 m
distance
Since speed = _
time
units o distance
Units o speed = __
units o time
metres
= _
(pronounced metres per second)
seconds
m
= _
s
= m s 1
O the many ways o writing this unit, the last way (m s 1 ) is the
best.
Sometimes particular combinations o undamental units
are so common that they are given a new derived name. For
example, the unit o orce is a derived unit it turns out to be
kg m s - 2 . This unit is given a new name the newton (N) so that
1 N = 1 kg m s - 2 .
The great thing about SI is that, so long as the numbers that are
substituted into an equation are in SI units, then the answer
will also come out in SI units. You can always play sae by
converting all the numbers into proper SI units. Sometimes,
however, this would be a waste o time.
There are some situations where the use o SI becomes
awkward. In astronomy, or example, the distances involved
1 ly = 9.5 1 0 1 5 m
1 pc = 3.1 1 0 1 6 m
There are also some units (or example the hour) which are so
common that they are oten used even though they do not orm
part o SI. Once again, beore these numbers are substituted
into equations they need to be converted. Some common unit
conversions are given on page 3 o the IB data booklet.
The table below lists the SI derived units that you will meet.
SI derived unit
SI base unit
Alternative SI unit
newton (N)
kg m s
-
pascal (Pa)
kg m- 1 s - 2
N m- 2
-
-2
hertz (Hz)
s
joule (J)
kg m2 s - 2
Nm
watt (W)
kg m s
J s- 1
coulomb (C)
As
volt (V)
-1
2
-3
-
kg m s
-3
ohm ()
kg m s
-3
weber (Wb)
kg m2 s - 2 A- 1
Vs
tesla (T)
kg s
Wb m- 2
becquerel (Bq)
s- 1
2
2
-2
A
-1
A
-1
WA- 1
A
-2
VA- 1
-
PREfIxES
To avoid the repeated use o scientifc notation, an alternative is to use one o the list o agreed prefxes given on page 2 in the IB data
booklet. These can be very useul but they can also lead to errors in calculations. It is very easy to orget to include the conversion actor.
1W
For example, 1 kW = 1 000 W. 1 mW = 1 0 - 3 W (in other words, ____
)
1 000
2
M E A S U R E M E N T A N D U N C E R TA I N T I E S
Estimation
ORDERS Of MAgNITUDE
It is important to develop a eeling or some o the numbers
that you use. When using a calculator, it is very easy to make
a simple mistake (eg by entering the data incorrectly) . A good
way o checking the answer is to rst make an estimate beore
resorting to the calculator. The multiple-choice paper (paper 1 )
does not allow the use o calculators.
Approximate values or each o the undamental SI units are
given below.
1 kg
A packet o sugar, 1 litre o water. A person would be
about 50 kg or more
1 m
Distance between ones hands with arms outstretched
1 s
Duration o a heart beat (when resting it can easily
double with exercise)
1 amp
Current fowing rom the mains electricity when a
computer is connected. The maximum current to a
domestic device would be about 1 0 A or so
1 kelvin 1 K is a very low temperature. Water reezes at 273 K
and boils at 373 K. Room temperature is about 300 K
1 mol
1 2 g o carbon1 2. About the number o atoms o
carbon in the lead o a pencil
The same process can happen with some o the derived units.
1 m s- 1
Walking speed. A car moving at 30 m s - 1 would be ast
1 ms
Quite a slow acceleration. The acceleration o gravity
is 1 0 m s - 2
-2
1 N
A small orce about the weight o an apple
1 V
Batteries generally range rom a ew volts up to 20 or
so, the mains is several hundred volts
1 Pa
A very small pressure. Atmospheric pressure is about
1 0 5 Pa
1 J
A very small amount o energy the work done
liting an apple o the ground
POSSIblE REASONAblE ASSUMPTIONS
Everyday situations are very complex. In physics we oten simpliy a problem by making simple assumptions. Even i we know
these assumptions are not absolutely true they allow us to gain an understanding o what is going on. At the end o the calculation
it is oten possible to go back and work out what would happen i our assumption turned out not to be true.
The table below lists some common assumptions. Be careul not to assume too much! Additionally we oten have to assume that
some quantity is constant even i we know that in reality it is varying slightly all the time.
Assumption
Example
Object treated as point particle
Mechanics: Linear motion and translational equilibrium
Friction is negligible
Many mechanics situations but you need to be very careul
No thermal energy (heat) loss
Almost all thermal situations
Mass o connecting string, etc. is negligible
Many mechanics situations
Resistance o ammeter is zero
Circuits
Resistance o voltmeter is innite
Circuits
Internal resistance o battery is zero
Circuits
Material obeys Ohms law
Circuits
Machine 1 00% ecient
Many situations
Gas is ideal
Thermodynamics
Collision is elastic
Only gas molecules have perectly elastic collisions
Object radiates as a perect black body
Thermal equilibrium, e.g. planets
SCIENTIfIC NOTATION
SIgNIfICANT fIgURES
Numbers that are too big or too small or decimals are oten
written in scientifc notation:
Any experimental measurement should be quoted with its
uncertainty. This indicates the possible range o values or
the quantity being measured. At the same time, the number
o signifcant fgures used will act as a guide to the amount
o uncertainty. For example, a measurement o mass which
is quoted as 23.456 g implies an uncertainty o 0.001 g
(it has ve signicant gures) , whereas one o 23.5 g implies
an uncertainty o 0.1 g (it has three signicant gures) .
a 1 0b
where a is a number between 1 and 1 0 and b is an integer.
e.g. 1 53.2 = 1 .532 1 0 2 ; 0.00872 = 8.72 1 0 - 3
A simple rule or calculations (multiplication or division) is to
quote the answer to the same number o signicant digits as
the LEAST precise value that is used.
For a more complete analysis o how to deal with uncertainties
in calculated results, see page 5.
M E A S U R E M E N T A N D U N C E R TA I N T I E S
3
Uncertainties and error in experimenta measurement
Systematic and random errors can oten be recognized rom a
graph o the results.
quantity A
ERRORS RANDOM AND SySTEMATIC (PRECISION
AND ACCURACy)
An experimental error just means that there is a dierence
between the recorded value and the perect or correct value.
Errors can be categorized as random or systematic.
Repeating readings does not reduce systematic errors.
perfect results
random error
systematic error
Sources o random errors include
The readability o the instrument.
The observer being less than perect.
quantity B
The eects o a change in the surroundings.
Sources o systematic errors include
An instrument with zero error. To correct or zero error the
value should be subtracted rom every reading.
Perect results, random and systematic errors o two
proportional quantities.
An instrument being wrongly calibrated.
The observer being less than perect in the same way every
measurement.
An accurate experiment is one that has a small systematic
error, whereas a precise experiment is one that has a small
random error.
true value
measured
true
value
value
probability
that result has a
certain value
measured
value
value
value
(a)
(b)
Two examples illustrating the nature o experimental results:
(a) an accurate experiment o low precision
(b) a less accurate but more precise experiment.
ESTIMATINg ThE UNCERTAINTy RANgE
An uncertainty range applies to any
experimental value. The idea is that,
instead o just giving one value that
implies perection, we give the likely
range or the measurement.
1.
Estimating rom frst principles
All measurement involves a readability
error. I we use a measuring cylinder to
fnd the volume o a liquid, we might
think that the best estimate is 73 cm3 ,
but we know that it is not exactly this
value (73.000 000 000 00 cm3 ) .
Uncertainty range is 5 cm3 . We say
volume = 73 5 cm3 .
cm 3
100
90
80
70
60
50
40
30
20
10
Normally the uncertainty range due to
readability is estimated as below.
Device
Example
Uncertainty
Analogue
scale
Rulers, meters with
moving pointers
(hal the smallest
scale division)
gRAPhICAl REPRESENTATION Of UNCERTAINTy
Digital scale
In many situations the best method o presenting and
analysing data is to use a graph. I this is the case, a neat way
o representing the uncertainties is to use error bars. The
graphs below explains their use.
Top-pan balances,
digital meters
(the smallest scale
division)
2.
quantity C
quantity A
Since the error bar represents the uncertainty range, the bestft line o the graph should pass through ALL o the rectangles
created by the error bars.
Estimating uncertainty range rom several repeated
measurements
I the time taken or a trolley to go down a slope is measured fve
times, the readings in seconds might be 2.01 , 1 .82, 1 .97, 2.1 6 and
1 .94. The average o these fve readings is 1 .98 s. The deviation o
the largest and smallest readings can be calculated (2.1 6 - 1 .98
= 0.1 8; 1 .98 - 1 .82 = 0.1 6). The largest value is taken as the
uncertainty range. In this example the time is 1 .98 s 0.1 8 s. It
would also be appropriate to quote this as 2.0 0.2 s.
SIgNIfICANT fIgURES IN UNCERTAINTIES
quantity E
quantity B
mistake
assumed
quantity F
4
quantity D
The best ft line is
included by all the error
bars in the upper two
graphs. This is not true in
the lower graph.
M E A S U R E M E N T A N D U N C E R TA I N T I E S
In order to be cautious when quoting uncertainties, fnal values
rom calculations are oten rounded up to one signifcant
fgure, e.g. a calculation that fnds the value o a orce to
be 4.264 N with an uncertainty o 0.362 N is quoted as
4.3 0.4 N. This can be unnecessarily pessimistic and it is also
acceptable to express uncertainties to two signifcant fgures.
For example, the charge on an electron is 1 .6021 76565 1 0 - 1 9 C
0.000000035 1 0 - 1 9 C. In data booklets this is sometimes
expressed as 1 .6021 76565(35) 1 0 - 1 9 C.
Uncertainties in cacuated resuts
MAThEMATICAl REPRESENTATION Of UNCERTAINTIES
Then the ractional uncertainty is
For example i the mass o a block was measured as 1 0 1 g
and the volume was measured as 5.0 0.2 cm3 , then the ull
calculations or the density would be as ollows.
p
_
p ,
which makes the percentage uncertainty
mass
10
Best value or density = ______
= __
= 2.0 g cm- 3
5
volume
11
The largest possible value o density = ___
= 2.292 g cm- 3
4.8
9
The smallest possible value o density = ___
= 1 .731 g cm- 3
5.2
p
_
p 1 00% .
In the example above, the ractional uncertainty o the density is
0.1 5 or 1 5%.
Thus equivalent ways o expressing this error are
density = 2.0 0.3 g cm- 3
Rounding these values gives density = 2.0 0.3 g cm- 3
We can express this uncertainty in one o three ways using
absolute, fractional or percentage uncertainties.
I a quantity p is measured then the absolute uncertainty would
be expressed as p.
MUlTIPlICATION, DIvISION OR POwERS
Whenever two or more quantities are multiplied or divided
and they each have uncertainties, the overall uncertainty
is approximately equal to the addition o the percentage
(ractional) uncertainties.
Using the same numbers rom above,
m = 1 g
(
1 g
m
_
_
m = 10 g
)
= 0.1 = 1 0%
V = 0.2 cm3
(
)
0.2 cm3 = 0.04 = 4%
V = _
_
5 cm3
V
The total % uncertainty in the result = (1 0 + 4) %
= 14 %
1 4% o 2.0 g cm- 3 = 0.28 g cm- 3 0.3 g cm- 3
So density = 2.0 0.3 g cm- 3 as beore.
OR density = 2.0 g cm- 3 1 5%
Working out the uncertainty range is very time consuming.
There are some mathematical short-cuts that can be used.
These are introduced in the boxes below.
ab
In symbols, i y = _
c
y
b _
a
c [note this is ALWAYS added]
_ _
Then _
y = a + b + c
Power relationships are just a special case o this law.
I y = an
|
|
y
a (always positive)
_
Then ___
y = n a
For example i a cube is measured to be 4.0 0.1 cm in length
along each side, then
0.1
% Uncertainty in length = _
= 2.5 %
4.0
3
3
Volume = (length) = (4.0) = 64 cm3
% Uncertainty in [volume] =
=
=
=
% uncertainty in [(length) 3 ]
3 (% uncertainty in [length] )
3 ( 2.5 % )
7.5 %
Absolute uncertainty = 7.5% o 64 cm3
= 4.8 cm3 5 cm3
Thus volume o cube = 64 5 cm3
OThER MAThEMATICAl OPERATIONS
Oter unctions
I the calculation involves mathematical operations other than
multiplication, division or raising to a power, then one has to
fnd the highest and lowest possible values.
There are no short-cuts possible. Find the highest and lowest
values.
In symbols
sin
Addition or subtraction
Whenever two or more quantities are added or subtracted and
they each have uncertainties, the overall uncertainty is equal to
the addition o the absolute uncertainties.
e.g. uncertainty o sin i = 60 5
1
0.91
0.87
0.82
I y = a b
y = a + b (note ALWAYS added)
uncertainty of thickness in a pipe wall
external radius of pipe
= 6.1 cm 0.1 cm ( 2% )
55 60 65
= 60 5
i
best value to sin = 0.87
max. sin = 0.91
internal radius of pipe
= 5.3 cm 0.1 cm ( 2% )
thickness o pipe wall = 6.1 - 5.3 cm
min. sin = 0.82
sin = 0.87 0.05
worst value used
= 0.8 cm
uncertainty in thickness = (0.1 + 0.1 ) cm
= 0.2 cm
= 25%
M E A S U R E M E N T A N D U N C E R TA I N T I E S
5
Uncertainties in graphs
UNCERTAINTy IN SlOPES
I the gradient o the graph has been used to calculate a
quantity, then the uncertainties o the points will give rise
to an uncertainty in the gradient. Using the steepest and the
shallowest lines possible (i.e. the lines that are still consistent
with the error bars) the uncertainty range or the gradient is
obtained. This process is represented below.
best t line
steepest gradient
quantity a
quantity a
ERROR bARS
Plotting a graph allows one to visualize all the readings at
one time. Ideally all o the points should be plotted with
their error bars. In principle, the size o the error bar could
well be dierent or every single point and so they should be
individually worked out.
shallowest
gradient
quantity b
In practice, it would oten take too much time to add all the
correct error bars, so some (or all) o the ollowing short-cuts
could be considered.
Rather than working out error bars or each point use the
worst value and assume that all o the other error bars are
the same.
Only plot the error bar or the worst point, i.e. the point
that is urthest rom the line o best ft. I the line o best ft
is within the limits o this error bar, then it will probably be
within the limits o all the error bars.
Only plot the error bars or the frst and the last points.
These are oten the most important points when
considering the uncertainty ranges calculated or the
gradient or the intercept (see right) .
Only include the error bars or the axis that has the worst
uncertainty.
quantity b
UNCERTAINTy IN INTERCEPTS
I the intercept o the graph has been used to calculate a
quantity, then the uncertainties o the points will give rise
to an uncertainty in the intercept. Using the steepest and the
shallowest lines possible (i.e. the lines that are still consistent
with the error bars) we can obtain the uncertainty in the
result. This process is represented below.
quantity a
A ull analysis in order to determine the uncertainties in the
gradient o a best straight-line graph should always make
use of the error bars for all of the data points.
best value
for intercept
maximum value
of intercept
minimum value
of intercept
quantity b
6
M E A S U R E M E N T A N D U N C E R TA I N T I E S
vectors and scaars
DIffERENCE bETwEEN vECTORS AND SCAlARS
REPRESENTINg vECTORS
I you measure any quantity, it must have a number AND a
unit. Together they express the magnitude o the quantity.
Some quantities also have a direction associated with them. A
quantity that has magnitude and direction is called a vector
quantity whereas one that has only magnitude is called a
scalar quantity. For example, all orces are vectors.
In most books a bold letter is used to represent a vector
whereas a normal letter represents a scalar. For example F
would be used to represent a orce in magnitude AND
direction. The list below shows some other recognized methods.
F,
F or F
The table lists some common quantities. The frst two quantities
in the table are linked to one another by their defnitions (see
page 9). All the others are in no particular order.
Vectors
Scalars
Displacement
Distance
Velocity
Speed
Acceleration
Mass
Force
Energy (all orms)
Momentum
Temperature
Electric feld strength
Potential or potential
dierence
Magnetic feld strength
Density
Gravitational feld strength
Area
Although the vectors used in many o the given examples are
orces, the techniques can be applied to all vectors.
Vectors are best shown in
diagrams using arrows:
pull
the relative magnitudes
o the vectors involved
are shown by the relative
length o the arrows
friction
normal
reaction
weight
the direction o the
vectors is shown by the
direction o the arrows.
ADDITION / SUbTRACTION Of vECTORS
I we have a 3 N and a 4 N orce, the overall orce (resultant
orce) can be
3N
anything between
=
7N
4N
1 N and 7 N
depending on
5N
3N
the directions
=
involved.
4N
COMPONENTS Of vECTORS
It is also possible to split one vector into two (or more) vectors.
This process is called resolving and the vectors that we get are
called the components o the original vector. This can be a very
useul way o analysing a situation i we choose to resolve all the
vectors into two directions that are at right angles to one another.
Fvertical
F
F
The way to take
3N
the directions
into account
is to do a scale
3N
diagram and use
the parallelogram
law o vectors.
This process is the same as
adding vectors in turn the
tail o one vector is drawn
starting rom the head o
the previous vector.
3N
4N
=
4N
1N
=
b
a+b
a
Parallelogram o vectors
Fhorizontal
Splitting a vector into components
forces
Push
Surface
force
Weight
TRIgONOMETRy
Vector problems can always be solved using scale diagrams,
but this can be very time consuming. The mathematics
o trigonometry oten makes it much easier to use the
mathematical unctions o sine or cosine. This is particularly
appropriate when resolving. The diagram below shows how to
calculate the values o either o these components.
Av
A
A v = Asin
These mutually perpendicular directions are totally
independent o each other and can be analysed separately. I
appropriate, both directions can then be combined at the end
to work out the fnal resultant vector.
components
PV
PH
SH
SV
A H = Acos
AH
See page 1 4 or an example.
W
Pushing a block along a rough surace
M E A S U R E M E N T A N D U N C E R TA I N T I E S
7
Ib Questions measurement and uncertainties
An object is rolled rom rest down an inclined plane. The
distance travelled by the object was measured at seven dierent
times. A graph was then constructed o the distance travelled
against the (time taken) 2 as shown below.
distance travelled/ (cm)
1.
3.
9
8
4.
7
6
A. 0.1 m
C. 1 .0 m
B. 0.2 m
D. 2.0 m
In order to determine the density o a certain type o wood,
the ollowing measurements were made on a cube o the
wood.
Mass
5
= 493 g
Length o each side = 9.3 cm
4
The percentage uncertainty in the measurement o mass is
0.5% and the percentage uncertainty in the measurement
o length is 1 .0% .
3
2
The best estimate or the uncertainty in the density is
1
A. 0.5%
C. 3.0%
0
0.0
B. 1 .5%
D. 3.5%
a) (i)
0.1
0.2
0.3
0.4
0.5
(time taken) 2 / s 2
What quantity is given by the gradient o such
a graph?
5.
[2]
(ii) Explain why the graph suggests that the collected
data is valid but includes a systematic error.
[2]
(iii) Do these results suggest that distance is proportional
to (time taken) 2 ? Explain your answer.
[2]
(iv) Making allowance or the systematic error, calculate
the acceleration o the object.
[2]
b) The ollowing graph shows that same data ater the uncertainty
ranges have been calculated and drawn as error bars.
distance travelled/ (cm)
A stone is dropped down a well and hits the water 2.0 s ater
it is released. Using the equation d = __12 g t2 and taking
g = 9.81 m s - 2 , a calculator yields a value or the depth d o
the well as 1 9.62 m. I the time is measured to 0.1 s then
the best estimate o the absolute error in d is
9
8
Astronauts wish to determine the gravitational acceleration
on Planet X by dropping stones rom an overhanging cli.
Using a steel tape measure they measure the height o the
cli as s = 7.64 m 0.01 m. They then drop three similar
stones rom the cli, timing each all using a hand-held
electronic stopwatch which displays readings to onehundredth o a second. The recorded times or three drops are
2.46 s, 2.31 s and 2.40 s.
a) Explain why the time readings vary by more
than a tenth o a second, although the stopwatch
gives readings to one hundredth o a second.
[1 ]
b) Obtain the average time t to all, and write it in
the orm (value uncertainty) , to the appropriate
number o signifcant digits.
[1 ]
c) The astronauts then determine the gravitational
2s
acceleration a g on the planet using the ormula ag = __
.
t
Calculate ag rom the values o s and t, and determine the
uncertainty in the calculated value. Express the result in
the orm
ag = (value uncertainty) ,
to the appropriate number o signifcant digits.
[3]
7
2
6
5
4
3
HL
2
6.
1
0
0.0
0.1
0.2
0.3
0.4
0.5
(time taken) 2 / s 2
Add two lines to show the range o the possible
acceptable values or the gradient o the graph.
2.
This question is about fnding the relationship between the
orces between magnets and their separations.
In an experiment, two magnets were placed with their Northseeking poles acing one another. The orce o repulsion, f,
and the separation o the magnets, d, were measured and the
results are shown in the table below.
[2]
Separation d/m
The lengths o the sides o a rectangular plate are measured, and
the diagram shows the measured values with their uncertainties.
50 0.5 mm
25 0.5 mm
Which one o the ollowing would be the best estimate o the
percentage uncertainty in the calculated area o the plate?
A. 0.02%
C. 3%
B. 1 %
D. 5%
8
Force o repulsion f/N
0.04
4.00
0.05
1 .98
0.07
0.74
0.09
0.32
a) Plot a graph o log (orce) against log (distance) .
[3]
b) The law relating the orce to the separation is
o the orm
f = kdn
(i)
Use the graph to fnd the value o n.
(ii) Calculate a value or k, giving its units.
I B Q U E S T I o N S M E A S U R E M E N T A N D U N C E R TA I N T I E S
[2]
[3]
2 m e ch an i cs
m
Definitions
These technical terms should not be conused with their everyday use. In particular one should note that
Vector quantities always have a direction associated with them.
Generally, velocity and speed are NOT the same thing. This is particularly important i the object is not going in a straight line.
The units o acceleration come rom its denition. (m s - 1 ) s = m s - 2 .
The denition o acceleration is precise. It is related to the change in velocity (not the same thing as the change in speed) .
Whenever the motion o an object changes, it is called acceleration. For this reason acceleration does not necessarily mean
constantly increasing speed it is possible to accelerate while at constant speed i the direction is changed.
A deceleration means slowing down, i.e. negative acceleration i velocity is positive.
Symbol
Displacement
Velocity
s
v or u
Defnition
Example
The distance moved in a
particular direction.
The displacement rom London to
Rome is 1 .43 1 0 6 m southeast.
The rate o change o
displacement.
change o displacement
velocity = ________________
time taken
Speed
v or u
The rate o change o distance.
distance gone
speed = __________
time taken
Acceleration
a
The rate o change o velocity.
change o velocity
acceleration = _____________
time taken
Vector
The average velocity during a fight
rom London to Rome is 1 60 m s - 1
southeast.
m s- 1
Vector
The average speed during a fight
rom London to Rome is 1 60 m s - 1
m s- 1
Scalar
The average acceleration o a plane
on the runway during take-o is
3.5 m s- 2 in a orwards direction. This
means that on average, its velocity
changes every second by 3.5 m s- 1
m s- 2
Vector
It should be noticed that the average value (over a period
o time) is very dierent to the instantaneous value (at one
particular time) .
In the example below, the positions o a sprinter are shown at
dierent times ater the start o a race.
The average speed over the whole race is easy to work out. It
is the total distance (1 00 m) divided by the total time (1 1 .3 s)
giving 8.8 m s - 1 .
t = 0.0 s
t = 2.0 s
But during the race, her instantaneous speed was changing all the
time. At the end o the rst 2.0 seconds, she had travelled 1 0.04 m.
This means that her average speed over the rst 2.0 seconds was
5.02 m s- 1 . During these rst two seconds, her instantaneous
speed was increasing she was accelerating. I she started at rest
(speed = 0.00 m s- 1 ) and her average speed (over the whole two
seconds) was 5.02 m s- 1 then her instantaneous speed at 2 seconds
must be more than this. In act the instantaneous speed or this
sprinter was 9.23 m s- 1 , but it would not be possible to work this
out rom the inormation given.
d = 28.21 m
d = 47.89 m
d = 69.12 m
t = 4.0 s
t = 6.0 s
t = 8.0 s
frames of reference
I two things are moving in the same
straight line but are travelling at dierent
speeds, then we can work out their
relative velocities by simple addition or
subtraction as appropriate. For example,
imagine two cars travelling along a
straight road at dierent speeds.
I one car (travelling at 30 m s - 1 )
overtakes the other car (travelling at
25 m s - 1 ) , then according to the driver o
the slow car, the relative velocity o the
ast car is +5 m s - 1 .
Vector or
scalar?
m
instantaneous vs average
start
d = 0.00 m d = 10.04 m
SI
Unit
In technical terms what we are doing is
moving rom one rame o reerence
into another. The velocities o 25 m s - 1
and 30 m s - 1 were measured according
30 m s -1
nish
d = 100.00 m
t = 11.3 s
to a stationary observer on the side o the
road. We moved rom this rame o reerence
into the drivers rame o reerence.
gap between the cars
increases
by 5 m s - 1
25 m s - 1
observer
one car overtaking another, as seen by an
observer on the side o the road.
one car overtaking another, as seen by the
driver o the slow car.
m ech an i cs
9
g
the use of graphs
2.
Graphs are very useul or representing the changes that
happen when an object is in motion. There are three possible
graphs that can provide useul inormation
To make things simple at the beginning, the graphs are
normally introduced by considering objects that are just moving
in one particular direction. I this is the case then there is
not much dierence between the scalar versions (distance or
speed) and the vector versions (displacement or velocity) as the
directions are clear rom the situation. More complicated graphs
can look at the component o a velocity in a particular direction.
displacementtime or distancetime graphs
velocitytime or speedtime graphs
accelerationtime graphs.
There are two common methods o determining particular physical
quantities rom these graphs. The particular physical quantity
determined depends on what is being plotted on the graph.
1.
Finding the gradient of the line.
To be a little more precise, one could fnd either the gradient o
a straight-line section o the graph (this fnds an average
value) , or
the tangent to the graph at one point (this fnds an
instantaneous value) .
Finding the area under the line.
I the object moves orward then backward (or up then down) ,
we distinguish the two velocities by choosing which direction to
call positive. It does not matter which direction we choose, but
it should be clearly labelled on the graph.
Many examination candidates get the three types o graph
muddled up. For example a speedtime graph might be
interpreted as a distancetime graph or even an acceleration
time graph. Always look at the axes o a graph very careully.
velocitytime graphs
accelerationtime graphs
The gradient o a velocitytime
graph is the acceleration
The area under a displacementtime
graph does not represent anything
useul
The area under a velocitytime
graph is the displacement
The gradient o an acceleration
time graph is not oten useul (it
is actually the rate o change o
acceleration)
e
e
10 .0
= 20 m s-1
rst 4 seconds
at con stan t
speed
2 0 .0
ob ject is s lo win g
d own a ccelera tion
20
= 1
= - 20 m s-2
10 .0
spee d = 2 0 = 5 m s - 1
4
0 1 .0 2 .0 3 .0 4.0 5 .0 6 .0 7.0 8.0
0 1 .0 2 .0 3 .0 4.0 5 .0 6 .0 7.0 8 .0
tim e / s
h igh e st p oin t a t t = 0 .9 s
velocity / m s -1
4.0
2 .0
object re tu rn s to
h a n d a t t = 1 .8 s
1 .0
level o f
ha n d a s zero
d ispla cem en t
2 .0
tim e / s
O bject is th ro wn vertica lly u pwa rd s.
tim e / s
a ccelera tion is co n sta n tly
in crea s in g, ra te o f ch a n ge
o f ve lo city is in crea s in g
in itia l u pw a rd
ve lo city is + ve m a x. h eigh t = a rea u n d er gra ph
1
= 0 .9 9 .0 m = 4.0 5 m
9 .0
2
- 9 .0
a ccelera tio n =
= - 10 m s - 2
0 .9
- 9 .0
1 .0
2 .0 tim e / s
in s ta n ta n eou s
d own wa rd
velo city = ze ro
velo city is
a t h igh e st p o in t
ne ga tive
( t = 0 .9 s )
O bject is th ro wn vertica l ly u p wa rd s.
object a t co n s ta n t
a ccelera tion o f 2 0 m s - 2
ve lo city s till ch a n gin g a ll
th e tim e
ra te o f
a ccele ra tion
is d ecrea s in g,
b u t velo city
co n tin u e s
to in crea se
2 0 .0
10 .0
d is ta n ce tra vel led in rs t 4 se con d s
= a rea u n d er gra p h
= 1 4 2 0 m = 40 m
2
O bject m ove s with co n sta n t a ccelera tion ,
then con sta n t velo city , then d ecelerate s.
O bje ct m ove s a t con sta n t sp eed , s top s then re tu rn s.
displacement / m
e
object's ve lo city is in crea s in g
object a t con s ta n t
20
a cce lera tion =
4
sp eed ( = 2 0 m s - 1 )
= 5 m s - 2 a ccelera tion is zero
acceleration / m s -2
2 0 .0
object re tu rn s a t a
fa s ter sp eed
20
spe ed =
1
0 1 .0 2 .0 3 .0 4.0 5 .0 6 .0 7.0 8.0 tim e / s
1
4 2 0 = 40 m s - 1
2
O b ject m o ve s with in crea s in g, th en co n s ta n t,
th en d ecrea s in g a ccelera tion .
Cha n ge in velo city =
acceleration / m s - 2
object s ta tion a ry
for 3 se con d s
spe ed = 0 m s - 1
The area under an acceleration
time graph is the change in velocity
velocity / m s 1
displacement / m
Displacementtime graphs
The gradient o a displacementtime
graph is the velocity
+ 10 .0
1 .0
10 .0
2 .0 tim e / s
ch a n ge in velo city = a rea u n d er gra p h
= - 10 .0 1 .8 m s - 1
= - 18 m s - 1
( cha n ge from + 9 .0 to - 9 .0 m s - 1 )
O b ject is th ro wn ve rtica lly u pwa rd s.
example of equation of uniform motion
A car accelerates uniormly rom rest. Ater 8 s it has travelled 1 20 m. Calculate: (i) its average acceleration
speed ater 8 s
1 at2
(i) s = ut + _
2
1 a 8 2 = 32 a
1 20 = 0 8 + _
2
a = 3.75 m s 2
10
m ech an i cs
(ii) v2 =
=
=
v =
u 2 + 2 as
0 + 2 3.75 1 20
900
30 m s 1
(ii) its instantaneous
uy d
practical
calculations
equations of
uniform motion
In order to determine how the
velocity (or the acceleration)
o an object varies in real
situations, it is oten necessary
to record its motion. Possible
laboratory methods include.
These equations can only be
used when the acceleration
is constant dont orget to
check i this is the case!
A strobe light gives out very
brie fashes o light at xed
time intervals. I a camera
is pointed at an object and
the only source o light is
the strobe light, then the
developed picture will have
captured an objects motion.
t = 0.0 s
t = 0.1 s
t = 0.2 s
t = 0.3 s
t = 0.4 s
a
acceleration (const)
t
time taken
s
distance travelled
u+v
s = _ t
2
)
tk
A ticker timer can be
arranged to make dots on
a strip o paper at regular
intervals o time (typically
every tieth o a second) . I
the piece o paper is attached
to an object, and the object
is allowed to all, the dots on
the strip will have recorded
the distance moved by the
object in a known time.
downwards +ve
v = u + at
20
5
2 .0
3 .0 tim e / s
1 .0
2 .0
a ccelera tion / m s -2
3 .0 tim e / s
30
20
10
1 at2
s = ut + _
2
1 at2
s = vt - _
2
The rst equation is derived
rom the denition o
acceleration. In terms o
these symbols, this denition
would be
(v - u)
a= _
t
This can be rearranged to
give the rst equation.
v = u + at
(1 )
The second equation comes
rom the denition o
average velocity.
average velocity = _s
t
Since the velocity is changing
uniormly we know that this
average velocity must be
given by
or
(u + v)
_s = _
t
2
downwards +ve
v2 = u 2 + 2as
(v + u)
average velocity = _
2
t = 0.5 s
d ispla cem en t / m
45
1 .0
velocity / m s -1
The ollowing equations link
these dierent quantities.
(
downwards +ve
initial velocity
nal velocity
The other equations o
motion can be derived by
using these two equations
and substituting or one o
the variables (see previous
page or an example o
their use) .
(2)
2 .0
3 .0 tim e / s
In the absence o air resistance, all alling objects have the
SAME acceleration o ree-all, INDEPENDENT o their mass.
Air resistance will (eventually) aect the motion o all
objects. Typically, the graphs o a alling object aected by air
resistance become the shapes shown below.
d ispla cem en t / m
stra igh t lin e a s
velocity becom es
con sta n t
20
5
1 .0
velocity / m s -1
2 .0
3 .0 tim e / s
23
20
term in a l
velocity of
23 m s -1
10
This can be rearranged to
give
(u + v) t
s=_
2
10
1 .0
downwards +ve
s hhy
u
v
Taking down as positive, the graphs o the motion o any object
in ree-all are
downwards +ve
Alternatively, two light gates
and a timer can be used to
calculate the average velocity
between the two gates. Several
light gates and a computer can
be joined together to make
direct calculations o velocity
or acceleration.
The list o variables to
be considered (and their
symbols) is as ollows
downwards +ve
lh
A light gate is a device that
senses when an object cuts
through a beam o light. The
time or which the beam is
broken is recorded. I the
length o the object that
breaks the beam is known,
the average speed o the
object through the gate can
be calculated.
falling objects
A very important example o uniormly accelerated motion is
the vertical motion o an object in a uniorm gravitational
feld. I we ignore the eects o air resistance, this is known as
being in ree-all.
1 .0
2 .0
3 .0 tim e / s
a ccelera tion = zero
a ccelera tion / m s -2
a t term in a l velocity
1 .0
2 .0
3 .0 tim e / s
As the graphs show, the velocity does not keep on rising. It
eventually reaches a maximum or terminal velocity. A
piece o alling paper will reach its terminal velocity in a much
shorter time than a alling book.
m ech an i cs
11
p
components of projectile motion
hz
I two children are throwing and catching a tennis ball between
them, the path o the ball is always the same shape. This motion is
known as projectile motion and the shape is called a parabola.
There are no orces in the horizontal direction, so there is no
horizontal acceleration. This means that the horizontal velocity
must be constant.
ball travels at a constant horizontal velocity
v3
v2
v1
path taken by ball
is a parabola
vH
vH
dH
vH
vH v4
vH
v5
vH
dH
dH
dH
dH
v6
v
The only orces acting during its fight are gravity and riction.
In many situations, air resistance can be ignored.
There is a constant vertical orce acting down, so there is
a constant vertical acceleration. The value o the vertical
acceleration is 1 0 m s - 2 , which is the acceleration due to gravity.
It is moving horizontally and vertically at the same time
but the horizontal and vertical components o the motion are
independent o one another. Assuming the gravitional orce is
constant, this is always true.
v2
vertical
velocity
v3
vH
vH v4
v1
vH
vH
v5
vH
changes
vH
v6
mathematics of parabolic motion
example
The graphs o the components o parabolic motion are shown below.
A projectile is launched horizontally rom the top o
a cli.
y-d
a y / m s -2
a x / m s -2
x-d
0
t /s
in itia l h orizon ta l velo city
uH
0
g
t /s
vy / m s -1
vx / m s -
h
ux
0
t /s
0
h eigh t
o f cli
uy
slope = - g
t /s
x
0
y/m
x/m
uH
slope = u x
t /s
0
m a xim u m
heigh t
v
vertical motion
u=0
v=?
a = 1 0 m s- 2
s=h
t =?
t /s
Once the components have been worked out, the actual velocities (or
displacements) at any time can be worked out by vector addition.
The solution o any problem involving projectile motion is as ollows:
use the angle o launch to resolve the initial velocity into components.
the time o fight will be determined by the vertical component o
velocity.
the range will be determined by the horizontal component (and the
time o fight) .
the velocity at any point can be ound by vector addition.
Useul short-cuts in calculations include the ollowing acts:
or a given speed, the greatest range is achieved i the launch angle is 45.
i two objects are released together, one with a horizontal velocity
and one rom rest, they will both hit the ground together.
12
m ech an i cs
s = ut +
so
h=0+
t2 =
vf
horizontal motion
u = uH
v = uH
a=0
s=x
t= ?
1 2
at
2
1
1 0 t2
2
2h
10
2h
s
10
Since v = u + at
t=
v = 0 + 10
2h
m
10
x = uH t
2h
m
10
The nal velocity vf is the vector addition o v and u H .
=
20h
m s -1
= uH
fd d -
fluiD resistance
When an object moves through a fuid (a liquid or a gas) , there will be a rictional fuid resistance that aects the objects motion.
An example o this eect is the terminal velocity that is reached by a ree-alling object, e.g. a spherical mass alling through a liquid
or a parachutist alling towards the Earth. See page 1 1 or how the motion graphs will be altered in these situations.
Modelling the precise eect o fuid resistance on moving objects is complex but simple predictions are possible. The Engineering
Physics option (see page 1 67) introduces a mathematical analysis o the rictional drag orce that acts on a perect sphere when it
moves through a fuid. Key points to note are that:
Viscous drag acts to oppose motion through a fuid
The drag orce is dependent on:
Relative velocity o the object with respect to the fuid
The shape and size o the object (whether the object is aerodynamic or not)
The fuid used (and a property called its viscosity) .
For example page 1 2 shows how, in the absence o fuid resistance, an object that is in projectile motion will ollow a parabolic
path. When fuid resistance is taken into account, the vertical and the horizontal components o velocity will both be reduced. The
eect will be a reduced range and, in the extreme, the horizontal velocity can be reduced to near zero.
parabolic path (no uid resistance)
path (with uid resistance)
experiment to Determine free-fall acceleration
All experiments to determine the ree-all acceleration or an object are based on the use o a constant acceleration equation with
recorded measurements o displacement and time. Some experimental set-ups will be more sophisticated and use more equipment
than others. This increased use o technology potentially brings greater precision but can introduce more complications. Simple
equipment oten means that, with a limited time available or experimentation, it is easier or many repetitions to be attempted.
I an object ree-alls a height, h, rom rest in a time, t, the acceleration, g, can be calculated using s = ut + __12 at2 which rearranges
2h
to give = __
. Rather than just calculating a single value, a more reliable value comes rom taking a series o measurement o the
t
dierent times o all or dierent heights h = __12 gt2 . A graph o h on the y-axis against t2 on the x-axis will give a straight line graph
that goes through the origin with a gradient equal to __12 g, making g twice the gradient.
2
Possible set-ups include:
Set-up
Comments
Direct measurement o a alling object,
e.g. ball bearing with a stop watch and
a metre ruler
Very simple set-up meaning many repetitions easily achieved so random error can be
eliminated. I height o all is careully controlled, great precision is possible even though
equipment is standard. For a simple everyday object such as a ball bearing, the eect o
air resistance will be negligible in the laboratory whereas the eect o air resistance on a
Ping-Pong ball will be signicant.
Electromagnet release and electronic
timing version o the above
The increased precision o the timing can improve accuracy but set-up will take longer.
Introduction o technology can mean that systematic errors are harder to identiy.
Motion o alling object automatically
recording on ticker-tape attached to
alling object
Physical record allows detailed analysis o motion and thus allows the objects whole
all to be considered (not just the overall time taken) and or the data to be graphically
analysed. Addition o moving paper tape introduces riction to the motion, however.
Distance sensor and data logger
All measurements can be automated and very precise. Sotware can be programmed
to perorm all the calculations and to plot appropriate graphs. Experimenter needs to
understand how to operate the data logger and associated sotware.
Video analysis o alling object
Capturing a visual record o the objects all against a known scale, allows detailed
measurements to be taken. Timing inormation rom the video recording needed, which
oten involves ICT.
m ech an i cs
13
f d -d d
forces what they are anD what they Do
In the examples below, a orce (the kick) can cause deormation
(the ball changes shape) or a change in motion (the ball
gains a velocity) . There are many dierent types o orces,
but in general terms one can describe any orce as the cause
o a deormation or a velocity change. The SI unit or the
measurement o orces is the newton (N) .
(a) deformation
(b) change in velocity
A (resultant) orce causes a CHANGE in velocity. I the
(resultant) orce is zero then the velocity is constant.
Remember a change in velocity is called an acceleration, so we
can say that a force causes an acceleration. A (resultant)
orce is NOT needed or a constant velocity (see page 1 6) .
The act that a orce can cause deormation is also important,
but the deormation o the ball was, in act, not caused by just
one orce there was another one rom the wall.
One orce can act on only one object. To be absolutely
precise the description o a orce should include
kick
kick
kick causes
deformation of football
kick causes a change in
motion of football
Eect o a orce on a ootball
its magnitude
its direction
the object on which it acts (or the part o a large object)
the object that exerts the orce
the nature o the orce
A description o the orce shown in the example would thus
be a 50 N push at 20 to the horizontal acting ON the ootball
FROM the boot.
Different types of forces
forces as vectors
The ollowing words all describe the orces (the pushes or
pulls) that exist in nature.
Since orces are vectors, vector mathematics must be used to fnd
the resultant orce rom two or more other orces. A orce can also
be split into its components. See page 7 or more details.
Gravitational force
Electrostatic force
Magnetic force
Normal reaction
Friction
Tension
Compression
Upthrust
Lift
One way o categorizing these orces is whether they result
rom the contact between two suraces or whether the orce
exists even i a distance separates the objects.
The origin o all these everyday orces is either gravitational or
electromagnetic. The vast majority o everyday eects that we
observe are due to electromagnetic orces.
(a) by vector mathematics
example: block being pushed on rough surface
force diagram:
S, surface force
P, push force
resultant W
S
force
W,
P
weight
(b) by components
example: block sliding down a smooth slope
measuring forces
The simplest experimental method or measuring the size o a
orce is to use the extension o a spring. When a spring is in
tension it increases in length. The dierence between the natural
length and stretched length is called the extension o a spring.
R, reaction
original
length
extension
= 5.0 cm
2N
W, weight
extension
= 15.0 cm
resultant down
slope = W sin
component into slope
resultant into
= W cos
slope = W cos - R
= zero
component down slope
= W sin
Vector addition
6N
free-boDy Diagrams
In a free-body diagram
extension / cm
15.0
mathematically,
F x
F = kx
10.0
spring constant
(units N m -1 )
5.0
2.0
4.0
6.0
one object (and ONLY one object) is chosen
all the orces on that object are shown and labelled.
For example, i we considered the simple situation o a book
resting on a table, we can construct ree-body diagrams or
either the book or the table.
8.0 force / N
situation:
Hookes law
Hookes law states that up to the elastic limit, the extension, x,
o a spring is proportional to the tension orce, F. The constant
o proportionality k is called the spring constant. The SI
units or the spring constant are N m- 1 . Thus by measuring
the extension, we can calculate the orce.
14
m ech an i cs
free-body diagram
free-body diagram
for book:
for table:
P,
push from
RT, reaction from table
book
RE , reaction
from Earths
RE
W
weight of table surface
w, weight of book
gravitational pull of Earth gravitational pull of Earth
n f
newtons irst law
Newtons frst law o motion states that an object continues in uniorm motion in a straight line or at rest unless a resultant external orce
acts. On frst reading, this can sound complicated but it does not really add anything to the description o a orce given on page 1 4. All it
says is that a resultant orce causes acceleration. No resultant orce means no acceleration i.e. uniorm motion in a straight line.
bk
lg hvy u
R
P, pull from person
R, reaction from ground
W
W, weight of suitcase
sin ce
a ccelera tion = zero
resu lta n t force = zero
R - W = zero
If the suitcase is too heavy to lift, it is not moving:
acceleration = zero
P+ R= W
c vg gh
P
phu
R
R
F, air friction
F
W
F is force forwards, due to engine
P is force backwards due to air resistance
At all times force up (2R) = force down (W) .
If F > P the car accelerates forwards.
If F = P the car is at constant velocity (zero acceleration) .
If F < P the car decelerates ( i.e. there is negative
acceleration and the car slows down) .
parachutist
free-falling
downwards
p h mvg ud
If W > F the parachutist accelerates downwards.
As the parachutist gets faster, the air friction increases until
W= F
The parachutist is at constant velocity
(the acceleration is zero) .
lift moving upwards
W, weight
R
2
R
2
W
The total force up from the oor of the lift = R.
The total force down due to gravity = W.
If R > W the person is accelerating upwards.
If R = W the person is at constant velocity
( acceleration = zero) .
If R < W the person is decelerating ( acceleration is
negative) .
m ech an i cs
15
e
equilibrium
I the resultant orce on an object is zero then it is said to
be in translational equilibrium (or just in equilibrium) .
Mathematically this is expressed as ollows:
F = zero
Translational equilibrium does NOT mean the same thing as
being at rest. For example i the child in the previous example
is allowed to swing back and orth, there are times when she is
instantaneously at rest but he is never in equilibrium.
From Newtons rst law, we know that the objects in the
ollowing situations must be in equilibrium.
1.
An object that is constantly at rest.
2.
An object that is moving with constant (uniorm) velocity
in a straight line.
Since orces are vector quantities, a zero resultant orce means
no orce IN ANY DIRECTION.
T
T
For 2-dimensional problems it is sucient to show that the
orces balance in any two non-parallel directions. I this is the
case then the object is in equilibrium.
T
W
tension, T
P, pull
At the end of the
swing the forces
are not balanced
but the child is
instantaneously
at rest.
W
W
Forces are not
balanced in the centre
as the child is in circular
motion and is
accelerating (see page 65) .
weight, W
if in equilibrium:
T sin = P ( since no resultant horizontal force)
T cos = W (since no resultant vertical force)
Different types of forces
Name o orce
Description
Gravitational orce
The orce between objects as a result o their masses. This is sometimes reerred to as the weight o the
object but this term is, unortunately, ambiguous see page 1 9.
Electrostatic orce
The orce between objects as a result o their electric charges.
Magnetic orce
The orce between magnets and/or electric currents.
Normal reaction
The orce between two suraces that acts at right angles to the suraces. I two suraces are smooth then
this is the only orce that acts between them.
Friction
The orce that opposes the relative motion o two suraces and acts along the suraces. Air resistance or
drag can be thought o as a rictional orce technically this is known as fuid riction.
Tension
When a string (or a spring) is stretched, it has equal and opposite orces on its ends pulling outwards.
The tension orce is the orce that the end o the string applies to another object.
Compression
When a rod is compressed (squashed) , it has equal and opposite orces on its ends pushing inwards.
The compression orce is the orce that the ends o the rod applies to another object. This is the opposite
o the tension orce.
Upthrust
This is the upward orce that acts on an object when it is submerged in a fuid. It is the buoyancy orce
that causes some objects to foat in water (see page 1 64) .
Lit
This orce can be exerted on an object when a fuid fows over it in an asymmetrical way. The shape o
the wing o an aircrat causes the aerodynamic lit that enables the aircrat to fy (see page 1 66) .
16
m ech an i cs
n d
newtons seconD law of motion
examples of newtons seconD law
Newtons frst law states that a resultant orce causes
an acceleration. His second law provides a means o
calculating the value o this acceleration. The best
way o stating the second law is use the concept
o the momentum o an object. This concept is
explained on page 23.
1.
A correct statement o Newtons second law
using momentum would be the resultant orce is
proportional to the rate o change o momentum. I
we use SI units (and you always should) then the law
is even easier to state the resultant orce is equal to
the rate o change o momentum. In symbols, this is
expressed as ollows
12 N
3 kg
no friction between block and surface
I a mass o 3 kg is accelerated in a straight line by a resultant orce o
1 2 N, the acceleration must be 4 m s - 2 . Since
F = ma,
F
12
_
-2
a= _
m = 3 =4ms .
2.
acceleration = 1.5 m s -2
Use o F = ma in
a slightly more
complicated situation
12 N
3 kg
friction force
I a mass o 3 kg is
accelerated in a straight line by a orce o 1 2 N, and the resultant
acceleration is 1 .5 m s - 2 , then we can work out the riction that must
have been acting. Since
p
In SI units, F = _
t
dp
or, in ull calculus notation, F = _
dt
F = ma
p is the symbol or the momentum o a body.
resultant orce = 3 1 .5
= 4.5 N
Until you have studied what this means this will
not make much sense, but this version o the law is
given here or completeness.
An equivalent (but more common) way o stating
Newtons second law applies when we consider the
action o a orce on a single mass. I the amount
o mass stays constant we can state the law as
ollows. The resultant orce is proportional to
the acceleration. I we also use SI units then the
resultant orce is equal to the product o the mass
and the acceleration.
Use o F = ma in a simple
situation
This resultant orce = orward orce - riction
thereore, riction = orward orce - resultant orce
= 1 2 - 4.5 N
= 7.5 N
3.
Use o F = ma in a
2-dimensional situation
normal reaction
friction
( max. 8.0 N)
3 kg
In symbols, in SI units,
F = m a
rultt for
urd
wto
urd
klogr
30 N
30
lrto
urd
- 2
A mass o 3 kg eels a gravitational pull towards the Earth o 30 N.
What will happen i it is placed on a 30 degree slope given that the
maximum riction between the block and the slope is 8.0 N?
n orm a l rea ction
friction
Note:
The F = ma version o the law only applies i we
use SI units or the equation to work the mass
must be in kilograms rather than in grams.
F is the resultant orce. I there are several orces
acting on an object (and this is usually true) then
one needs to work out the resultant orce beore
applying the law.
This is an experimental law.
There are no exceptions Newtons laws apply
throughout the Universe. (To be absolutely precise,
Einsteins theory o relativity takes over at very
large values o speed and mass.)
The F = ma version o the law can be used whenever
the situation is simple or example, a constant
orce acting on a constant mass giving a constant
acceleration. I the situation is more difcult (e.g. a
changing orce or a changing mass) then one needs to
dp
use the F = _ version.
dt
3 kg
com pon en t
in to slope
30
30 N
30
com pon en t d own
th e slope
into slope: normal reaction = component into slope
The block does not accelerate into the slope.
down the slope:
component down slope = 30 N sin 30
= 15 N
maximum riction orce up slope = 8 N
resultant orce down slope = 1 5 - 8
=7N
F = ma
F
acceleration down slope = _
m
7 = 2.3 m s - 2
=_
3
m ech an i cs
17
n ird
statement o the law
In symbols,
Newtons second law is an experimental law that allows us to
calculate the effect that a force has. Newtons third law highlights
the fact that forces always come in pairs. It provides a way of
checking to see if we have remembered all the forces involved.
It is very easy to state. When two bodies A and B interact, the
force that A exerts on B is equal and opposite to the force that
B exerts on A. Another way of saying the same thing is that
for every action on one object there is an equal but opposite
reaction on another object.
FA B = - FB A
Key points to notice include
The two forces in the pair act on different objects this
means that equal and opposite forces that act on the same
object are NOT Newtons third law pairs.
Not only are the forces equal and opposite, but they must be
of the same type. In other words, if the force that A exerts on
B is a gravitational force, then the equal and opposite force
exerted by B on A is also a gravitational force.
examples o the law
rc b rr-kr
push of
B on A
push of
A on B
A
2.0 m s - 1
B
1.5 m s - 1
a rr-kr u f r
If one roller-skater
pushes another, they
both feel a force. The
forces must be equal
and opposite, but the
acceleration will be
dierent (since they
have dierent
masses) .
The person with the
smaller mass will
gain the greater
velocity.
A
push of wall
on girl
2.5 m s - 1
The force on the
girl causes her
to accelerate
backwards.
B
a bk b n ird
push of girl
on wall
The mass of the
wall (and
Earth) is so
large that the
force on it does
not eectively
cause any
acceleration.
a ccrig cr
R, reaction from table
W, weight
These two forces are not third
law pairs. There must be another
force (on a dierent object) that
pairs with each one:
R
W
EARTH
F, push forward from the ground on the car
In order to accelerate, there must be a forward force on the car.
The engine makes the wheels turn and the wheels push on the
ground.
force from car on ground = - force from ground on car
If the table pushes
upwards on the book
with force R, then the
book must push down on
the table with force R.
18
If the Earth pulls the book
down with force W, then the
book must pull the Earth up
with force W.
m ech an i cs
mass and
weight
Mass and weight are two very dierent things. Unortunately
their meanings have become muddled in everyday language.
Mass is the amount o matter contained in an object (measured
in kg) whereas the weight o an object is a orce (measured in N) .
R
2M
M
W
weight, W
new weight = 2W
Double the mass means double the weight
To make things worse, the term weight can be ambiguous
even to physicists. Some people choose to defne weight as the
gravitational orce on an object. Other people defne it to be the
reading on a supporting scale. Whichever defnition you use,
you weigh less at the top o a building compared with at the
bottom the pull o gravity is slightly less!
situation:
acceleration upwards
I an object is taken to the Moon, its mass would be the same,
but its weight would be less (the gravitational orces on the
Moon are less than on the Earth) . On the Earth the two terms
are oten muddled because they are proportional. People talk
about wanting to gain or lose weight what they are actually
worried about is gaining or losing mass.
Although these two defnitions are the same i the object is
in equilibrium, they are very dierent in non-equilibrium
situations. For example, i both the object and the scale
were put into a lit and the lit accelerated upwards then the
defnitions would give dierent values.
If the lift is accelerating
upwards:
R> W
The sae thing to do is to avoid using the term weight i at all
possible! Stick to the phrase gravitational orce or orce o
gravity and you cannot go wrong.
Gravitational orce = m g
On the surace o the Earth, g is approximately 1 0 N kg- 1 ,
whereas on the surace o the moon, g 1 .6 N kg- 1
Weight can be dened as either
(a) the pull of gravity, W or
(b) the force on a supporting
scale R.
OR
R
W
Two dierent defnitions o weight
m ech an i cs
19
sd
factors affecting friction static
anD Dynamic
push = zero
friction
friction, F = zero
F= 0 N
block
P = 0 N stationary
P= 5 N
block
stationary
F= 5 N
push
P = 10 N
block
stationary
F = 10 N ( = Fmax)
block accelerates
Friction arises from the
unevenness of the surfaces.
increasing push force
Friction is the orce that opposes the relative motion o two
suraces. It arises because the suraces involved are not perectly
smooth on the microscopic scale. I the suraces are prevented
rom relative motion (they are at rest) then this is an example
o static riction. I the suraces are moving, then it is called
dynamic riction or kinetic riction.
P = 15 N
F= 9 N
The value o Fm a x depends upon
the nature o the two suraces in contact.
push causes
motion to
RIGHT
the normal reaction orce between the two suraces. The
maximum rictional orce and the normal reaction orce are
proportional.
I the two suraces are kept in contact by gravity, the value o
Fm a x does NOT depend upon the area o contact
friction opposes motion,
acting to LEFT
A key experimental act is that the value o static riction
changes depending on the applied orce. Up to a certain
maximum orce, Fm a x , the resultant orce is zero. For example, i
we try to get a heavy block to move, any value o pushing orce
below Fm a x would ail to get the block to accelerate.
Once the object has started moving, the maximum value o
riction slightly reduces. In other words,
Fk < Fm a x
For two suraces moving over one another, the dynamic
rictional orce remains roughly constant even i the speed
changes slightly.
coefficient of friction
example
Experimentally, the maximum rictional orce and the normal
reaction orce are proportional. We use this to defne the
coefcient o riction, .
I a block is placed on a slope, the angle o the slope can be
increased until the block just begins to slide down the slope.
This turns out to be an easy experimental way to measure the
coefcient o static riction.
coecient of friction =
reaction, R
P
pull forward
F
frictional force
W
gravitational attraction
Fmax = R
R, reaction
component of W down
slope ( W sin )
The coefcient o riction is defned rom the maximum value
that riction can take
Fm a x = R
where R = normal reaction orce
It should be noted that
since the maximum value or dynamic riction is less than
the maximum value or static riction, the values or the
coefcients o riction will be dierent
d < s
the coefcient o riction is a ratio between two orces it
has no units.
i the suraces are smooth then the maximum riction is
zero i.e. = 0.
the coefcient o riction is less than 1 unless the suraces
are stuck together.
Ff s R and Ff = d R
20
m ech an i cs
friction F
component of W into
slope (W cos )
W
If balanced,
F = W sin
R = W cos
is increased.
When block just starts moving,
F = Fmax
static =
=
Fmax
R
W sin
W cos
= tan
w
when is work Done?
Definition of work
Work is done when a orce moves its point o application in
the direction o the orce. I the orce moves at right angles to
the direction o the orce, then no work has been done.
Work is a scalar quantity. Its defnition is as ollows.
1) before
after
v
at rest
force
block now
moving
work has
been done
force
distance
2) before
block now
higher up
work has
been done
after
F
work done = Fs cos
s
Work done = F s cos
I the orce and the displacement are in the same direction,
this can be simplifed to
Work done = orce distance
From this defnition, the SI units or work done are N m. We
defne a new unit called the joule: 1 J = 1 N m.
examples
force
distance
(1) lifting vertically
small distance
force
large force
3) before
force
spring has been
compressed
work has
been done
after
force
The task in the second case would be easier to perorm (it
involves less orce) but overall it takes more work since work
has to be done to overcome riction. In each case, the useul
work is the same.
I the orce doing work is not constant (or example, when a
spring is compressed) , then graphical techniques can be used.
distance
4) before
(2) pushing along a rough slope
is ta n ce
la rge d
r fo rce
s m a l le
original length
after
FA
book supported by shelf
no work is done
Fmax
after
constant
velocity v
friction-free su rface
v
friction-free su rface
object continues at constant velocity
no work is done
xmax
The total work done is
the area under the
orcedisplacement
graph.
force
5) before
x
Fmax
x
In the examples above the work done has had dierent results.
In 1 ) the orce has made the object move aster.
In 2) the object has been lited higher in the gravitational feld.
In 3) the spring has been compressed.
In 4) and 5) , NO work is done. Note that even though
the object is moving in the last example, there is no orce
moving along its direction o action so no work is done.
F = kx
0
total work done
= area under graph
= 1 k x2
2
xmax
Useul equations or the work done include:
extension
work done when liting something vertically = mgh
where m represents mass (in kg)
g represents the Earths gravitational feld strength
(1 0 N kg- 1 ) h represents the height change (in m)
work done in compressing or extending a spring = __12 k x2
m ech an i cs
21
e d
concepts of energy anD work
K 0 J
P = 1000 J
Energy and work are linked together. When you do work on an object, it gains
energy and you lose energy. The amount o energy transerred is equal to the
work done. Energy is a measure o the amount o work done. This means that the
units o energy must be the same as the units o work joules.
energy transformations conservation of energy
In any situation, we must be able to account or the changes in energy. I it is lost
by one object, it must be gained by another. This is known as the principle o
conservation o energy. There are several ways o stating this principle:
K = 250 J
P = 750 J
Overall the total energy o any closed system must be constant.
Energy is neither created nor destroyed, it just changes orm.
K = 250 J
P = 750 J
There is no change in the total energy in the Universe.
energy types
Kinetic energy
Gravitational potential
Elastic potential energy
Radiant energy
Electrostatic potential
Thermal energy
Nuclear energy
Solar energy
Chemical energy
Electrical energy
Internal energy
Light energy
Equations or the frst three types o energy are given below.
Kinetic energy =
=
__1 mv
2
p2
___
2m
2
where m is the mass (in kg) , v is the velocity (in m s - 1 )
where p is the momentum (see page 23) (in kg m s- 1 ), and m is
the mass (in kg)
Gravitational potential energy = mgh where m represents mass (in kg), g represents
the Earths gravitational feld (1 0 N kg- 1 ) , h represents the height change (in m)
K = 500 J
P = 500 J
K = 750 J
P = 250 J
K = 1000 J
P = 0 J
Elastic potential energy =
the extension (in m)
__1 k x
2
2
where k is the spring constant (in N m- 1 ) , x is
power anD efficiency
examples
1. Power
Power is defned as the RATE at which
energy is transerred. This is the same as
the rate at which work is done.
energy transerred
Power = __
time taken
work
done
__
Power =
time taken
The SI unit or power is the joule per
second (J s- 1 ) . Another unit or power is
defned - the watt (W) . 1 W = 1 J s - 1 .
I something is moving at a constant
velocity v against a constant rictional
orce F, the power P needed is P = F v
1.
2. Efciency
Depending on the situation, we can
categorize the energy transerred (work
done) as useul or not. In a light bulb, the
useul energy would be light energy, the
wasted energy would be thermal energy
(and non-visible orms o radiant energy).
We defne efciency as the ratio o
useul energy to the total energy
transerred. Possible orms o the
equation include:
useul work OUT
Efciency = __
total work IN
useul energy OUT
__
Efciency =
total energy IN
useul power OUT
__
Efciency =
total power IN
Since this is a ratio it does not have any
units. Oten it is expressed as a percentage.
22
m ech an i cs
A grasshopper (mass 8 g) uses its
hindlegs to push or 0.1 s and as a
result jumps 1 .8 m high. Calculate
(i) its take o speed, (ii) the
power developed.
(i) PE gained = mgh
1 mv2
KE at start = _
2
1 mv2 = mgh (conservation o
_
2
energy)
=
v = 2gh
2 1 0 1 .8
= 6 m s- 1
mgh
(ii) Power = _
t
0.008 1 0 1 .8
= __
0.1
1 .4 W
2.
A 60W lightbulb has an efciency o
1 0%. How much energy is wasted
every hour?
Power wasted = 90% o 60W
= 54W
Energy wasted = 54 60 60J
= 1 90 kJ
m d
Definitions linear momentum anD impulse
conservation of momentum
Linear momentum (always given the symbol p) is defned as
the product o mass and velocity.
The law o conservation o linear momentum states that the total
linear momentum o a system o interacting particles remains
constant provided there is no resultant external force.
Momentum = mass velocity
p=mv
The SI units or momentum must be kg m s . Alternative
units o N s can also be used (see below) . Since velocity is
a vector, momentum must be a vector. In any situation,
particularly i it happens quickly, the change o momentum
p is called the impulse (p = F t) .
-1
use of momentum in newtons seconD law
Newtons second law states that the resultant orce is
proportional to the rate o change o momentum.
Mathematically we can write this as
(fnal momentum - initial momentum)
p
F = ____ = _
time taken
t
Example 1
A jet o water leaves a hose and hits a wall where its velocity
is brought to rest. I the hose cross-sectional area is 25 cm2 ,
the velocity o the water is 50 m s - 1 and the density o the
water is 1 000 kg m- 3 , what is the orce acting on the wall?
ve lo city = 5 0 m s - 1
50 m
d e n sity o f
wa te r = 1 0 0 0 kg m - 3
cro ss-se ctio n a l
a re a = 2 5 cm 2 = 0 .0 0 2 5 m 2
To see why, we start by imagining two isolated particles A and
B that collide with one another.
The orce rom A onto B, FA B will cause Bs momentum to
change by a certain amount.
I the time taken was t, then the momentum change (the
impulse) given to B will be given by p B = FA B t
By Newtons third law, the orce rom B onto A, FB A will be
equal and opposite to the orce rom A onto B, FA B = - FB A .
Since the time o contact or A and B is the same, then
the momentum change or A is equal and opposite to the
momentum change or B, pA = - FA B t.
This means that the total momentum (momentum o A
plus the momentum o B) will remain the same. Total
momentum is conserved.
This argument can be extended up to any number o
interacting particles so long as the system o particles is still
isolated. I this is the case, the momentum is still conserved.
elastic anD inelastic collisions
The law o conservation o linear momentum is not enough to
always predict the outcome ater a collision (or an explosion) .
This depends on the nature o the colliding bodies. For
example, a moving railway truck, mA , velocity v, collides with
an identical stationary truck mB . Possible outcomes are:
( a ) ela stic collision
a t re st
n ew velocity = v
In one second, a jet o water 50 m long hits the wall. So
volume o water hitting wall = 0.0025 50 = 0.1 25 m3
every second
mass o water hitting wall = 0.1 25 1 000 = 1 25 kg
every second
( b) tota lly in ela stic collision
momentum o water hitting wall = 1 25 50 = 6250 kg m s - 1
every second
This water is all brought to rest,
change in momentum, p = 6250 kg m s - 1
p
6250 = 6250 N
orce = _ = _
1
t
Example 2
The graph below shows the variation with time o the orce on
a ootball o mass 500 g. Calculate the fnal velocity o the ball.
force/N
The football was given an impulse of approximately
100 0.01 = 1 N s during this 0.01 s.
100
Area under graph is the total
90
impulse given to the
80
ball 5 N s p = mv
70
p
v =
60
m
5Ns
50
=
0.5 kg
40
= 10 m s-1
30
nal
20
velocity v= 10 m s-1
10
0.00 0.02 0.04 0.06 0.08 0.10
time/s
mB
mA
n ew velocity = v
2
mA
( c) in ela stic collision
n ew velocity = v
4
mA
mB
n ew velocity = 3 v
4
mB
In (a) , the trucks would have to have elastic bumpers. I this
were the case then no mechanical energy at all would be lost
in the collision. A collision in which no mechanical energy
is lost is called an elastic collision. In reality, collisions
between everyday objects always lose some energy the
only real example o elastic collisions is the collision between
molecules. For an elastic collision, the relative velocity o
approach always equals the relative velocity o separation.
In (b) , the railway trucks stick together during the collision
(the relative velocity o separation is zero) . This collision
is what is known as a totally inelastic collision. A large
amount o mechanical energy is lost (as heat and sound) , but
the total momentum is still conserved.
In energy terms, (c) is somewhere between (a) and (b) . Some
energy is lost, but the railway trucks do not join together. This
is an example o an inelastic collision. Once again the total
momentum is conserved.
Linear momentum is also conserved in explosions.
m ech an i cs
23
ib questons mechancs
1.
A. 1 :
2
2.
6.
Two identical objects A and B all rom rest rom dierent
heights. I B takes twice as long as A to reach the ground,
what is the ratio o the heights rom which A and B ell?
Neglect air resistance.
B. 1 :2
C. 1 :4
D. 1 :8
A trolley is given an initial push along a horizontal foor to
get it moving. The trolley then travels orward along the foor,
gradually slowing. What is true o the horizontal orce(s) on
the trolley while it is slowing?
A. There is a orward orce and a backward orce, but the
orward orce is larger.
B. There is a orward orce and a backward orce, but the
backward orce is larger.
C. There is only a orward orce, which diminishes with time.
D. There is only a backward orce.
3.
A mass is suspended by
cord rom a ring which
is attached by two urther
cords to the ceiling and
the wall as shown.
The cord rom the ceiling
makes an angle o less
than 45 with the vertical
as shown. The tensions in
the three cords are labelled
R, S and T in the diagram.
S
5.
The vehicles collide head-on and become entangled together.
a) During the collision, how does the orce exerted by
the car on the truck compare with the orce exerted
by the truck on the car? Explain.
[2]
b) In what direction will the entangled vehicles move ater
collision, or will they be stationary? Support
your answer, reerring to a physics principle.
[2]
c) Determine the speed (in km h- 1 ) o the combined
wreck immediately ater the collision.
[3]
d) How does the acceleration o the car compare with the
acceleration o the truck during the collision? Explain.
[2]
e) Both the car and truck drivers are wearing seat belts.
Which driver is likely to be more severely jolted in the
collision? Explain.
[2]
T
f) The total kinetic energy o the system decreases as a result
o the collision. Is the principle o conservation o energy
violated? Explain.
[1 ]
R
7. a) A net orce o magnitude F acts on a body. Dene the
impulse I o the orce.
How do the tensions R, S
and T in the three cords
compare in magnitude?
4.
A car and a truck are both travelling at the speed limit o
60 km h- 1 but in opposite directions as shown. The truck has
twice the mass o the car.
[1 ]
b) A ball o mass 0.0750 kg is travelling horizontally with a
speed o 2.20 m s - 1 . It strikes a vertical wall and rebounds
horizontally.
A. R > T > S
B. S > R > T
C. R = S = T
D. R = S > T
ball mass
0.0750 kg
A 24 N orce causes a 2.0 kg mass to accelerate at 8.0 m s- 2 along
a horizontal surace. The coecient o dynamic riction is:
A. 0.0
B. 0.4
C. 0.6
D. 0.8
2.20 ms -1
An athlete trains by dragging a heavy load across a rough
horizontal surace.
Due to the collision with the wall, 20 % o the balls initial
kinetic energy is dissipated.
(i)
F
Show that the ball rebounds rom the wall with a
speed o 1 .97 m s - 1 .
[2]
(ii) Show that the impulse given to the ball by the
wall is 0.31 3 N s.
25
[2]
c) The ball strikes the wall at time t = 0 and leaves the
wall at time t = T.
The athlete exerts a orce o magnitude F on the load at an
angle o 25 to the horizontal.
The sketch graph shows how the orce F that the wall
exerts on the ball is assumed to vary with time t.
a) Once the load is moving at a steady speed, the average
horizontal rictional orce acting on the load is 470 N.
Calculate the average value o F that will enable the
load to move at constant speed.
F
[2]
b) The load is moved a horizontal distance o 2.5 km in 1.2 hours.
Calculate
(i)
the work done on the load by the orce F.
(ii) the minimum average power required to move
the load.
0
[2]
Explain, in terms o energy changes, why the minimum
average power required is greater than in (b) (ii) .
[2]
i B Q u esti o n s m ech an i cs
t
The time T is measured electronically to equal 0.0894 s.
c) The athlete pulls the load uphill at the same speed as in
part (a) .
24
T
[2]
Use the impulse given in (b) (ii) to estimate the average
value o F.
[4]
33 T h e r m a l p h Ys i C s
T cct
TemperaTure and heaT flow
Kelvin and Celsius
Hot and cold are just labels that identiy
the direction in which thermal energy
(sometimes known as heat) will be
naturally transerred when two objects
are placed in thermal contact. This
leads to the concept o the hotness o
an object. The direction o the natural
ow o thermal energy between two
objects is determined by the hotness
o each object. Thermal energy
naturally ows rom hot to cold.
Most o the time, there are only two sensible temperature scales to chose
between the Kelvin scale and the Celsius scale.
Heat is not a substance that ows
rom one object to another. What
has happened is that thermal energy
has been transerred. Thermal energy
(heat) reers to the non-mechanical
transer o energy between a system
and its surroundings.
There is an easy relationship between a temperature T as measured on the Kelvin
scale and the corresponding temperature t as measured on the Celsius scale. The
approximate relationship is
T (K) = t (C) + 273
700 K
630 K
600 K
COLD
mercury boils
400 K
373 K
300 K
273 K
400 C
357 C
300 C
500 K
200 C
100 C
water boils
200 K
water freezes
mercury freezes
carbon dioxide freezes
100 K
oxygen boils
0K
HOT
Notice the size of the units is
identical on each scale.
Celsius scale
This means that the size o the units used on each scale is identical, but they have
dierent zero points.
Kelvin scale
The temperature o an object is a
measure o how hot it is. In other
words, i two objects are placed in
thermal contact, then the temperature
dierence between the two objects will
determine the direction o the natural
transer o thermal energy. Thermal
energy is naturally transerred down
the temperature dierence rom
high temperature to low temperature.
Eventually, the two objects would
be expected to reach the same
temperature. When this happens,
they are said to be in thermal
equilibrium.
In order to use them, you do not need to understand the details o how either o
these scales has been defned, but you do need to know the relation between them.
Most everyday thermometers are marked with the Celsius scale and temperature is
quoted in degrees Celsius (C) .
hydrogen boils
0 C
-100 C
-200 C
-273 C
The Kelvin scale is an absolute thermodynamic temperature scale and a measurement on
this scale is also called the absolute temperature.
Zero Kelvin is called absolute zero (see page 29) .
direction of transfer of thermal energy
examples: Gases
For a given sample o a gas, the pressure, the volume and the
temperature are all related to one another.
The pressure, P, is the orce per unit area rom the gas acting
at 90 on the container wall.
F
p= _
A
The SI units o pressure are N m- 2 or Pa (Pascals) .
1 Pa = 1 N m- 2
Gas pressure can also be measured in atmospheres
(1 atm 1 0 5 Pa)
In order to investigate how these quantities are interrelated, we
choose:
one quantity to be the independent variable (the thing we
alter and measure)
another quantity to be the dependent variable (the second
thing we measure) .
The third quantity needs to be controlled (i.e. kept constant) .
The specifc values that will be recorded also depend on the
mass o gas being investigated and the type o gas being used
so these need to be controlled as well.
The volume, V, o the gas is measured in m3 or cm3
(1 m3 = 1 0 6 cm3 )
The temperature, t, o the gas is measured in C or K
T H E R M A L P H YS I C S
25
ht t gy
miCrosCopiC vs maCrosCopiC
KineTiC TheorY
When analysing something physical, we have a choice.
Molecules are arranged in dierent ways depending on the
phase o the substance (i.e. solid, liquid or gas) .
The macroscopic point o view considers the system as a
whole and sees how it interacts with its surroundings.
The microscopic point o view looks inside the system to
see how its component parts interact with each other.
So ar we have looked at the temperature o a system in a
macroscopic way, but all objects are made up o atoms and
molecules.
According to kinetic theory these particles are constantly in
random motion hence the name. See below or more details.
Although atoms and molecules are dierent things (a molecule
is a combination o atoms) , the dierence is not important at
this stage. The particles can be thought o as little points o
mass with velocities that are continually changing.
inTernal enerGY
I the temperature o an object changes then it must have
gained (or lost) energy. From the microscopic point o view,
the molecules must have gained (or lost) this energy.
The two possible orms are kinetic energy and potential energy.
speed in a random direction
molecule has KE
v
solids
Macroscopically, solids have a fxed volume and a fxed shape.
This is because the molecules are held in position by bonds.
However the bonds are not absolutely rigid. The molecules
vibrate around a mean (average) position. The higher the
temperature, the greater the vibrations.
Each molecule vibrates
around a mean
position.
Bonds
between
molecules
The molecules in a solid are
held close together by the
intermolecular bonds.
liquids
A liquid also has a fxed volume but its shape can change.
The molecules are also vibrating, but they are not completely
fxed in position. There are still strong orces between the
molecules. This keeps the molecules close to one another, but
they are ree to move around each other.
F
equilibrium
position
resultant force back towards equilibrium
position due to neighbouring molecules
molecule has PE
The molecules have kinetic energy because they are
moving. To be absolutely precise, a molecule can have
either translational kinetic energy (the whole molecule is
moving in a certain direction) or rotational kinetic energy
(the molecule is rotating about one or more axes) .
The molecules have potential energy because o the
intermolecular orces. I we imagine pulling two
molecules urther apart, this would require work against
the intermolecular orces.
The total energy that the molecules possess (random kinetic
plus inter molecule potential) is called the internal energy
o a substance. Whenever we heat a substance, we increase its
internal energy.
Bonds between
neighbouring
molecules; these can
be made and broken,
allowing a molecule to
move.
Each molecule is free
to move throughout the
liquid by moving around
its neighbours.
Gases
A gas will always expand to fll the container in which it is
put. The molecules are not fxed in position, and any orces
between the molecules are very weak. This means that the
molecules are essentially independent o one another, but
they do occasionally collide. More detail is given on page 31 .
Molecules in random
motion; no xed bonds
between molecules so
they are free to move
Temperature is a measure of the average kinetic energy
of the molecules in a substance.
I two substances have the same temperature, then their
molecules have the same average kinetic energy.
heaT and worK
Many people have conused ideas about heat and work. In
answers to examination questions it is very common to read,
or example, that heat rises when what is meant is that the
transer o thermal energy is upwards.
same temperature
v
same average
kinetic energy
V
m
M
molecules with large
mass moving with
lower average speed
molecules with small
mass moving with
higher average speed
When a orce moves through a distance, we say that work
is done. Work is the energy that has been transmitted rom
one system to another rom the macroscopic point o view.
When work is done on a microscopic level (i.e. on
individual molecules) , we say that heating has taken
place. Heat is the energy that has been transmitted. It can
either increase the kinetic energy o the molecules or their
potential energy or, o course, both.
In both cases energy is being transerred.
26
T H E R M A L P H YS I C S
scfc t ccty
deiniTions and miCrosCopiC explanaTion
In theory, i an object could be heated up with no energy loss,
then the increase in temperature T depends on three things:
meThods o measurinG heaT CapaCiTies and
speCiiC heaT CapaCiTies
The are two basic ways to measure heat capacity.
the energy given to the object Q,
1.
the mass, m, and
The experiment would be set up as below:
Electrical method
the substance rom which the object is made.
heater ( placed in object)
1000 J
1000 J
mass m
substance X
mass m
substance Y
V
voltmeter
dierent temperature
change
ammeter A
small temperature
change since
more molecules
variable power supply
ItV
the specifc heat capacity c = _
.
m(T2 - T1 )
Sources o experimental error
large temperature
change since fewer
molecules
loss o thermal energy rom the apparatus.
Two dierent blocks with the same mass and same energy
input will have a dierent temperature change.
We defne the thermal capacity C o an object as the energy
required to raise its temperature by 1 K. Dierent objects
(even dierent samples o the same substance) will have
dierent values o heat capacity. Specifc heat capacity is
the energy required to raise a unit mass o a substance
by 1 K. Specifc here just means per unit mass.
the container or the substance and the heater will also be
warmed up.
it will take some time or the energy to be shared
uniormly through the substance.
2.
Method o mixtures
The known specifc heat capacity o one substance can be used
to fnd the specifc heat capacity o another substance.
before
temperature TA (hot)
In symbols,
Thermal capacity
Specifc heat
capacity
Q
C = _ (J K- 1 or J C - 1 )
T
Q
c = _ (J kg- 1 K- 1 or J kg- 1 C - 1 )
(m T)
Q = mcT
mix together
Note
A particular gas can have many dierent values o specifc heat
capacity it depends on the conditions used see page 1 61.
These equations reer to the temperature dierence
resulting rom the addition o a certain amount o energy. In
other words, it generally takes the same amount o energy
to raise the temperature o an object rom 25 C to 35 C as
it does or the same object to go rom 402 C to 41 2 C. This
is only true so long as energy is not lost rom the object.
temperature
I an object is raised above room temperature, it starts to
lose energy. The hotter it becomes, the greater the rate at
which it loses energy.
temperature TB
(cold)
mass m A
mass m B
temperature Tmax
after
Procedure:
measure the masses o the liquids mA and m B .
measure the two starting temperatures TA and TB .
increase in
temperature if no
energy is lost
mix the two liquids together.
record the maximum temperature o the mixture Tm a x .
I no energy is lost rom the system then,
energy lost by hot substance cooling down = energy
gained by cold substance heating up
increase in
temperature
in a real situation
m A cA (TA - Tm a x ) = m B cB (Tm a x - TB )
time
Temperature change o an object being heated at a
constant rate
Again, the main source o experimental error is the loss o
thermal energy rom the apparatus; particularly while the
liquids are being transerred. The changes o temperature o
the container also need to be taken into consideration or a
more accurate result.
T H E R M A L P H YS I C S
27
p (tt) tt ltt t
meThods of measurinG
The two possible methods or measuring latent heats shown
below are very similar in principle to the methods or
measuring specifc heat capacities (see previous page) .
temperature C
/
definiTions and miCrosCopiC view
When a substance changes phase, the temperature remains
constant even though thermal energy is still being transerred.
500
400
1.
molten lead
A method or measuring the specifc latent heat o
vaporization o water
set-up
300
liquid and
solid mix
electrical circuit
heater
solid
to electrical
circuit
200
100
V
voltmeter
A ammeter
water
1 2 3 4 5 6 7 8 9 10 11 12 13 14
time / min
heater
Cooling curve or molten lead (idealized)
The amount o energy associated with the phase change is
called the latent heat. The technical term or the change
o phase rom solid to liquid is usion and the term or the
change rom liquid to gas is vaporization.
The energy given to the molecules does not increase their
kinetic energy so it must be increasing their potential energy.
Intermolecular bonds are being broken and this takes energy.
When the substance reezes bonds are created and this process
releases energy.
It is a very common mistake to think that the molecules must
speed up during a phase change. The molecules in water
vapour at 1 00 C must be moving with the same average
speed as the molecules in liquid water at 1 00 C.
The specifc latent heat o a substance is defned as the
amount o energy per unit mass absorbed or released during a
change o phase.
beaker
variable power supply
The amount o thermal energy provided to water at its boiling
point is calculated using electrical energy = I t V. The mass
vaporized needs to be recorded.
ItV
The specifc latent heat L = _
.
(m 1 - m 2 )
Sources o experimental error
Loss o thermal energy rom the apparatus.
Some water vapour will be lost beore and ater timing.
2.
A method or measuring the specifc latent heat o
usion o water
Providing we know the specifc heat capacity o water, we can
calculate the specifc latent heat o usion or water. In the
example below, ice (at 0 C) is added to warm water and the
temperature o the resulting mix is measured.
ice
water
In symbols,
temperature/C
Q
Specifc latent heat L = _ (J kg- 1 ) Q = ML
M
In the idealized situation o no energy loss, a constant rate
o energy transer into a solid substance would result in a
constant rate o increase in temperature until the melting
point is reached:
mix together
mass: m ice
temp.: 0 C
mass: m water
temp.: T C
liquid
solid and liquid mix
mass: m water + m ice
temp.: T mix
solid
I no energy is lost rom the system then,
energy supplied/J
energy lost by water cooling down = energy gained by ice
Phase-change graph with temperature vs energy
m w a te r cw a te r (Tw a te r - Tm ix ) = m ice L u s io n + m ice cw a te r Tm ix
In the example above, the specifc heat capacity o the liquid is
less than the specifc heat capacity o the solid as the gradient
o the line that corresponds to the liquid phase is greater than
the gradient o the line that corresponds to the solid phase.
A given amount o energy will cause a greater increase in
temperature or the liquid when compared with the solid.
Sources o experimental error
Loss (or gain) o thermal energy rom the apparatus.
I the ice had not started at exactly zero, then there would
be an additional term in the equation in order to account
or the energy needed to warm the ice up to 0 C.
Water clinging to the ice beore the transer.
28
T H E R M A L P H YS I C S
Th g 1
Gas laws
(a) constant volume
graph extrapolates pressure / Pa
back to -273 C
-300
-100
100 temp. / C
(2) constant pressure
0
100 temp. / C
(3) constant temperature
volume / m 3
-200
-100
(c) constant temperature
absolute temperature / K
0
absolute temperature / K
pressure / Pa
(b) constant pressure
graph extrapolates
back to -273 C
(1) constant volume
volume / m 3
-200
pressure / Pa
-300
The trends can be seen more clearly i this inormation is
presented in a slightly dierent way.
pressure / Pa
For the experimental methods shown below, the graphs below
outline what might be observed.
volume / m 3
1
-3
volume / m
Points to note:
Although pressure and volume both vary linearly with
Celsius temperature, neither pressure nor volume is
proportional to Celsius temperature.
A dierent sample o gas would produce a dierent straightline variation or pressure (or volume) against temperature
but both graphs would extrapolate back to the same low
temperature, - 273 C. This temperature is known as
absolute zero.
As pressure increases, the volume decreases. In act they are
inversely proportional.
experimenTal invesTiGaTions
1 . Temperature t as the independent variable; P as the
dependent variable; V as the control.
temperature t measured
pressure gauge
to measure P
surface of water
air in
ask
xed volume of air
water (or oil) bath
Fixed volume o gas is trapped in the ask. Pressure is
measured by a pressure gauge.
Temperature o gas altered by temperature o bath time is
needed to ensure bath and gas at same temperature.
2. Temperature t as the independent variable; V as the
dependent variable; P as the control.
temperature t measured
capillary tube
scale to measure V
(length and volume)
surface of water
water bath
bead of
sulfuric acid
gas (air)
zero of scale
volume V
From these graphs or a fxed mass o gas we can say that:
p
1 . At constant V, p T or _ = constant (the pressure law)
T
V = constant (Charless law)
2. At constant p, V T or _
T
1 or p V = constant (Boyles law)
3. At constant T, p _
V
These relationships are known as the ideal gas laws. The
temperature is always expressed in Kelvin (see page 25) . These
laws do not always apply to experiments done with real gases.
A real gas is said to deviate rom ideal behaviour under certain
conditions (e.g. high pressure) .
Volume o gas is trapped in capillary tube by bead o
concentrated suluric acid.
Concentrated suluric acid is used to ensure gas
remains dry.
Heating gas causes it to expand moving bead.
Pressure remains equal to atmospheric.
Temperature o gas altered by temperature o bath; time
is needed to ensure bath and gas at same temperature.
3. P as the independent variable; V as the dependent variable;
t as the control.
zero of scale
scale to
measure V
(length
and volume)
trapped air
pressure gauge
to measure p
air
oil column
pump
surface of oil
oil
Volume o gas measured against calibrated scale.
Increase o pressure orces oil column to compress gas.
Temperature o gas will be altered when volume is
changed; time is needed to ensure gas is always at room
temperature.
T H E R M A L P H YS I C S
29
T g 2
equaTion of sTaTe
definiTions
The three ideal gas laws can be combined together to produce
one mathematical relationship.
pV
_
= constant
T
The concepts o the mole, molar mass and the Avogadro
constant are all introduced so as to be able to relate the mass
o a gas (an easily measurable quantity) to the number o
molecules that are present in the gas.
This constant will depend on the mass and type o gas.
Ideal gas
An ideal gas is one that ollows the gas
laws or all values o o P, V and T (see
page 29) .
Mole
The mole is the basic SI unit or
amount o substance. One mole o
any substance is equal to the amount
o that substance that contains the
same number o particles as 0.01 2 kg o
carbon1 2 ( 1 2 C) . When writing the unit
it is (slightly) shortened to the mol.
Avogadro
constant, NA
This is the number o atoms in 0.01 2 kg
o carbon1 2 ( 1 2 C) . It is 6.02 1 0 2 3 .
Molar mass
The mass o one mole o a substance
is called the molar mass. A simple rule
applies. I an element has a certain mass
number, A, then the molar mass will be
A grams.
N
n=_
NA
number o atoms
number o moles = __
Avogadro constant
I we compare the value o this constant or dierent masses
o dierent gases, it turns out to depend on the number o
molecules that are in the gas not their type. In this case
we use the defnition o the mole to state that or n moles o
ideal gas
pV
_
= a universal constant.
nT
The universal constant is called the molar gas constant R.
The SI unit or R is J mol- 1 K- 1
R = 8.31 4 J mol- 1 K- 1
pV
Summary: _ = R
Or p V = n R T
nT
example
a) What volume will be occupied by 8 g o helium (mass
number 4) at room temperature (20 C) and atmospheric
pressure (1 .0 1 0 5 Pa)
8 = 2 moles
n = _
4
T = 20 + 273 = 293 K
ideal Gases and real Gases
nRT
2 8.31 4 293 = 0.049 m3
__
V = _
p =
1 .0 1 0 5
b) How many atoms are there in 8 g o helium (mass number 4)?
8 = 2 moles
n = _
4
number o atoms = 2 6.02 1 0 2 3
= 1 .2 1 0 2 4
linK beTween The maCrosCopiC and
miCrosCopiC
The equation o state or an ideal gas, pV = nRT, links the three
macroscopic properties o a gas (p, V and T) . Kinetic theory
(page 26) describes a gas as being composed o molecules in
random motion and or this theory to be valid, each o these
macroscopic properties must be linked to the microscopic
behaviour o molecules.
A detailed analysis o how a large number o randomly
moving molecules interact beautiully predicts another
ormula that allows the links between the macroscopic and
the microscopic to be identifed. The derivation o the ormula
only uses Newtons laws and a handul o assumptions. These
assumptions describe rom the microscopic perspective what we
mean by an ideal gas.
An ideal gas is a one that ollows the gas laws or all values
o p, V and T and thus ideal gases cannot be liquefed. The
microscopic description o an ideal gas is given on page 3 1 .
Real gases, however, can approximate to ideal behaviour
providing that the intermolecular orces are small enough
to be ignored. For this to apply, the pressure/density o the
gas must be low and the temperature must be moderate.
Equating the right-hand side o this ormula with the righthand side o the macroscopic equation o state or an ideal gas
shows that:
2 N__
EK
nRT = _
3
N , so
But n = _
NA
N RT = _
2 N__
_
EK
NA
3
__
3 _
R T
EK = _
2 NA
R (the molar gas constant) and NA (Avogadro constant) are
fxed numbers so this equation shows that the absolute
temperature is proportional to the average KE per molecule
__
The detail o this derivation is not required by the IB syllabus but
the assumptions and the approach are outlined on the ollowing
page. The result o this derivation is that the pressure and
volume o the idealized gas are related to just two quantities:
2 N__
pV = _
EK
3
The number o molecules present, N
__
The average random kinetic energy per molecule, EK.
30
T H E R M A L P H YS I C S
T EK
R
R
The ratio __
is called the Boltzmanns constant kB . kB = __
N
N
A
__
3 k T= _
3 _
R T
EK = _
2 B
2 NA
A
mc g
KineTiC model of an ideal Gas
before
wall
Assumptions:
When a molecule bounces o the
walls o a container its momentum
changes (due to the change in
direction momentum is a vector) .
Newtons laws apply to molecular
behaviour
there are no intermolecular orces
except during a collision
There must have been a orce on the
molecule rom the wall (Newton II) .
the molecules are treated as points
the molecules are in random motion
the collisions between the molecules
are elastic (no energy is lost)
A single molecule hitting the walls o the
container.
after
wall
there is no time spent in these
collisions.
There must have been an equal and
opposite orce on the wall rom the
molecule (Newton III) .
Each time there is a collision between
a molecule and the wall, a orce is
exerted on the wall.
The pressure o a gas is explained as
ollows:
The average o all the microscopic
orces on the wall over a period o
time means that there is eectively a
constant orce on the wall rom the gas.
This orce per unit area o the wall is
what we call pressure.
result
F
P= _
A
overall force
on wall
overall force
on molecule
The pressure o a gas is a result o
collisions between the molecules and the
walls o the container.
Since the temperature o a gas is a measure
o the average kinetic energy o the
molecules, as we lower the temperature o
a gas the molecules will move slower. At
absolute zero, we imagine the molecules
to have zero kinetic energy. We cannot go
any lower because we cannot reduce their
kinetic energy any urther!
pressure law
Charless law
boYles law
Macroscopically, at a constant
volume the pressure o a gas is
proportional to its temperature in
kelvin (see page 29) . Microscopically
this can be analysed as ollows
Macroscopically, at a constant pressure,
the volume o a gas is proportional to
its temperature in kelvin (see page 29) .
Microscopically this can be analysed as
ollows
Macroscopically, at a constant
temperature, the pressure o a gas is
inversely proportional to its volume
(see page 29) . Microscopically this can
be seen to be correct.
I the temperature o a gas goes up,
the molecules have more average
kinetic energy they are moving
aster on average.
A higher temperature means aster
moving molecules (see let) .
The constant temperature o gas
means that the molecules have a
constant average speed.
Fast moving molecules will have a
greater change o momentum when
they hit the walls o the container.
Faster moving molecules hit the
walls with a greater microscopic
orce (see let) .
Thus the microscopic orce rom
each molecule will be greater.
I the volume o the gas increases,
then the rate at which these
collisions take place on a unit area
o the wall must go down.
The molecules are moving aster so
they hit the walls more oten.
The average orce on a unit area o
the wall can thus be the same.
For both these reasons, the total
orce on the wall goes up.
Thus the pressure remains the same.
Thus the pressure goes up.
low temperature
low temperature
high temperature
The microscopic orce that each
molecule exerts on the wall will
remain constant.
Increasing the volume o the
container decreases the rate with
which the molecules hit the wall
average total orce decreases.
I the average total orce decreases
the pressure decreases.
high pressure
low pressure
high temperature
constant volume
low pressure
high pressure
Microscopic justifcation o the
pressure law
constant
pressure
low volume
high volume
Microscopic justifcation o
Charless law
constant
temperature
low volume
high volume
Microscopic justifcation o Boyles law
T H E R M A L P H YS I C S
31
ib questons thermal physcs
The following information relates to questions 1 and 2 below.
b) An electrical heater or swimming pools has the ollowing
inormation written on its side:
temperature
A substance is heated at a constant rate o energy transer.
A graph o its temperature against time is shown below.
50 Hz
(i)
P
N
L
O
2.3 kW
Estimate how many days it would take this
heater to heat the water in the swimming pool.
(ii) Suggest two reasons why this can only be an
approximation.
6.
M
[4]
[2]
a) A cylinder ftted with a piston contains 0.23 mol o
helium gas.
K
piston
time
1.
helium gas
Which regions o the graph correspond to the substance
existing in a mixture o two phases?
The ollowing data are available or the helium with the
piston in the position shown.
A. KL, MN and OP
B. LM and NO
Volume
= 5.2 1 0 - 3 m3
C. All regions
Pressure
= 1 .0 1 0 5 Pa
D. No regions
2.
Temperature = 290 K
In which region o the graph is the specifc heat capacity o
the substance greatest?
(i)
A. KL
(ii) State the assumption made in the calculation
in (a) (i) .
B. LM
7.
C. MN
D. OP
3.
B. the gas molecules repel each other more strongly
C. the average velocity o gas molecules hitting the wall is greater
She records her measurements as ollows:
D. the requency o collisions with gas molecules with the
walls is greater
A lead bullet is fred into an iron plate, where it deorms and
stops. As a result, the temperature o the lead increases by an
amount T. For an identical bullet hitting the plate with twice
the speed, what is the best estimate o the temperature increase?
A. T
(1 )
This question is about determining the specifc latent heat o
usion o ice.
A student determines the specifc latent heat o usion o
ice at home. She takes some ice rom the reezer, measures
its mass and mixes it with a known mass o water in an
insulating jug. She stirs until all the ice has melted and
measures the fnal temperature o the mixture. She also
measured the temperature in the reezer and the initial
temperature o the water.
When the volume o a gas is isothermally compressed to
a smaller volume, the pressure exerted by the gas on the
container walls increases. The best microscopic explanation
or this pressure increase is that at the smaller volume
A. the individual gas molecules are compressed
4.
Use the data to calculate a value or the universal
gas constant.
(2)
Mass o ice used
mi
0.1 2 kg
Initial temperature o ice
Ti
-1 2 C
Initial mass o water
mw
0.40 kg
Initial temperature o water
Tw
22 C
Final temperature o mixture
T
1 5 C
B. 2 T
The specifc heat capacities o water and ice are
cw = 4.2 kJ kg- 1 C - 1 and ci = 2.1 kJ kg- 1 C - 1
C. 2 T
D. 4 T
5.
a)
Set up the appropriate equation, representing energy
transers during the process o coming to thermal
equilibrium, that will enable her to solve or the
specifc latent heat L i o ice. Insert values into the
equation rom the data above, but do not solve
the equation.
[5]
Clearly show any estimated values. The ollowing
inormation will be useul:
b)
Explain the physical meaning o each energy transfer
term in your equation (but not each symbol) .
[4]
Specifc heat capacity o water
41 86 J kg- 1 K- 1
c)
Density o water
1 000 kg m- 3
State an assumption you have made about the
experiment, in setting up your equation in (a) .
[1 ]
d)
Why should she take the temperature o
the mixture immediately ater all the ice
has melted?
[1 ]
e)
Explain rom the microscopic point o view, in terms
o molecular behaviour, why the temperature o the
ice does not increase while it is melting.
[4]
In winter, in some countries, the water in a swimming pool
needs to be heated.
a) Estimate the cost o heating the water in a typical swimming
pool rom 5 C to a suitable temperature or swimming. You
may choose to consider any reasonable size o pool.
Cost per kW h o electrical energy
(i)
Estimated values
(ii) Calculations
32
$0.1 0
[4]
[7]
I B Q u E S T I o n S T H E R M A L P H YS I C S
4 w aV e s
Oo
DefinitiOns
simple HarmOnic mOtiOn (sHm)
Many systems involve vibrations or oscillations;
an object continually moves to-and-ro about a
fxed average point (the mean position) retracing
the same path through space taking a fxed
time between repeats. Oscillations involve the
interchange o energy between kinetic and potential.
Simple harmonic motion is defned as the motion that takes place
when the acceleration, a, o an object is always directed towards, and
is proportional to, its displacement rom a fxed point. This acceleration
is caused by a restoring orce that must always be pointed towards
the mean position and also proportional to the displacement rom the
mean position.
Mass moving
between two
horizontal springs
Mass moving on
a vertical spring
Simple pendulum
Buoy bouncing
up and down in
water
An oscillating
ruler as a result
o one end being
displaced while
the other is fxed
Kinetic
energy
Moving
mass
Moving
mass
Moving
pendulum
bob
Moving
buoy
Moving
sections
o the
ruler
Potential
energy store
Elastic potential
energy in the
springs
Elastic potential
energy in the
springs and
gravitational
potential
energy
Gravitational
potential
energy o bob
Gravitational
PE o buoy and
water
Elastic PE o
the bent ruler
a -x or a = - (constant) x
The negative sign signifes that the acceleration is always pointing back
towards the mean position.
acceleration a / m s -2
A
displacement x / m
-A
Points to note about SHM:
The time period T does not depend on the amplitude A. It is
isochronous.
Not all oscillations are SHM, but there are many everyday examples
o natural SHM oscillations.
Defnition
Displacement, The instantaneous distance (SI
x
measurement: m) o the moving
object rom its mean position (in a
specifed direction)
Amplitude, A The maximum displacement (SI
measurement: m) rom the mean
position
Frequency, f The number o oscillations
completed per unit time. The SI
measurement is the number o
cycles per second or Hertz (Hz).
Period, T
The time taken (SI measurement: s)
1
or one complete oscillation. T = __
f
Phase
dierence,
F -x or F = - (constant) x
Since F = ma
This is a measure o how in step
dierent particles are. I moving
together they are in phase. is
measured in either degrees () or
radians (rad) . 360 or 2 rad is one
complete cycle so 1 80 or rad is
completely out o phase by hal a
cycle. A phase dierence o 90 or
/2 rad is a quarter o a cycle.
object oscillates between extremes
example Of sHm: mass between twO springs
sim ple ha rm onic m otion
displacem en t velocity a ccelera tion
aga inst ti m e aga inst ti m e aga inst ti m e
la rge d isplacem en t to right
m a xim um
displa cem en t
zero velocity
m a xim um
acceleration
right
zero
velocity
mass m
la rge force to left
left
sm a ll d isplacem en t to right
right
small velocity
to left
mass m
sm a ll force to left
left
zero
displa cem en t
m a xim um
velocity
zero
acceleration
right
large velocity
to left
mass m
zero net force
left
sm a ll d isplacem en t to left
right
small velocity
to left
mass m
left
sm a ll force to right
la rge d isplacem en t to left
m a xim um
displa cem en t
zero velocity
m a xim um
acceleration
right
zero velocity
mass m
displacement, x
la rge force to right
left
amplitude, A
mean position
W AV E S
33
gh of ho oo
acceleratiOn, VelOcity anD Displacement During sHm
acceleration leads velocity by 90
velocity leads displacement by 90
acceleration and displacement are
1 80 out of phase
displacement
displacement lags velocity by 90
velocity
T
4
3T
4
T
2
velocity lags acceleration by 90
time
T
acceleration
energy cHanges During simple HarmOnic mOtiOn
During SHM, energy is interchanged between KE and PE. Providing there are no resistive forces which dissipate this energy, the
total energy must remain constant. The oscillation is said to be undamped.
Energy in SHM is proportional to:
the mass m
the (amplitude) 2
the (frequency) 2
E
tot
p
Graph showing the
variation with distance, x
of the energy during SHM
k
x0
- x0
x
tot
k
Graph showing the
variation with time, t
of the energy
during SHM
p
t
T
4
34
W AV E S
T
2
3T
4
T
tv v
intrODuctiOn rays anD waVe frOnts
transVerse waVes
Light, sound and ripples on the surace o a pond are all
examples o wave motion.
Suppose a stone is thrown into a pond. Waves spread out as
shown below.
They all transer energy rom one place to another.
situation
They do so without a net motion o the medium through
which they travel.
They all involve oscillations (vibrations) o one sort or
another. The oscillations are SHM.
(1) wave front diagram
(2) ray diagram
A continuous wave involves a succession o individual
oscillations. A wave pulse involves just one oscillation.
Two important categories o wave are transverse and
longitudinal (see below) . The table gives some examples.
direction of
energy ow
The ollowing pages analyse some o the properties that are
common to all waves.
Example of energy transfer
Water ripples
(Transverse)
A foating object gains an up and
down motion.
Sound waves
(Longitudinal)
The sound received at an ear makes
the eardrum vibrate.
Light wave
(Transverse)
The back o the eye (the retina) is
stimulated when light is received.
Earthquake waves
(Both T and L)
Buildings collapse during an
earthquake.
Waves along a
stretched rope
(Transverse)
A sideways pulse will travel down
a rope that is held taut between two
people.
Compression waves
down a spring
(Longitudinal)
A compression pulse will travel
down a spring that is is held taut
between two people.
cross-section through water
wave pattern moves
out from centre
wave pattern at a given
instant of time
water surface moves wave pattern slightly
up and down
later in time
centre of pond
edge of pond
The top o the wave is known as the crest, whereas the
bottom o the wave is known as the trough.
Note that there are several aspects to this wave that can be
studied. These aspects are important to all waves.
The movement o the wave pattern. The wave fronts
highlight the parts o the wave that are moving together.
The direction o energy transer. The rays highlight the
direction o energy transer.
The oscillations o the medium.
It should be noted that the rays are at right angles to the wave
ronts in the above diagrams. This is always the case.
lOngituDinal waVes
Sound is a longitudinal wave. This is because the
oscillations are parallel to the direction o energy transer.
o
This wave is an example o a transverse wave because the
oscillations are at right angles to the direction o energy
transer.
Transverse mechanical waves cannot be propagated through
fuids (liquids or gases) .
view from above
(1) wave front diagram
(2) ray diagram
A point on the wave where everything is bunched together
(high pressure) is known as a compression. A point
where everything is ar apart (low pressure) is known as a
rarefaction.
displacement
of molecules
to the right
loudspeaker
distance
along wave
to the left
rarefaction rarefaction
wave moves
to right
situation
loudspeaker
cross-section through wave at one instant of time
direction
of energy
transfer
motion of air molecules in
same direction as energy
transfer
wave pattern moves
out from loudspeaker
variation of
pressure
rarefaction
v
compression compression
average
pressure
distance
along wave
Relationship between displacement and pressure graphs
W AV E S
35
wv chrcrc
DefinitiOns
waVe equatiOns
There are some useul terms that need to be dened in order to analyse wave
motion in more detail. The table below attempts to explain these terms and they are
also shown on the graphs.
There is a very simple relationship
that links wave speed, wavelength and
requency. It applies to all waves.
Because the graphs seem to be identical, you need to look at the axes o the graphs
careully.
The time taken or one complete
oscillation is the period o the wave, T.
The displacementtime graph on the let represents the oscillations or one point
on the wave. All the other points on the wave will oscillate in a similar manner,
but they will not start their oscillations at exactly the same time.
In this time, the wave pattern will have
moved on by one wavelength, .
The displacementposition graph on the right represents a snapshot o all the
points along the wave at one instant o time. At a later time, the wave will have
moved on but it will retain the same shape.
A
time / s
Term
Symbol
Displacement
Amplitude
Period
x
A
T
Frequency
Wavelength
f
A
Defnition
This measures the change that has taken place
as a result o a wave passing a particular point.
Zero displacement reers to the mean (or average)
position. For mechanical waves the displacement is
the distance (in metres) that the particle moves rom
its undisturbed position.
This is the maximum displacement rom the mean
position. I the wave does not lose any o its energy
its amplitude is constant.
This is the time taken (in seconds) or one complete
oscillation. It is the time taken or one complete wave
to pass any given point.
This is the number o oscillations that take place in one
second. The unit used is the hertz (Hz). A requency o
50 Hz means that 50 cycles are completed every second.
This is the shortest distance (in metres) along the
wave between two points that are in phase with
one another. In phase means that the two points
are moving exactly in step with one another. For
example, the distance rom one crest to the next
crest on a water ripple or the distance rom one
compression to the next one on a sound wave.
c
This is the speed (in m s - 1 ) at which the wave ronts
pass a stationary observer.
Intensity
I
The intensity o a wave is the power per unit area
that is received by the observer. The unit is W m- 2 .
The intensity o a wave is proportional to the square
o its amplitude: I A2 .
The period and the requency o any wave are inversely related. For example, i the
1
requency o a wave is 1 00 Hz, then its period must be exactly ___
o a second.
1 00
36
W AV E S
c = f
In words,
velocity = requency wavelength
example
Wave speed
In symbols,
1
T = __
f
1 =f
Since __
T
position / m
-
distance
c = ________
= __
T
time
A stone is thrown onto a still water
surace and creates a wave. A small
foating cork 1 .0 m away rom the
impact point has the ollowing
displacementtime graph (time is
measured rom the instant the stone
hits the water) :
displacement/cm
T
+
displacement / x
displacement / x
The graphs can be used to represent longitudinal AND transverse waves because
the y-axis records only the value o the displacement. It does NOT speciy the
direction o this displacement. So, i this displacement were parallel to the
direction o the wave energy, the wave would be a longitudinal wave. I this
displacement were at right angles to the direction o the wave energy, the wave
would be a transverse wave.
This means that the speed o the wave
must be given by
2
1
time/s
0
1.4
1.5
1.6 1.7
1.8
-1
-2
a) the amplitude o the wave:
2 cm
.........................................................
b) the speed o the wave:
d = ____
1.0 = 0.67 m s - 1
c = __
t
1.5
.........................................................
c)
the requency o the wave:
1 = ____
1 = 3.33 Hz
f = __
T
0.3
.........................................................
d) the wavelength o the wave:
0.666 = 0.2 m
= __c = ______
3.33
f
.........................................................
eo pu
electrOmagnetic waVes
Visible light is one
part o a much larger
spectrum o similar
waves that are all
electromagnetic.
10 21
frequency
wavelength
3 10 15 Hz
10 -7 m
radium
10 - 12
10 20
-rays
10 - 11
10 19
10 - 10
X-rays
10 18
X-ray tube
10 - 9
10 17
10 - 8
UV
the sun
10 16
10 - 7
10 15
UV
VISIBLE
10 - 6
10 14
IR
10 13
1 10 15 H z
10 12
light bulb
10 - 5
10 - 4
10 - 3
10 11
wavelength / m
Although all
electromagnetic
waves are identical
in their nature, they
have very dierent
properties. This is
because o the huge
range o requencies
(and thus energies)
involved in the
electromagnetic
spectrum.
10 - 13
frequency f / Hz
These oscillating felds
propagate (move)
as a transverse wave
through space. Since
no physical matter
is involved in this
propagation, they
can travel through a
vacuum. The speed
o this wave can be
calculated rom basic
electric and magnetic
constants and it
is the same or all
electromagnetic waves,
3.0 1 08 m s- 1 .
10 22
electric heater
10 - 2
microwaves
Violet
Indigo
VISIBLE
Charges that are
accelerating generate
electromagnetic
felds. I an electric
charge oscillates,
it will produce a
varying electric and
magnetic feld at
right angles to one
another.
possible
source
/m
f / H3
See page 1 32
(option A) or more
details.
Blue
Green
Yellow
Orange
Red
10 10
10 - 1
microwave oven
10 9
1
short radio waves
10 8
10 1
10 7
standard broadcast
10 2
TV broadcast aerial
10 6
10 3
10 5
3 10 14 Hz
10 -6 m
long radio
waves
10 4
10 4
10 5
10 3
radio broadcast aerial
W AV E S
37
ivgg d o od xlly
1. Direct metHODs
The most direct method to measure the speed o sound is to record the time taken t or sound to cover a known distance d: speed c = __dt . In
air at normal pressures and temperatures, sound travels at approximately 330 m s- 1 . Given the much larger speed o light (3 1 0 8 m s- 1 ),
a possible experiment would be to use a stop watch to time the dierence between seeing an event (e.g. the ring o a starting pistol or a
race or seeing two wooden planks being hit together) and hearing the same event some distance away (1 00 m or more).
Echoes can be used to put the source and observer o the sound in the same place. Standing a distance d in ront o a tall wall (e.g.
the side o a building that is not surrounded by other buildings) can allow the echo rom a pulse o sound (e.g. a single clap o the
hands) to be heard. With practice, it is possible or an experimenter to adjust the requency o clapping to synchronize the sound o
the claps with their echoes. When this is achieved, the requency o clapping f can be recorded (counting the number o claps in
a given time) and the time period T between claps is just T = __1f . In this time, the sound travels to the wall and back. The speed o
sound is thus c = 2df.
In either o the above situations a more reliable result will be achieved i a range o distances, rather than one single value is used.
A graph o distance against time will allow the speed o sound to be calculated rom the gradient o the best-t straight line (which
should go through the origin) .
Timing pulses o sound over smaller distances requires small time intervals to be recorded with precision. It is possible to automate
the process using electronic timers and / or data loggers. This equipment would allow, or example, the speed o a sound wave
along a metal rod or through water to be investigated.
2. inDirect metHODs
Since c = f , the speed o sound can be calculated i we measure a sounds requency and wavelength.
Frequency measurement
a) A microphone and a cathode ray oscilloscope (CRO) [page 1 1 6] can display a graph o the oscillations o a sound wave.
Appropriate measurements rom the graph allow the time period and hence the requency to be calculated.
b) Stroboscopic techniques (e.g. fashing light o known requency) can be used to measure the requency o the vibrating object
(e.g. a tuning ork) that is the source o the sound.
c)
Frequency o sound can be controlled at source using a known requency source (e.g. a standard tuning ork) or a calibrated
electronic requency generator.
d) Comparisons can also be made between the unknown requency and a known requency.
Wavelength measurement
a) The intererence o waves (see page 40) can be employed to nd the path dierence between consecutive positions o
destructive intererence. The path dierence between these two situations will be .
S
path A
1
2
*
D
source of
frequency f
path B
detector (microphone
and cathode ray oscilloscope (CRO) )
b) Standing waves (see page 48) in a gas can be employed to nd the location o adjacent nodes. The positions in an enclosed tube
can be revealed either:
in the period pattern made by dust in the tube
electronically using a small movable microphone.
c)
A resonance tube (see page 49) allows the column length or dierent maxima to be recorded. The length distance between
adjacent maxima will be __
.
2
3. factOrs tHat affect tHe speeD Of sOunD
Factors include:
Nature o material
Density
Temperature (or an ideal gas, c
T)
Humidity (or air) .
38
W AV E S
i
intensity
The sound intensity, I, is the amount of energy that a sound wave brings to a unit area every second. The units of sound
intensity are W m- 2 .
It depends on the amplitude of the sound. A more intense sound (one that is louder) must have a larger amplitude.
Intensity (amplitude) 2
displacement of a particle
when a sound wave passes
This relationship between intensity and amplitude is true for all waves.
louder sound
of same pitch
time
amplitude, A
I A2
inVerse square law Of raDiatiOn
The surface area A of a sphere of radius r is calculated using:
As the distance of an observer from a point source of light
increases, the power received by the observer will decrease
as the energy spreads out over a larger area. A doubling of
distance will result in the reduction of the power received to a
quarter of the original value.
area 4A
A = 4r2
If the point source radiates a total power P in all directions,
then the power received per unit area (the intensity I) at a
distance r away from the point source is:
P
I = _____
4r2
For a given area of receiver, the intensity of the received
radiation is inversely proportional to the square of the distance
from the point source to the receiver. This is known as the
inverse square law and applies to all waves.
area A
I x- 2
d
d
waVefrOnts anD rays
As introduced on page 35, waves can be described in terms of
the motion of a wavefront and/or in terms of rays.
rays spreading out
A ray is the path taken by the wave energy as it travels out
from the source.
A wavefront is a surface joining neighbouring points
where the oscillations are in phase with one another. In two
dimensions, the wavefront is a line and in one dimension, the
wavefront is a point.
wavefront
point source
of wave energy
W AV E S
39
soo
interference Of waVes
When two waves o the same type meet, they interfere
and we can work out the resulting wave using the principle
o superposition. The overall disturbance at any point and
at any time where the waves meet is the vector sum o the
disturbances that would have been produced by each o the
individual waves. This is shown below.
I the waves have the same amplitude and the same
requency then the intererence at a particular point can be
constructive or destructive.
graphs
wave 1 displacement (at P)
A
y 1 / unit
(a) wave 1
time
wave 2 displacement (at P)
0
t/s
A
time
(b) wave 2
y 2 / unit
time
time
resultant displacement (at P)
0
t/s
zero result
2A
(c) wave 1 + wave 2 = wave 3
y / unit
time
0
t/s
constructive
time
destructive
Wave superposition
tecHnical language
examples Of interference
Constructive intererence takes place when the two waves
are in step with one another they are said to be in phase.
There is a zero phase difference between them. Destructive
intererence takes place when the waves are exactly out o
step they are said to be out of phase. There are several
dierent ways o saying this. One could say that the phase
dierence is equal to hal a cycle or 1 80 degrees or radians.
Water waves
A ripple tank can be used to view the intererence o water
waves. Regions o large-amplitude waves are constructive
intererence. Regions o still water are destructive intererence.
Intererence can take place i there are two possible routes or
a ray to travel rom source to observer. I the path dierence
between the two rays is a whole number o wavelengths, then
constructive intererence will take place.
path dierence = n constructive
1 ) destructive
path dierence = (n + __
2
n = 0, 1 , 2, 3 . . .
For constructive or destructive intererence to take place, the
sources o the waves must be phase linked or coherent.
Sound
It is possible to analyse any noise in terms o the component
requencies that make it up. A computer can then generate
exactly the same requencies but o dierent phase. This
antisound will interere with the original sound. An observer in
a particular position in space could have the overall noise level
reduced i the waves superimposed destructively at that position.
Light
The colours seen on the surace o a soap bubble are a result
o constructive and destructive intererence o two light
rays. One ray is refected o the outer surace o the bubble
whereas the other is refected o the inner surace.
superpOsitiOn Of waVe pulses
Whenever wave pulses meet, the principle o superposition applies: At any instant in time, the net displacement that results rom
dierent waves meeting at the same point in space is just the vector sum o the displacements that would have been produced by
each individual wave. yo v e ra ll = y1 + y2 + y3 etc.
a) i)
b) i)
A
A
pulse P
A
pulse Q
pulse P
pulse Q
A
ii)
P + Q = 2A
P+ Q = A- A= 0
ii)
iii)
iii)
A
A
pulse Q
pulse P
A
A
pulse Q
40
W AV E S
pulse P
poo
pOlarizeD ligHt
brewsters law
Light is part o the electromagnetic spectrum. It is made up o
oscillating electric and magnetic elds that are at right angles
to one another (or more details see page 1 32) . They are
transverse waves; both elds are at right angles to the direction
o propagation. The plane of vibration o electromagnetic
waves is dened to be the plane that contains the electric eld
and the direction o propagation.
A ray o light incident on the boundary between two media
will, in general, be refected and reracted. The refected ray
is always partially plane-polarized. I the refected ray and
the reracted ray are at right angles to one another, then the
refected ray is totally plane-polarized. The angle o incidence
or this condition is known as the polarizing angle.
direction
of
magnetic
eld
oscillation
reected ray is totally
plane-polarized
plane vibration of
direction of electric EM wave containing
electric oscillations
eld oscillation
incident ray
is unpolarized
transmitted ray
is partially polarized
represents electric eld oscillation into the paper
represents electric eld oscillation in the plane of the paper
There are an innite number o ways or the elds to be oriented.
Light (or any EM wave) is said to be unpolarized i the plane
o vibration varies randomly whereas plane-polarized light
has a xed plane o vibration. The diagrams below represent the
electric elds o light when being viewed head on.
unpolarized light: over a
period of time, the electric
eld oscillates in
random directions
A mixture o polarized light and unpolarized light is partially
plane-polarized. I the plane o polarization rotates uniormly
the light is said to be circularly polarized.
Most light sources emit unpolarized light whereas radio waves, radar
and laboratory microwaves are oten plane-polarized as a result o
the processes that produce the waves. Light can be polarized as a
result o refection or selective absorption. In addition, some crystals
exhibit double refraction or birefringence where an unpolarized
ray that enters a crystal is split into two plane-polarized beams that
have mutually perpendicular planes o polarization.
medium 1 ( vacuum)
medium 2 ( water)
r
distance
along
the wave
polarized light: over a
period of time, the electric
eld only oscillates in
one direction
i
i + r = 90
Brewsters law relates the reractive index o medium 2, n, to
the incident angle i:
sin
sin
n = _____i = _____i = tan i
sin r
cos i
A polarizer is any device that produces plane-polarized light
rom an unpolarized beam. An analyser is a polarizer used to
detect polarized light.
Polaroid is a material which preerentially absorbs any light in
one particular plane o polarization allowing transmission only
in the plane at 90 to this.
a)
b)
transparent
indicates the
preferred directions
zero transmission
maluss law
Optically actiVe substances
When plane-polarized light is incident on an analyser, its
preerred direction will allow a component o the light to be
transmitted:
An optically active substance is one that rotates the
plane o polarization o light that passes through it. Many
solutions (e.g. sugar solutions o dierent concentrations) are
optically active.
0
analysers
preferred
direction
0 cos
transmitted component of
electric eld after analyser
= 0 cos
plane-polarized light
seen head-on with
electric eld amplitude, 0
analyser
plane of
vibration
rotated
through
angle
original
plane of
vibration
optically active
substance
The intensity o light is proportional to the (amplitude) 2 .
I is transmitted intensity o light in W m- 2
Transmitted intensity I E
I0 is incident intensity o light in W m- 2
2
I E0 2 cos 2 as expressed by Maluss law:
I = I0 cos 2
is the angle between the plane o vibration and the analysers
preerred direction
W AV E S
41
u o oo
pOlarOiD sunglasses
cOncentratiOn Of sOlutiOns
Polaroid is a material containing long chain molecules. The
molecules selectively absorb light that have electric elds
aligned with the molecules in the same way that a grid o
wires will selectively absorb microwaves.
For a given optically active solution, the angle through
which the plane o polarization is rotated is proportional to:
The length of the solution through which the planepolarized light passes.
The concentration of the solution.
grid
viewed
head on
electric
eld
absorption
A polarimeter is a device that measures or a given solution.
It consists o two polarizers (a polarizer and an analyser) that
are initially aligned. The optically active solution is introduced
between the two and the analyser is rotated to nd the
maximum transmitted light.
stress analysis
grid
viewed
head on
electric
eld
transmission
Glass and some plastics become bireringent (see page 41 )
when placed under stress. When polarized white light is
passed through stressed plastics and then analysed, bright
coloured lines are observed in the regions o maximum stress.
When worn normally by a person standing up, Polaroid dark
glasses allow light with vertically oscillating electric elds to
be transmitted and absorb light with horizontally oscillating
electric elds.
The absorption will mean that the overall light intensity is
reduced.
Light that has reected from horizontal surfaces will be
horizontally plane-polarized to some extent.
Polaroid sunglasses will preferentially absorb reected light,
reducing glare rom horizontal suraces.
liquiD-crystal Displays (lcDs)
LCDs are used in a wide variety o dierent applications that
include calculator displays and computer monitors. The liquid
crystal is sandwiched between two glass electrodes and is
bireringent. One possible arrangement with crossed polarizers
surrounding the liquid crystal is shown below:
reector
polarizers
With no liquid crystal between the electrodes, the second
polarizer would absorb all the light that passed through the
rst polarizer. The screen would appear black.
The liquid crystal has a twisted structure and, in the absence
o a potential dierence, causes the plane o polarization to
rotate through 90.
This means that light can pass through the second polarizer,
reach the refecting surace and be transmitted back along its
original direction.
With no pd between the electrodes, the LCD appears light.
liquid
crystal
electrodes
etched into glass
light enters the
LCD from the front
furtHer pOlarizatiOn examples
Only transverse waves can be polarized. Page 41 has
concentrated on the polarization o light but all EM waves that
are transverse are able, in principle, to be polarized.
Sound waves, being longitudinal waves, cannot be polarized.
The nature of radio and TV broadcasts means that the signal is
oten polarized and aerials need to be properly aligned i they
are to receive the maximum possible signal strength.
42
W AV E S
A pd across the liquid crystal causes the molecules to
align with the electric eld. This means less light will be
transmitted and this section o the LCD will appear darker.
The extent to which the screen appears grey or black can be
controlled by the pd
Coloured lters can be used to create a colour image.
A picture can be built up from individual picture elements.
Microwave radiation (with a typical wavelength of a few
cm) can be used to demonstrate wave characteristics in the
laboratory. Polarization can be demonstrated using a grid o
conducting wires. I the grid wires are aligned parallel to the
plane o vibration o the electric eld the microwaves will
be absorbed. Rotation o the grid through 90 will allow the
microwaves to be transmitted.
wv bhvou fo
relectiOn anD transmissiOn
relectiOn O twO-DimensiOnal plane waVes
In general, when any wave meets the boundary between two
dierent media it is partially refected and partially transmitted.
The diagram below shows what happens when plane waves
are refected at a boundary. When working with rays, by
convention we always measure the angles between the rays
and the normal. The normal is a construction line that is
drawn at right angles to the surace.
incident ray
normal
incident
ray
reected ray
medium (1)
medium (2)
incident normal
reected
angle i
angle r
reected
ray
transmitted ray
surface
medium (2) is optically
denser than medium (1)
Law of reection: i = r
law O relectiOn
The location and nature o optical images
can be worked out using ray diagrams and
the principles o geometric optics. A ray is
a line showing the direction in which light
energy is propagated. The ray must always
be at right angles to the waveront. The
study o geometric optics ignores the wave
and particle nature o light.
light leaves in
all directions
diuse reection
light leaves in
one direction
mirror reection
types O relectiOn
When a single ray o light strikes a smooth mirror it produces a single refected
ray. This type o perect refection is very dierent to the refection that takes
place rom an uneven surace such as the walls o a room. In this situation, a
single incident ray is generally scattered in all directions. This is an example o a
diffuse refection.
rays spreading out
wavefront
point source
of light
We see objects by receiving light that has come rom them. Most objects do
not give out light by themselves so we cannot see them in the dark. Objects
become visible with a source o light (e.g. the Sun or a light bulb) because diuse
refections have taken place that scatter light rom the source towards our eyes.
The surfaces of the
picture scatter the
light in all directions.
Light from the
central bulb sets
o in all directions.
When a mirror refection takes place, the
direction o the refected ray can be predicted
using the laws o refection. In order to
speciy the ray directions involved, it is usual
to measure all angles with respect to an
imaginary construction line called the normal.
For example, the incident angle is always taken
as the angle between the incident ray and the
normal. The normal to a surace is the line at
right angles to the surace as shown below.
normal
incident ray
i
r
reected ray
The laws o refection are that:
t he incident angle is equal to the refected
angle
An observer sees the painting by
receiving this scattered light.
Our brains are able to work out the location o the object by assuming that rays
travel in straight lines.
t he incident ray, the refected ray and
the normal all lie in the same plane (as
shown in the diagram) .
The second statement is only included in
order to be precise and is oten omitted. It
should be obvious that a ray arriving at a
mirror (such as the one represented above)
is not suddenly refected in an odd direction
(e.g. out o the plane o the page) .
W AV E S
43
s d v d
refractiVe inDex anD snells law
examples
Reraction takes place at the boundary between two media.
In general, a wave that crosses the boundary will undergo a
change o direction. The reason or this change in direction is
the change in wave speed that has taken place.
1. Parallel-sided block
A ray will always leave a parallel-sided block travelling in a
parallel direction to the one with which it entered the block.
The overall eect o the block has been to move the ray
sideways. An example o this is shown below.
As with refection, the ray directions are always specied by
considering the angles between the ray and the normal. I a
ray travels into an optically denser medium (e.g. rom air into
water) , then the ray o light is reracted towards the normal.
I the ray travels into an optically less dense medium then the
ray o light is reracted away from the normal.
incident ray
glass
normal
ray leaves block
parallel to incident ray
refracted
ray
incident ray
less dense medium
more dense medium
2. Ray travelling between two media
I a ray goes between two dierent media, the two individual
reractive indices can be used to calculate the overall reraction
using the ollowing equation
n
sin
_____
n 1 sin 1 = n 2 sin 2 or __
n = sin
1
2
2
1
n1 reractive index o medium 1
1 angle in medium 1
ray refracted
towards normal
normal
refracted
ray
more dense medium
incident ray
n2 reractive index o medium 2
2 angle in medium 2
Suppose a ray o light is shone into a sh tank that contains
water. The reraction that takes place would be calculated as
shown below:
1 st reraction:
air
water
sin a
= _____
(n air = 1.0)
(n water = 1.3) n glass
glass
sin b
(n glass = 1.6)
2nd reraction:
c
less dense medium
a
nglass sin b =
nw ate r sin c
b
nglass
sin c
_____
_____
nw ate r = sin b
ray refracted away
from the normal
Snells law allows us to work out the angles involved.
When a ray is reracted between two dierent media,
the ratio
sin(angle o incidence)
________________
sin(angle o reraction)
is a constant.
The constant is called the reractive index n between the two
media. This ratio is equal to the ratio o the speeds o the
waves in the two media.
sin i
_____
=n
sin r
I the reractive index or a particular substance is given as a
particular number and the other medium is not mentioned
then you can assume that the other medium is air (or to be
absolutely correct, a vacuum) . Another way o expressing this
is to say that the reractive index o air can be taken to
be 1 .0.
For example the reractive index or a type o glass might be
given as
ng la s s = 1 .34
This means that a ray entering the glass rom air with an
incident angle o 40 would have a reracted angle given by
Overall the reraction is rom incident angle a to reracted
angle c.
sin a = _____
sin a _____
sin b
i.e.
no ve rall = _____
sin c
sin c
sin b
= nw a te r
refractiOn Of plane waVes
The reason or the change in direction in reraction is the change
in speed o the wave.
normal
medium 1
( e.g. air)
i
medium 2 (e.g. glass)
wavelength smaller
since speed reduced
boundary
r
refracted ray
Snells law (an experimental law o reraction) states that
sin 40
sin r = _______
= 0.4797
1 .34
r = 28.7
ngla s s
44
n g la s s
sin a ir
Va ir
_______
= ____
= _____
n a ir = sin
Vg la s s
g la s s
W AV E S
sin i
the ratio ____
= constant, or a given requency.
sin r
The ratio is equal to the ratio o the speeds in the dierent media
n1
sin 2
V2 speed o wave in medium 2
___
______
___
n 2 = sin = V1 speed o wave in medium 1
1
ro d
tOtal internal reflectiOn anD critical angle
examples
In general, both refection and reraction can happen at the
boundary between two media.
1.
What a sh sees under water
entire world outside water is
visible in an angle of twice the
critical angle
It is, under certain circumstances, possible to guarantee
complete (total) refection with no transmission at all.
This can happen when a ray meets the boundary and it is
travelling in the denser medium.
n1< n2
partial
transmission
1
grazing
emergence
n1
n2
c
2
critical angle c
At greater than the critical angle, the surface
of the water acts like a mirror. Objects inside
the water are seen by reection.
total
reection
1 2
3
2.
O
source
partial reection
Ray1
This ray is partially refected and partially reracted.
Ray2
This ray has a reracted angle o nearly 90. The
critical ray is the name given to the ray that has a
reracted angle o 90. The critical angle is the angle
o incidence c or the critical ray.
Ray3
Prismatic refectors
A prism can be used in place o a mirror. I the light strikes the
surace o the prism at greater than the critical angle, it must
be totally internally refected.
Prisms are used in many optical devices. Examples include:
periscopes the double refection allows the user to see
over a crowd.
binoculars the double refection means that the
binoculars do not have to be too long
This ray has an angle o incidence greater than the critical
angle. Reraction cannot occur so the ray must be totally
refected at the boundary and stay inside medium 2. The
ray is said to be totally internally refected.
SLR cameras the view through the lens is refected up to
the eyepiece.
The critical angle can be worked out as ollows. For the critical ray,
periscope
n1 sin 1 = n 2 sin 2
1 = 90
2 = c
1
sin c = ___
n2
binoculars
The prism arrangement delivers the
image to the eyepiece the right way
up. By sending the light along the
instrument three times, it also allows
the binoculars to
be shorter.
eyepiece
lens
metHODs fOr Determining refractiVe inDex
experimentally
Part of ray identied
in several positions
objective lens
glass
block
Part of ray in
block can be inferred
from measurements
1.
Locate paths taken by dierent rays either by sending a ray
through a solid and measuring its position or aligning objects
by eye. Uncertainties in angle measurement are dependent on
protractor measurements. (See diagrams on let)
2.
Use a travelling microscope to measure real and apparent
depth and apply ollowing ormula:
Part of
ray identied
ray heading towards
centre will not be
refracted entering the block
semi-circular
glass block
real depth o object
n = _______________________
apparent depth o object
3.
Very accurate measurements o angles o reraction can be
achieved using a prism o the substance and a custom piece
o equipment call a spectrometer.
centre
W AV E S
45
Dfo
DiractiOn
There are some important points to note rom these diagrams.
When waves pass through apertures they tend to spread out.
Waves also spread around obstacles. This wave property is called
diffraction.
D
iraction becomes relatively more important when the
wavelength is large in comparison to the size o the aperture
(or the object) .
geometric
shadow region
d>
T
he wavelength needs to be o the same order o magnitude
as the aperture or diraction to be noticeable.
practical signiicance O DiractiOn
d
geometric
shadow region
d
d
d>
geometric
shadow
region
d
diraction more
important with
smaller
obstacles
geometric
shadow
region
d
The electron microscope resolves items that cannot be
resolved using a light microscope. The electrons have
an eective wavelength that is much smaller than the
wavelength o visible light (see page 1 27) .
geometric
shadow
region
Radio telescopes the size of the dish limits the maximum
resolution possible. Several radio telescopes can be linked
together in an array to create a virtual radio telescope with
a greater diameter and with a greater ability to resolve
astronomical objects. (See page 1 81 )
d < geometric
d
shadow region
Diraction eects mean that it is impossible ever to see atoms
because they are smaller than the wavelength o visible light,
meaning that light will diract around the atoms. It is, however,
possible to image atoms using smaller wavelengths. Practical
devices where diraction needs to be considered include:
CDs and DVDs the maximum amount of information that
can be stored depends on the size and the method used or
recording inormation.
d geometric
shadow
region
Whenever an observer receives inormation rom a source
o electromagnetic waves, diraction causes the energy to
spread out. This spreading takes place as a result o any
obstacle in the way and the width o the device receiving the
electromagnetic radiation. Two sources o electromagnetic
waves that are angularly close to one another will both spread
out and interere with one another. This can aect whether or
not they can be resolved (see page 1 01 ) .
geometric
shadow region
examples O DiractiOn
Diraction provides the reason why we can hear something
even i we can not see it.
d = width o obstacle/gap
Diraction - wave energy is received in geometric shadow region.
basic ObserVatiOns
Diraction is a wave eect. The objects involved (slits, apertures,
etc.) have a size that is o the same order o magnitude as the
wavelength.
intensity
There is a central maximum
intensity.
Other maxima occur roughly
halfway between the minima.
As the angle increases,
the intensity of the
maxima decreases.
1st minimum
Diraction of a single slit.
46
W AV E S
angle
I you look at a distant street light at night and then squint
your eyes the light spreads sideways this is as a result o
diraction taking place around your eyelashes! (Needless to
say, this explanation is a simplifcation.)
to-o o v
principles Of tHe twO-sOurce interference
pattern
Two-source interference is simply another application of
the principle of superposition, for two coherent sources having
roughly the same amplitude.
Two sources are coherent if:
they have the same frequency
matHematics
The location of the light and dark fringes can be mathematically
derived in one of two ways. The derivations do not need to be
recalled.
Method 1
The simplest way is to consider two parallel rays setting off from
the slits as shown below.
there is a constant phase relationship between the two sources.
parallel rays
regions where waves are in phase:
constructive interference
destructive
interference
S1
d
S2
S2
S1
Two dippers in water moving together are coherent sources. This
forms regions of water ripples and other regions with no waves.
Two loudspeakers both connected to the same signal generator
are coherent sources. This forms regions of loud and soft sound.
A set-up for viewing two-source interference with light
is shown below. It is known as Youngs double slit
experiment. A monochromatic source of light is one that
gives out only one frequency. Light from the twin slits (the
sources) interferes and patterns of light and dark regions,
called fringes, can be seen on the screen.
region in which
superposition occurs
Set-up 1
separation
monochromatic of slits
light source
S0
S1
S2
source
twin source
slit
slits (less than 5 mm)
0.1 m
1m
possible
screen
positions
Set-up 2
The use of a laser makes the set-up easier.
laser
double
slit
screen
The experiment results in a regular pattern of light and dark
strips across the screen as represented below.
intensity distribution
intensity
view seen
fringe width, d
path dierence
p = dsin
If these two rays result in a bright patch, then the two rays
must arrive in phase. The two rays of light started out in phase
but the light from source 2 travels an extra distance. This extra
distance is called the path difference.
Constructive interference can only happen if the path
difference is a whole number of wavelengths. Mathematically,
Path difference = n
[where n is an integer e.g. 1 , 2, 3 etc.]
From the geometry of the situation
Path difference = d sin
In other words n = d sin
Method 2
If a screen is used to make the fringes visible, then the rays
from the two slits cannot be absolutely parallel, but the
physical set-up means that this is effectively true.
p
sin = __
s
X
__
tan =
P
D
If is small sin tan
p
S1
X
__
__
X so
s = D
Xs
p = ___
D
s
For constructive
N 2
interference:
p
p = n
S2
Xn s
D
n = ____
D
nD
Xn = _____
s
D
fringe separation d = Xn + 1 - Xn = ___
s
D
___
s=
d
This equation only applies when the angle is small.
Example
Laser light of wavelength 450 nm is shone on two slits that
are 0.1 mm apart. How far apart are the fringes on a screen
placed 5.0 m away?
D
4.5 1 0 - 7 5 = 0.0225 m = 2.25 cm
______________
d = _____
s =
1 .0 1 0 - 4
dark bright dark bright
W AV E S
47
nur d produco of d (ory) v
stanDing waVes
A special case o intererence occurs when two waves meet
that are:
o f the same amplitude
o f the same frequency
There are some points on the rope that are always at rest.
These are called the nodes. The points where the maximum
movement takes place are called antinodes. The resulting
standing wave is so called because the wave pattern remains
xed in space it is its amplitude that changes over time. A
comparison with a normal (travelling) wave is given below.
travelling in opposite directions.
Stationary wave
Normal (travelling)
wave
Amplitude
All points on the
wave have dierent
amplitudes. The
maximum amplitude
is 2A at the antinodes.
It is zero at the nodes.
All points on the
wave have the same
amplitude.
Frequency
All points oscillate
with the same
requency.
All points oscillate
with the same
requency.
Wavelength
This is twice the
distance rom one
node (or antinode)
to the next node (or
antinode) .
This is the shortest
distance (in metres)
along the wave
between two points
that are in phase with
one another.
Phase
All points between
one node and the
next node are moving
in phase.
All points along a
wavelength have
dierent phases.
Energy
Energy is not
transmitted by the
wave, but it does have
an energy associated
with it.
Energy is transmitted
by the wave.
In these conditions a standing wave will be ormed.
The conditions needed to orm standing waves seem quite
specialized, but standing waves are in act quite common. They
oten occur when a wave refects back rom a boundary along
the route that it came. Since the refected wave and the incident
wave are o (nearly) equal amplitude, these two waves can
interere and produce a standing wave.
Perhaps the simplest way o picturing a standing wave would
be to consider two transverse waves travelling in opposite
directions along a stretched rope. The series o diagrams below
shows what happens.
resultant wave
wave 1 moves
wave 2 moves
displacement
a)
total
distance
wave 1
wave 2
displacement
b)
total
wave 1
distance
wave 2
displacement
c)
wave 2
total
distance
wave 1
d)
displacement
total
wave 2
distance
wave 1
displacement
e)
wave 2
Although the example let involved transverse waves on a rope,
a standing wave can also be created using sound or light waves.
All musical instruments involve the creation o a standing
sound wave inside the instrument. The production o laser light
involves a standing light wave. Even electrons in hydrogen
atoms can be explained in terms o standing waves.
A standing longitudinal wave can be particularly hard to
imagine. The diagram below attempts to represent one
example a standing sound wave.
total
wave 1
antinode
zero
movement
max.
movement
node
distance
antinode
Production o standing waves
antinode
total displacement
antinode
etc.
distance
A longitudinal standing wave
node
node
node
A standing wave the pattern remains xed
48
node
W AV E S
node
etc.
bod odo
bOunDary cOnDitiOns
The boundary conditions o the system speciy the conditions
that must be met at the edges (the boundaries) o the system
when standing waves are taking place. Any standing wave that
meets these boundary conditions will be a possible resonant
mode o the system.
As beore, the boundary conditions determine the standing
waves that can exist in the tube. A closed end must be a
displacement node. An open end must be an antinode. Possible
standing waves are shown or a pipe open at both ends and a
pipe closed at one end.
A
N
A
N
A
' = l
N = node
A = antinode
0 = 2l,
1 st harmonic = f0
l
N = node
A = a n tin o d e
1 st ha rm on ic
freq u en cy = f0
0 = 2l
l
1. Transverse waves on a string
I the string is xed at each end, the ends o the string cannot
oscillate. Both ends o the string would refect a travelling wave
and thus a standing wave is possible. The only standing waves
that t these boundary conditions are ones that have nodes at
each end. The diagrams below show the possible resonant modes.
f' = 2 f0
A
N
A
N
A
" =
N
2l
3
f" = 3 f0
A
' = l, f' = 2 f0
N
A
N
A
N
A
Harmonic modes or a pipe open at both ends
N
A
N
A
N
N = node
A = a n tin o d e
l
1 st ha rm on ic
freq u en cy = f0
0 = 4l
2
" = 3 l, f" = 3 f0
N
A
N
A
N
A
N
N
A
' =
l
f' = 3 f0
"' = 2 , f"' = 4 f0
N
A
N
A
N
A
N
A
N
4l
3
N
A
N
A
" =
Harmonic modes or a string
The resonant mode that has the lowest requency is called the
undamental or the frst harmonic. Higher resonant modes
are called harmonics. Many musical instruments (e.g. piano,
violin, guitar etc.) involve similar oscillations o metal strings.
2. Longitudinal sound waves in a pipe
A longitudinal standing wave can be set up in the column o air
enclosed in a pipe. As in the example above, this results rom
the refections that take place at both ends.
example
4l
3
f" = 5 f0
N
A
N
A
N
A
Harmonic modes or a pipe closed at one end
Musical instruments that involve a standing wave in a column o
air include the fute, the trumpet, the recorder and organ pipes.
resOnance tube
An organ pipe (open at one end) is 1 .2 m long.
Calculate its undamental requency.
The speed o sound is 330 m s - 1 .
l = 1 .2 m __ = 1 .2 m (rst harmonic)
4
= 4.8 m
Tuning fork of
known frequency
A
N
x
v = f
330
f = ____ 69 Hz
4.8
Resonance will occur
at dierent values of x .
The distance between
adjacent resonance lengths = 2
W AV E S
49
ib questons waves
1.
2.
b) On the diagram above
A surfer is out beyond the breaking surf in a deep-water
region where the ocean waves are sinusoidal in shape. The
crests are 20 m apart and the surfer rises a vertical distance of
4.0 m from wave trough to crest, in a time of 2.0 s. What is
the speed of the waves?
A. 1 .0 m s - 1
B. 2.0 m s - 1
C. 5.0 m s - 1
D. 1 0.0 m s - 1
(i)
A standing wave is established in air in a pipe with one closed
and one open end.
draw an arrow to indicate the direction in which
the marker is moving.
[1 ]
(ii) indicate, with the letter A, the amplitude of
the wave.
[1 ]
(iii) indicate, with the letter , the wavelength of the
wave.
[1 ]
(iv) draw the displacement of the string a time __T4 later,
where T is the period of oscillation of the wave.
Indicate, with the letter N, the new position of the
marker.
[2]
X
The wavelength of the wave is 5.0 cm and its speed is 1 0 cm s - 1 .
c) Determine
3.
The air molecules near X are
(i)
A. always at the centre of a compression.
(ii) how far the wave has moved in __T4 s.
the frequency of the wave.
[1 ]
[2]
B. always at the centre of a rarefaction.
Interference of waves
C. sometimes at the centre of a compression and sometimes
at the centre of a rarefaction.
d) By reference to the principle of superposition, explain
what is meant by constructive interference.
[4]
D. never at the centre of a compression or a rarefaction.
The diagram below (not drawn to scale) shows an
arrangement for observing the interference pattern
produced by the light from two narrow slits S 1 and S 2 .
This question is about sound waves.
A sound wave of frequency 660 Hz passes through air. The
variation of particle displacement withdistance along the
wave at one instant of time is shown below.
P
0.5
displacement /mm 0
S1
0
1 .0
2 .0
monochromatic
light source
distance / m
yn
dM
S2
O
x
0.5
double slit
b) Using data from the above graph, deduce for this sound
wave,
(i)
4.
the wavelength.
[1 ]
(iii) the speed.
[2]
e) (i)
The diagram below represents the direction of oscillation of a
disturbance that gives rise to a wave.
a transverse wave and
(iii) Deduce an expression for in terms of D and yn .
[1 ]
A wave travels along a stretched string. The diagram below
shows the variation with distance along the string of the
displacement of the string at a particular instant in time. A
small marker is attached to the string at the point labelled
M. The undisturbed position of the string is shown as a
dotted line.
direction of wave travel
M
50
I B Q u E S t I o n S W AV E S
[1 ]
For a particular arrangement, the separation of the slits is
1 .40 mm and the distance from the slits to the screen is
1 .50 m. The distance yn is the distance of the eighth bright
fringe from O and the angle = 2.70 1 0 - 3 rad.
[1 ]
(ii) a longitudinal wave.
State the condition in terms of the distance S 2 X
and the wavelength of the light , for there to be
a bright fringe at P.
[2]
(ii) Deduce an expression for in terms of S 2 X and d. [2]
a) By redrawing the diagram, add arrows to show the
direction of wave energy transfer to illustrate the
difference between
(i)
screen
The distance S 1 S 2 is d, the distance between the double slit
and screen is D and D d such that the angles and shown
on the diagram are small. M is the mid-point of S 1 S 2 and it is
observed that there is a bright fringe at point P on the screen,
a distance yn from point O on the screen. Light from S 2 travels
a distance S 2 X further to point P than light from S 1 .
[1 ]
(ii) the amplitude.
D
single slit
a) State whether this wave is an example of a longitudinal or
a transverse wave.
[1 ]
f) Using your answers to (e) to determine
5.
(i) the wavelength of the light.
[2]
(ii) the separation of the fringes on the screen.
[3]
A bright source of light is viewed through two polarisers
whose preferred directions are initially parallel. Calculate the
angle through which one sheet should be turned to reduce
the transmitted intensity to half its original value.
5 E lE Ctr i Ci ty an d m ag n E ti s m
Eecc ce C'
ConsErvation of ChargE
Coulombs law
Two types o charge exist positive and negative. Equal amounts
o positive and negative charge cancel each other. Matter that
contains no charge, or matter that contains equal amounts o
positive and negative charge, is said to be electrically neutral.
The diagram shows the orce between two point charges that
are ar away rom the infuence o any other charges.
Charges are known to exist because o the orces that exist
between all charges, called the electrostatic force: like
charges repel, unlike charges attract.
F
force
F
The directions o the orces are along the line joining the
charges. I they are like charges, the orces are away rom
each other they repel. I they are unlike charges, the orces
are towards each other they attract.
+
+
-
F
F
F
F
F
+
F
+
F
A very important experimental observation is that charge is
always conserved.
Charged objects can be created by riction. In this process
electrons are physically moved rom one object to another.
In order or the charge to remain on the object, it normally
needs to be an insulator.
before
neutral
comb
neutral hair
distance
r
q2
charge
q1
charge
F
force
Each charge must eel a orce o the same size as the orce on
the other one.
Experimentally, the orce is proportional to the size o both
charges and inversely proportional to the square o the
distance between the charges.
kq1 q 2
q1 q2
F= _
= k_
r2
r2
This is known as Coulombs law and the constant k is called
the Coulomb constant. In act, the law is oten quoted in
a slightly dierent orm using a dierent constant or the
medium called the permittivity, .
value of rst charge
value of second charge
force between
two point charges
q q
F= 1 2 2
4 0 r
constants
after
attraction
+ + positive hair
+
negative +
+
comb
-- +
+
distance between
the charges
permittivity of free
space (a constant)
1
k= _
4 0
I there are two or more charges near another charge, the
overall orce can be worked out using vector addition.
force on q A (due to q C )
electrons have been transferred
from hair to comb
The total charge beore any process must be equal to the
total charge aterwards. It is impossible to create a positive
charge without an equal negative charge. This is the law o
conservation o charge.
overall force on q A
(due to q B and q C)
qB
qA
force on q A (due to q B )
qC
Veo o of eeo foe
ConduCtors and insulators
Electrical conductors
Electrical insulators
A material that allows the fow o charge through it is called an
electrical conductor. I charge cannot fow through a material
it is called an electrical insulator. In solid conductors the fow
o charge is always as a result o the fow o electrons rom atom
to atom.
all metals
e.g. copper
aluminium
brass
graphite
plastics
e.g. polythene
nylon
acetate
rubber
dry wood
glass
ceramics
ElEctri ci ty an d M ag n Eti s M
51
Eecc fe
ElECtriC iElds dEinition
A charge, or combination o charges, is said to produce an
electric feld around it. I we place a test charge at any point
in the feld, the value o the orce that it eels at any point will
depend on the value o the test charge only.
A
A test charge placed at A
would feel this force.
In practical situations, the test charge needs to be small so that it
doesnt disturb the charge or charges that are being considered.
The defnition o electric feld, E, is
F
E=_
q 2 = orce per unit positive point test charge.
Coulombs law can be used to relate the electric feld around a
point charge to the charge producing the feld.
q1
E=_
4 0 r2
When using these equations you have to be very careul:
A test charge placed at B
would feel this force.
q1
not to muddle up the charge producing the feld and the
charge sitting in the feld (and thus eeling a orce)
not to use the mathematical equation or the feld around a
point charge or other situations (e.g. parallel plates) .
B
A test charge would eel a dierent orce at dierent points
around a charge q 1 .
rEprEsEntation o ElECtriC iElds
This is done using feld lines.
+
At any point in a feld:
two opposite
charges
the direction o feld is represented by the direction o the
feld lines closest to that point
the magnitude o the feld is represented by the number o
feld lines passing near that point.
The eld here
must be strong
as the eld lines
are close
together.
The direction of the
force here must be
as shown.
The eld here
must be
weak as the
eld lines
are far apart.
+
+
two like
charges
a negatively
charged
conducting
sphere
F
Field around a positive point charge
The resultant electric feld at any position due to a collection o
point charges is shown to the right.
The parallel feld lines between two plates mean that the electric
feld is uniorm.
Electric feld lines:
(radial eld)
two oppositely charged
parallel metal plates
+ + + + +
+ + + + +
begin on positive charges and end on negative charges
never cross
are close together when the feld is strong.
parallel eld lines
in the centre
Patterns o electric felds
52
ElEctri ci ty an d M ag n Eti s M
Eecc e ee eecc e feece
EnErgy diErEnCE in an ElECtriC iEld
ElECtriC potEntial diErEnCE
When placed in an electric feld, a charge eels a orce. This
means that i it moves around in an electric feld work will be
done. As a result, the charge will either gain or lose electric
potential energy. Electric potential energy is the energy that
a charge has as a result o its position in an electric feld. This
is the same idea as a mass in a gravitational feld. I we lit a
mass up, its gravitational potential energy increases. I the
mass alls, its gravitational potential energy decreases. In the
example below a positive charge is moved rom position A
to position B. This results in an increase in electric potential
energy. Since the feld is uniorm, the orce is constant. This
makes it very easy to calculate the work done.
In the example on the let, the actual energy dierence
between A and B depended on the charge that was moved.
I we doubled the charge we would double the energy
dierence. The quantity that remains fxed between A and B
is the energy dierence per unit charge. This is called the
potential difference, or pd, between the points.
force needed to move charge
q
+
B
+
A
q
distance d
Potential dierence
energy dierence
between two points = per unit charge moved
energy dierence
work done
= __ = __
charge
charge
W
V= _
q
The basic unit or potential dierence is the joule/coulomb,
J C - 1 . A very important point to note is that or a given
electric feld, the potential dierence between any two points
is a single fxed scalar quantity. The work done between these
two points does not depend on the path taken by the test
charge. A technical way o saying this is the electric feld is
conservative.
units
position of higher electric
potential energy
position of lower electric
potential energy
Charge moving in an electric feld
Change in electric potential energy = orce distance
= Eq d
See page 52 or a defnition o electric feld, E.
The smallest amount o negative charge available is the charge
on an electron; the smallest amount o positive charge is the
charge on a proton. In everyday situations this unit is ar too
small so we use the coulomb, C. One coulomb o negative
charge is the charge carried by a total o 6.25 1 0 1 8 electrons.
From its defnition, the unit o potential dierence (pd) is
J C - 1 . This is given a new name, the volt, V. Thus:
1 volt = 1 J C - 1
In the example above the electric potential energy at B is
greater than the electric potential energy at A. We would
have to put in this amount o work to push the charge rom A
to B. I we let go o the charge at B it would be pushed by the
electric feld. This push would accelerate it so that the loss in
electrical potential energy would be the same as the gain in
kinetic energy.
Voltage and potential dierence are dierent words or the
same thing. Potential dierence is probably the better name
to use as it reminds you that it is measuring the dierence
between two points.
When working at the atomic scale, the joule is ar too big to
use or a unit or energy. The everyday unit used by physicists
or this situation is the electronvolt. As could be guessed rom
its name, the electronvolt is simply the energy that would be
gained by an electron moving through a potential dierence
o 1 volt.
1 electronvolt = 1 volt 1 .6 1 0 - 1 9 C
B+
A
+
velocity v
= 1 .6 1 0 - 1 9 J
The normal SI prefxes also apply so one can measure energies
in kiloelectronvolts (keV) or megaelectronvolts (MeV) . The
latter unit is very common in particle physics.
Exmpe
A positive charge released at B will be
accelerated as it travels to point A.
gain in kinetic energy = loss in electric potential energy
1
__
mv2 = Eqd
2
mv2 = 2Eqd
v=
Calculate the speed o an electron accelerated in a vacuum by
a pd o 1 000 V (energy = 1 KeV) .
KE o electron = V e = 1 000 1 .6 1 0 - 1 9
= 1 .6 1 0 - 1 6 J
1 mv2 = 1 .6 1 0 - 1 6 J
_
2
v = 1 .87 1 0 7 m s - 1
2Eqd
_
m
ElEctri ci ty an d M ag n Eti s M
53
Eecc ce
ElECtriCal ConduCtion in a mEtal
CurrEnt
Whenever charges move we say that a current is owing. A current is the name
or moving charges and the path that they ollow is called the circuit. Without a
complete circuit, a current cannot be maintained or any length o time.
Current is defned as the rate o fow
o electrical charge. It is always given
the symbol, I. Mathematically the
defnition or current is expressed as
ollows:
charge owed
Current = __
time taken
Q
dQ
_
I=
or (in calculus notation) I = _
dt
t
1 coulomb
1 ampere =
1 second
Current ows THROUGH an object when there is a potential dierence ACROSS the
object. A battery (or power supply) is the device that creates the potential dierence.
By convention, currents are always represented as the ow o positive charge. Thus
conventional current, as it is known, ows rom positive to negative. Although
currents can ow in solids, liquids and gases, in most everyday electrical circuits
the currents ow through wires. In this case the things that actually move are the
negative electrons the conduction electrons. The direction in which they move is
opposite to the direction o the representation o conventional current. As they move
the interactions between the conduction electrons and the lattice ions means that
work needs to be done. Thereore, when a current ows, the metal heats up. The
speed o the electrons due to the current is called their drit velocity.
conventional current, I
1 A = 1 C s- 1
I a current ows in just one direction
it is known as a direct current.
A current that constantly changes
direction (frst one way then the other)
is known as an alternating current
or ac.
In SI units, the ampere is the base unit
and the coulomb is a derived unit
metal wire
positive ions
held in place
conduction electrons
drift velocity
Electrical conduction in a metal
It is possible to estimate the drit velocity o electrons using the generalized drit
speed equation. All currents are comprised o the movement o charge-carriers and
these could be positive or negative; not all currents involve just the movement o
electrons. Suppose that the number density o the charge-carriers (the number per
unit volume that are available to move) is n, the charge on each carrier is q and their
average speed is v.
In a time t,
the average distance moved by a charge-carrier = v t
so volume o charge moved past a point = A vt
so number o charge-carriers moved past a point = n Avt
so charge moved past a point, Q = nAvt q
Q
current I = _
t
I = nAvq
It is interesting to compare:
A typical drit speed o an electron: 1 0 - 4 m s - 1
(5A current in metal conductor o cross section 1 mm2 )
The speeds o the electrons due to their random motion: 1 0 6 m s - 1
The speed o an electrical signal down a conductor: approx. 3 1 0 8 m s - 1
54
__
ElEctri ci ty an d M ag n Eti s M
1 C=1 As
Eecc cc
ohms law ohmiC and non-ohmiC bEhaviour
rEsistanCE
The graphs below show how the current varies with potential dierence or some
typical devices.
Resistance is the mathematical ratio
between potential dierence and
current. I something has a high
resistance, it means that you would
need a large potential dierence across
it in order to get a current to ow.
(c) diode
potential
dierence
current
current
(b) lament lamp
current
(a) metal at constant
temperature
potential
dierence
potential
dierence
potential dierence
Resistance = __
current
V
In symbols, R = _
I
We defne a new unit, the ohm, , to
be equal to one volt per amp.
1 ohm = 1 V A- 1
I current and potential dierence are proportional (like the metal at constant temperature)
the device is said to be ohmic. Devices where current and potential dierence are not
proportional (like the flament lamp or the diode) are said to be non-ohmic.
Ohms law states that the current owing through a piece o metal is proportional to the potential dierence across it providing the
temperature remains constant.
In symbols,
V I [i temperature is constant]
A device with constant resistance (in other words an ohmic device) is called a resistor.
powEr dissipation
resistor. All this energy is going into heating up the resistor. In
symbols:
energy dierence
Since potential dierence = __
charge owed
charge owed
__
And current =
time taken
This means that potential dierence current
P = V I
Sometimes it is more useul to use this equation in a slightly
dierent orm, e.g.
P = V I but V = I R so
(energy dierence)
(charge owed)
energy dierence
= __ __ = __
time
(charge owed)
(time taken)
P = (I R) I
P = I2 R
Similarly
This energy dierence per time is the power dissipated by the
V2
P= _
R
CirCuits KirChoffs CirCuit laws
ExamplE
An electric circuit can contain many
dierent devices or components. The
mathematical relationship V = IR can be
applied to any component or groups o
components in a circuit.
A 1 .2 kW electric kettle is plugged into
the 250 V mains supply. Calculate
When analysing a circuit it is important
to look at the circuit as a whole. The
power supply is the device that is
providing the energy, but it is the whole
circuit that determines what current
ows through the circuit.
Two undamental conservation laws
apply when analysing circuits: the
conservation o electric charge and the
conservation o energy. These laws are
collectively known as Kirchos circuit
laws and can be stated mathematically as:
First law:
I = 0 (junction)
Second law: V = 0 (loop)
The frst law states that the algebraic
sum o the currents at any junction in
the circuit is zero. The current owing
into a junction must be equal to the
current owing out o a junction. In
the example (right) the unknown
current x = 5.5 + 2.7 3.4 = 4.8 A
3.4A
5.5A
2.7A
(i) the current drawn
(ii) its resistance
1 200
(i) I = _
= 4.8 A
250
x
The second law states that around
any loop, the total energy per unit
charge must sum to zero. Any source
o potential dierence within the loop
must be completely dissipated across
the components in the loop (potential
drop across the component) . Care
needs to be taken to get the sign o any
pd correct.
I the chosen loop direction is rom
the negative side o a battery to its
positive side, this is an increase in
potential and the value is positive
when calculating the sum.
I the direction around the loop is
in the same direction as the current
owing through the component, this
is a potential drop and the value is
negative when calculating the sum.
250
(ii) R = _
= 52
4.8
The example below shows one loop in a
larger circuit. Anti-clockwise consideration
o the loop means that:
1 2.0 - 5.3 - x + 2.7 - 3.2 = 0.
The potential dierence across the bulb,
x = 6.2 V
pd = +12v pd = -3.2v
+
-
+
+
pd = -5.3v
+- -
pd = +2.7v
pd = -x
An example o the use o Kirchos
circuit laws is shown on page 59.
ElEctri ci ty an d M ag n Eti s M
55
re ee d e
rEsistors in sEriEs
ElECtriCal mEtErs
A series circuit has components connected one ater another in a continuous chain.
The current must be the same everywhere in the circuit since charge is conserved.
The total potential dierence is shared among the components.
A current-measuring meter is called an
ammeter. It should be connected in
series at the point where the current
needs to be measured. A perect
ammeter would have zero resistance.
power supply
(24 V)
+ -
I
(2 A)
electrical
energy is
converted
into:
R1
(3 )
R2
(4 )
resistor
thermal energy
bulb
light energy
and thermal
energy
potential
dierence:
6V
R3
(5 )
M
motor
mechanical
energy
and thermal
energy
8V
I
(2 A)
10 V
(6 + 8 + 10 = 24 V)
pd of power supply
Total resistance = 3 + 4 + 5 = 1 2
rEsistors in parallEl
A parallel circuit branches and allows the charges more than one possible route
around the circuit.
Vtotal
I total
Itotal
V
I1
I2 + I3
R1
I1
I2 + I3
V
I2
I3
R2
V
M
R3
I2
I3
Example o a parallel circuit
Since the power supply fxes the potential dierence, each component has the same
potential dierence across it. The total current is just the addition o the currents in
each branch.
1 =_
1 +_
1 +_
1 - 1
_
Ito ta l = I1 + I2 + I3
Rto ta l
5
3
4
20 + 1 5 + 1 2 - 1
__
V
V
V
_
_
_
=
=
+
+
60
R1
R2
R3
47 - 1
=_
1 = _
1 +_
1 +_
1
_
60
Rto ta l
R1
R2
R3
60
Rto ta l = _
47
= 1 .28
56
ElEctri ci ty an d M ag n Eti s M
A meter that measures potential
dierence is called a voltmeter. It
should be placed in parallel with the
component or components being
considered. A perect voltmeter has
infnite resistance.
Example o a series circuit
We can work out what share they take
by looking at each component in turn,
e.g.
The potential dierence across the
resistor = I R1
The potential dierence across the
bulb = I R2
Rto ta l = R1 + R2 + R3
This always applies to a series circuit.
Note that V = IR correctly calculates the
potential dierence across each individual
component as well as calculating it across
the total.
pe e cc e
potEntial dividEr CirCuit
ExamplE
The example on the right is an example o a circuit involving
a potential divider. It is so called because the two resistors
divide up the potential dierence o the battery. You can
calculate the share taken by one resistor rom the ratio o
the resistances but this approach does not work unless the
voltmeters resistance is also considered. An ammeters internal
resistance also needs to be considered. One o the most
common mistakes when solving problems involving electrical
circuits is to assume the current or potential dierence remains
constant ater a change to the circuit. Ater a change, the only
way to ensure your calculations are correct is to start again.
In the circuit below the voltmeter has a resistance o 20 k.
Calculate:
A variable potential divider (a potentiometer) is oten the
best way to produce a variable power supply. When designing
the potential divider, the smallest resistor that is going to be
connected needs to be taken into account: the potentiometers
resistance should be signifcantly smaller.
A potentiometer has
3 terminals the 2 ends
and the central connection
(a) the pd across the 20 k resistor with the switch open
(b) the reading on the voltmeter with the switch closed.
6.0 V
10 k
20 k
V
20 k
20
(a) pd = _________
6.0 = 4.0 V
(20 + 10)
(b) resistance o 20 k resistor and voltmeter combination, R,
given by:
1 =_
1 k - 1
1 +_
_
R
20
20
R = 1 0 k
10
pd = _
6.0 = 3.0 V
(1 0 + 1 0)
sEnsors
A light-dependent resistor (LDR) , is a device whose
resistance depends on the amount o light shining on its
surace. An increase in light causes a decrease in resistance.
output voltage
In order to measure the VI characteristics o an unknown
resistor R, the two circuits (A and B) below are constructed.
Both will both provide a range o readings or the potential
dierence, V, across and current, I, through R. Providing that
R >> the resistance o the potentiometer, this circuit (circuit B)
is preerred because the range o readings is greater.
When light shines on the
LDR LDR, there will be a
decrease in pd across
the LDR.
pd Vtotal
Circuit B allows the potential dierence across R (and
hence the current through R) to be reduced down to zero.
Circuit A will not go below the minimum value achieved
when the variable resistor is at its maximum value.
Circuit B allows the potential dierence across R (and hence
the current through R) to be increased up to the maximum
value Vsu p p ly that can be supplied by the power supply in
regular intervals. The range o values obtainable by Circuit A
depends on a maximum o resistance o the variable resistor.
Circuit A variable resistor
variable resistor
A
Vsupply
R
When light shines on the
10 k LDR, there will be an
increase in pd across
the xed resistor.
A thermistor is a resistor whose value o resistance depends
on its temperature. Most are semi-conducting devices that
have a negative temperature coefcient (NTC) . This
means that an increase in temperature causes a decrease in
resistance. Both o these devices can be used in potential
divider circuits to create sensor circuits. The output potential
dierence o a sensor circuit depends on an external actor.
V
10 k
Circuit B potentiometer
pd Vtotal
A
Vsupply
When the temperature
of the thermistor increases,
there will be an increase in
pd across the xed
resistor.
R
V
When the temperature
of the thermistor
NTC
increases, there will be a
thermistor decrease
in pd across
the thermistor.
potentiometer
ElEctri ci ty an d M ag n Eti s M
57
re
rEsistivity
The resistivity, , o a material is defned in terms o its
resistance, R, its length l and its cross-sectional area A.
l
R= _
A
The units o resistivity must be ohm metres ( m) . Note that
this is the ohm multiplied by the metre, not ohms per metre.
Exmpe
The resistivity o copper is 3.3 1 0 - 7 m; the resistance o a
1 00 m length o wire o cross-sectional area 1 .0 mm2 is:
1 00 = 0.3
R = 3.3 1 0 - 7 _
1 0- 4
invEstigating rEsistanCE
The resistivity equation predicts that the resistance R o a substance will be:
a) Proportional to the length l o the substance
b) Inversely proportional to the cross-sectional area A o the substance.
These relationships can be predicted by considering resistors in series and in parallel:
a) Increasing l is like putting another resistor in series. Doubling l is the same as putting an identical resistor in series. R in series
with R has an overall resistance o 2R. Doubling l means doubling R. So R l. A graph o R vs I will be a straight line going
through the origin.
b) Increasing A is like putting another resistor in parallel. Doubling A is the same as putting an identical resistor in parallel. R in
R
1
1
parallel with R has an overall resistance o __
. Doubling A means halving R. So R __
. A graph o R vs __
will be a straight line
A
A
2
going through the origin.
To practically investigate these relationships, we have:
Independent variable:
Control variables:
Either l or A
A or l (depending on above choice) ;
Temperature;
Substance.
Data collection:
For each value o independent variable:
a range o values or V and I should be recorded
R can be calculated rom the gradient o a V vs I graph.
Data analysis
Values o R and the independent variable analysed graphically.
Possible sources o error/uncertainty include:
Temperature variation o the substance (particularly i currents are high) . Circuits should not be let connected.
d
The cross-sectional area o the wire is calculated by measuring the wires diameter, d, and using A = r2 = ___
. Several sets o
4
measurements should be taken along the length o the wire and the readings in a set should be mutually perpendicular.
2
The small value o the wires diameter will mean that the uncertainties generated using a ruler will be large. This will be improved
using a vernier calliper or a micrometer.
58
ElEctri ci ty an d M ag n Eti s M
Ee e Kcf'
KirCho CirCuit laws ExamplE
Great care needs to be taken when applying Kirchos laws to ensure that every term in the equation is correctly identifed as
positive or negative. The concept o em ( see page 60) as sources o electrical energy can be used along with V = IR to provide
an alternative statement o the second law which may help avoid conusion: Round any closed circuit, the sum o the ems is
equal to the sum o the products o current and resistance.
(em) = (IR)
Process to follow
Draw a ull circuit diagram.
It helps to set up the equations in symbols beore substituting numbers and units.
It helps to be as precise as possible. Potential dierence V is a dierence between two points in the circuit so speciy which two
points are being considered (use labels) .
Give the unknown currents symbols and mark their directions on the diagram. I you make a mistake and choose the wrong
direction or a current, the solution to the equations will be negative.
Use Kirchos frst law to identiy appropriate relationships between currents.
Identiy a loop to apply Kirchos second law. Go all around the loop in one direction (clockwise or anticlockwise) adding the
ems and I R in senses shown below:
With chosen direction around
loop in the direction shown, and
IR are both positive in the
Kircho equation:
emf
I
I
chosen direction around loop
R
I
I
(emf) = (IR)
(If chosen direction opposite to that
shown, values are negative)
The total number o dierent equations generated by Kirchos laws needs to be the same as the number o unknowns or the
problem to be able to be solved.
Use simultaneous equations to substitute and solve or the unknown values.
A new loop can be identifed to check that calculated values are correct.
Exmpe
6v
Sub (1 ) into (4)
20
A
B
i1
30i2 + 1 0(i1 + i2 ) = 5
i1
10
C
D
i3
i3
(3) 4
1 0i1 + 40i2 = 5
(5)
1 20i1 + 40i2 = 24
(6)
1 1 0i1 = 1 9
(6) - (5)
i2
i1 = 0.1 727 A
= 172.7 mA
i2
(3) 1 0i2 = 6 - 30i1
E
5v
= 0.81 82
F
30
Kircho 1 st law junction C(or D)
i1 + i2 = i3
= 81.8 mA
(1 )
Kircho 2nd law and ACDB
1 0i3 + 20i1 = 6
i2 = 0.081 82 A
i3 = 1 72.7 + 81 .8 mA
= 254.5 mA
(2)
Sub (1 ) into (2)
1 0 (i1 + i2 ) + 20i1 = 6
30i1 + 1 0i2 = 6
(3)
Kircho 2nd law and CEFD
- 30i2 - 1 0i3 = -5
(4)
ElEctri ci ty an d M ag n Eti s M
59
ie ece ce
ElECtromotivE forCE and intErnal rEsistanCE
perfect battery
(e m f) = 6 V
When a 6V battery is connected in a circuit some energy will be
used up inside the battery itsel. In other words, the battery has
some internal resistance. The TOTAL energy dierence per unit
charge around the circuit is still 6 volts, but some o this energy is
used up inside the battery. The energy dierence per unit charge
rom one terminal o the battery to the other is less than the total
made available by the chemical reaction in the battery.
internal resistance
r
terminals of battery
R
external resistance
For historical reasons, the TOTAL energy dierence per unit
charge around a circuit is called the electromotive force (emf) .
However, remember that it is not a orce (measured in newtons)
but an energy dierence per charge (measured in volts) .
e m = I Rto ta l
= I(r + R)
= Ir + IR
In practical terms, em is exactly the same as potential
dierence i no current fows.
IR = em - Ir
terminal p d, V lost volts
= I (R + r)
V = - Ir
An electric battery is a device consisting o one or more cells
joined together. In a cell, a chemical reaction takes place,
which converts stored chemical energy into electrical energy.
There are two dierent types o cell: primary and secondary.
A primary cell cannot be recharged. During the lietime o
the cell, the chemicals in the cell get used in a non-reversible
reaction. Once a primary cell is no longer able to provide
electrical energy, it is thrown away. Common examples
include zinccarbon batteries and alkaline batteries.
A secondary cell is designed to be recharged. The chemical
reaction that produces the electrical energy is reversible. A
reverse electrical current charges the cell allowing it to be
reused many times. Common examples include a leadacid
car battery, nickelcadmium and lithium-ion batteries.
The charge capacity o a cell is how much charge can fow beore
the cells stops working. Typical batteries have charge capacities that
are measured in Amp-hours (A h). 1 A h is the charge that fows
when a current o 1 A fows or one hour i.e. 1 A h = 3600 C.
disChargE CharaCtEristiCs
terminal voltage (V)
When current (and thus electrical energy) is drawn rom a cell,
the terminal potential dierence varies with time. A perect cell
would maintain its terminal pd throughout its lietime; real cells,
however, do not. The terminal potential dierence o a typical cell:
loses its initial value quickly,
has a stable and reasonably constant value or most o its lietime.
This is ollowed by a rapid decrease to zero (cell discharges) .
The graph below shows the discharge characteristics or one
particular type o leadacid car battery.
13
12
11
10
9
8
7
0
discharge characteristics
ambient temperature: 25 C
10.8
10.5
14.3A9.5A 5.6A
33A
55A
9.6
165A 110A
2 3
5 10 20 30 60 2 3
discharge time
To experimentally determine the internal resistance r o a cell
(and its em ) , the circuit below can be used:
battery
terminal
terminal pd, V
V
battery
terminal
internal
emf, resistance, r
external resistance, R
I
Procedure:
Vary external resistance R to get a number (ideally 1 0 or
more) o matching readings o V and I over as wide a range
as possible.
Repeat readings.
Do not leave current running or too long (especially at
high values o I) .
Take care that nothing overheats.
Data analysis:
The relevant equation, V = - Ir was introduced above.
A plot o V on the y-axis and I on the x-axis gives a straight
line graph with
gradient = - r
y-intercept =
rECharging sECondary CElls
In order to recharge a secondary cell, it is connected to an external
DC power source. The negative terminal o the secondary cell is
connected to the negative terminal o the power source and the
positive terminal o the power source with the positive terminal
o the secondary cell. In order or a charging current, I, to fow,
the voltage output o the power source must be slightly higher
than that o the battery. A large dierence between the power
source and the cell's terminal potential dierence means that the
charging process will take less time but risks damaging the cell.
secondary cell being charged
I
I
7.6
(min)
60
3.0A
dEtErmining intErnal rEsistanCE
ExpErimEntally
current, I
CElls and battEriEs
5 10 20
(h)
ElEctri ci ty an d M ag n Eti s M
- +
power source
(slightly higher pd)
mec orce fe
magnEtiC iEld linEs
There are many similarities between the magnetic orce and
the electrostatic orce. In act, both orces have been shown to
be two aspects o one orce the electromagnetic interaction
(see page 78) . It is, however, much easier to consider them as
completely separate orces to avoid conusion.
Page 52 introduced the idea o electric felds. A similar concept
is used or magnetic felds. A table o the comparisons between
these two felds is shown below.
Electric eld
Symbol
Magnetic eld
E
B
Caused by
Charges
Magnets (or electric
currents)
Aects
Charges
Magnets (or electric
currents)
Two types
o
Charge: positive and
negative
Pole: North and
South
Simple orce
rule:
Like charges repel,
unlike charges attract
Like poles repel,
unlike poles attract
In order to help visualize a magnetic feld we, once again, use
the concept o feld lines. This time the feld lines are lines o
magnetic feld also called fux lines. I a test magnetic North
pole is placed in a magnetic feld, it will eel a orce.
geographic North Pole
Earth
A magnet free to
move in all
S
directions would
line up pointing
along the eld
lines. A compass is
normally only free to
N
move horizontally, so it
ends up pointing along the
horizontal component of the eld.
geographic
The magnetic North pole of the
compass points towards the geographic South Pole
North Pole hence its name.
An electric current can also cause a magnetic feld. The
mathematical value o the magnetic felds produced in this way
is given on page 63. The feld patterns due to dierent currents
can be seen in the diagrams below.
The direction o the orce is shown by the direction o the
feld lines.
The strength o the orce is shown by how close the lines are
to one another.
A test South pole
here would feel a
force in the
opposite direction.
Force here
strong since
eld lines are
close together.
N
Overall force is in
direction shown
because a North
pole would feel a
repulsion and an
attraction as shown.
S
Force here
weak since
eld lines are
far apart.
rotate
thumb
(current direction)
I current
I
curl of ngers
gives direction of
eld lines
The feld lines are circular around the current.
The direction o the feld lines can be remembered with the righthand grip rule. I the thumb o the right hand is arranged to point
along the direction o a current, the way the fngers o the right
hand naturally curl will give the direction o the feld lines.
Field pattern o a straight wire carrying current
cross-section
current into
page
N
N
S
S
A small magnet placed in the feld would rotate until lined up
with the feld lines. This is how a compass works. Small pieces
o iron (iron flings) will also line up with the feld lines they
willbe induced to become little magnets.
current out
of page
Field pattern o a at circular coil
A long current-carrying coil is called a solenoid.
Field pattern o an isolated bar magnet
Despite all the similarities between electric felds and magnetic
felds, it should be remembered that they are very dierent.
For example:
A magnet does not eel a orce when placed in an electric
feld.
eld pattern of
solenoid is the same
as a bar magnet
cross-section
A positive charge does not eel a orce when placed
stationary in a magnetic feld.
Isolated charges exist whereas isolated poles do not.
The Earth itsel has a magnetic feld. It turns out to be
similar to that o a bar magnet with a magnetic South pole
near the geographic North Pole as shown below.
N
S
poles of solenoid can
be predicted using
right-hand grip rule
Field pattern or a solenoid
ElEctri ci ty an d M ag n Eti s M
61
mec ces
magnEtiC forCE on a CurrEnt
force on current
When a current-carrying wire is placed in a magnetic feld the
magnetic interaction between the two results in a orce. This
is known as the motor effect. The direction o this orce is at
right angles to the plane that contains the feld and the current
as shown below.
I
N
S
zero force
N
I
I
thumb
(force)
F
force (F)
eld ( B)
rst nger
(eld) B
second nger
(current) I
current (I)
F
I
N
S
force at right
angles to plane of
current and eld
lines
Flemings let-hand rule
Experiments show that the orce is proportional to:
the magnitude o the magnetic feld, B
the magnitude o the current, I
the length o the current, L, that is in the magnetic feld
F
I
N
S
S
force maximum
when current and
eld are at right
angles
the sine o the angle, , between the feld and current.
The magnetic feld strength, B is defned as ollows:
F = BIL sin
or
F
B=_
IL sin
A new unit, the tesla, is introduced. 1 T is defned to be equal to
1 N A- 1 m- 1 . Another possible unit or magnetic feld strength is
Wb m- 2 . Another possible term is magnetic ux density.
magnEtiC forCE on a moving ChargE
A single charge moving through a magnetic feld also eels a
orce in exactly the same way that a current eels a orce.
In this case the orce on a moving charge is proportional to:
Since the orce on a moving charge is always at right angles to
the velocity o the charge the resultant motion can be circular.
An example o this would be when an electron enters a region
where the magnetic feld is at right angles to its velocity as
shown below.
the magnitude o the magnetic feld, B
S
the magnitude o the charge, q
the velocity o the charge, v
the sine o the angle, , between the velocity o the charge
and the feld.
We can use these relationships to give an alternative defnition
o the magnetic feld strength, B. This defnition is exactly
equivalent to the previous defnition.
F = Bqv sin
or
F
B=_
qv sin
electron
F
F
r
F
N
F
An electron moving at right angles to a magnetic feld
62
ElEctri ci ty an d M ag n Eti s M
Exe e ec fe ue cue
The formulae used on this page do not need to be
remembered.
straight wirE
The feld pattern around a long straight wire shows that as
one moves away rom the wire, the strength o the feld gets
weaker. Experimentally the feld is proportional to:
the value o the current, I
the inverse o the distance away rom the wire, r. I the
distance away is doubled, the magnetic feld will halve.
two parallEl wirEs dEinition
o thE ampErE
Two parallel current-carrying wires provide a good example
o the concepts o magnetic feld and magnetic orce. Because
there is a current owing down the wire, each wire is
producing a magnetic feld. The other wire is in this feld so
it eels a orce. The orces on the wires are an example o a
Newtons third law pair o orces.
r
length l1
The feld also depends on the medium around the wire.
These actors are summarized in the equation:
length l2
I1
I
B= _
2r
I2
B1
B1
B1
F
r
B1
B1 = eld produced by I1
I1
=
2r
I
r
force felt by I2
= B1 I2 l2
force per unit
length of I2
I
=
Magnetic feld o a straight current
= B1 I2
The constant is called the permeability and changes i the
medium around the wire changes. Most o the time we consider
the feld around a wire when there is nothing there so we
use the value or the permeability o a vacuum, 0 . There is
almost no dierence between the permeability o air and the
permeability o a vacuum. There are many possible units or this
constant, but it is common to use N A- 2 or T m A- 1 .
Permeability and permittivity are related constants. In other
words, i you know one constant you can calculate the other.
In the SI system o units, the permeability o a vacuum is
defned to have a value o exactly 4 1 0 - 7 N A- 2 . See the
defnition o the ampere (right) or more detail.
length l1
I1 I2
2r
length l2
I1
I2
B2
force felt by I1 B
2
= B2 I1 l1
F
force per unit
length of I1
B2 I1 l1
l1
B2
r
B2
B2 = eld produced by I2
I2
=
2r
= B2 I1
The magnetic feld o a solenoid is very similar to the
magnetic feld o a bar magnet. As shown by the parallel feld
lines, the magnetic feld inside the solenoid is constant. It
might seem surprising that the feld does not vary at all inside
the solenoid, but this can be experimentally verifed near the
centre o a long solenoid. It does tend to decrease near the
ends o the solenoid as shown in the graph below.
axis
I
=
r
=
magnEtiC iEld in a solEnoid
B1 I2 l2
l2
I
magnetic eld along axis
constant eld
B
in centre
=
I1 I2
2r
I I
Magnitude o orce per unit length on either wire = ____
2r
1 2
This equation is experimentally used to defne the ampere.
The coulomb is then defned to be one ampere second. I we
imagine two infnitely long wires carrying a current o one
amp separated by a distance o one metre, the equation would
predict the orce per unit length to be 2 1 0- 7 N. Although it is
not possible to have infnitely long wires, an experimental setup can be arranged with very long wires indeed. This allows the
orces to be measured and ammeters to be properly calibrated.
The mathematical equation or this constant feld at the centre
o a long solenoid is
n I
B= _
l
( )
Thus the feld only depends on:
the current, I
distance
(n = number of turns, l = length)
n
the number o turns per unit length, _
l
the nature o the solenoid core,
It is independent o the cross-sectional area o the solenoid.
Variation o magnetic feld in a solenoid
ElEctri ci ty an d M ag n Eti s M
63
ib Questons electrcty and magnetsm
1.
2.
Which one o the feld patterns below could be produced by
two point charges?
A.
C.
B.
D.
Two long, vertical wires X and Y carry currents in the same
direction and pass through a horizontal sheet o card.
X
12 V battery
a) On the circuit above, add circuit symbols showing the
correct positions o an ideal ammeter and an ideal
voltmeter that would allow the V-I characteristics o this
lamp to be measured.
[2]
The voltmeter and the ammeter are connected correctly in
the circuit above.
Y
b) Explain why the potential dierence across the lamp
(i)
cannot be increased to 1 2 V.
[2]
(ii) cannot be reduced to zero.
Iron flings are scattered on the card. Which one o the
ollowing diagrams best shows the pattern ormed by the iron
flings? (The dots show where the wires X and Y enter the card.)
A.
[2]
An alternative circuit or measuring the V-I characteristic
uses a potential divider.
c) (i)
C.
Draw a circuit that uses a potential divider to
enable the V-I characteristics o the flament to
be ound.
(ii) Explain why this circuit enables the potential
dierence across the lamp to be reduced to
zero volts.
B.
3.
[3]
[3]
[2]
The graph below shows the V-I characteristic or two 1 2 V
flament lamps A and B.
D.
potential
dierence / V
12
lamp A
lamp B
This question is about the electric feld due to a charged
sphere and the motion o electrons in that feld.
The diagram below shows an isolated metal sphere in a
vacuum that carries a negative electric charge o 9.0 nC.
-
0
0
a) On the diagram draw arrows to represent the electric feld
pattern due to the charged sphere.
[3]
b) The electric feld strength at the surace o the sphere and
at points outside the sphere can be determined by assuming
that the sphere acts as though a point charge o magnitude
9.0 nC is situated at its centre. The radius o the sphere
is 4.5 1 0 - 2 m. Deduce that the magnitude o the feld
strength at the surace o the sphere is 4.0 1 0 4 V m- 1 .
[1 ]
0.5
1.0
current / A
d) State and explain which lamp has the greater power
dissipation or a potential dierence o 1 2 V.
[3]
The two lamps are now connected in series with a 1 2 V
battery as shown below.
12 V battery
An electron is initially at rest on the surace o the sphere.
c) (i) Describe the path ollowed by the electron as it leaves
the surace o the sphere.
[1 ]
(ii) Calculate the initial acceleration o the electron.
[3]
(iii) State and explain whether the acceleration o the
electron remains constant, increases or
decreases as it moves away rom the sphere.
[2]
(iv) At a certain point P, the speed o the electron is
6.0 1 0 6 m s - 1 . Determine the potential dierence
between the point P and the surace o the sphere. [2]
4.
In order to measure the voltage-current (V-I) characteristics
o a lamp, a student sets up the ollowing electrical circuit.
64
lamp A
e) (i)
lamp B
State how the current in lamp A compares with
that in lamp B.
[1 ]
(ii) Use the V-I characteristics o the lamps to deduce
the total current rom the battery.
[4]
(iii) Compare the power dissipated by the two lamps.
[2]
i B Q u Esti o n s ElEctri ci ty an d M ag n Eti s M
6 c i r c U l a r M o t i o n a n d g r av i t at i o n
Um u m
Mechanics of circUlar Motion
MatheMatics of circUlar Motion
The phrase uniorm circular motion
is used to describe an object that is
going around a circle at constant speed.
Most o the time this also means that
the circle is horizontal. An example o
uniorm circular motion would be the
motion o a small mass on the end o a
string as shown below.
The diagram below allows us to work out the direction o the centripetal
acceleration which must also be the direction o the centripetal orce. This direction
is constantly changing.
situation diagram
vector diagram
change in velocity directed in
towards centre of circle
vB
B
vA
vB
vA
A
The object is shown moving between two points A and B on a horizontal circle. Its velocity
has changed rom vA to vB . The magnitude o velocity is always the same, but the direction
has changed. Since velocities are vector quantities we need to use vector mathematics to
work out the average change in velocity. This vector diagram is also shown above.
mass moves at
constant speed
Example o uniorm circular motion
In this example, the direction o the average change in velocity is towards the centre
o the circle. This is always the case and thus true or the instantaneous acceleration.
For a mass m moving at a speed v in uniorm circular motion o radius r,
v2 [in towards the centre o the circle]
Centripetal acceleration ace n trip e ta l = __
r
A orce must have caused this acceleration. The value o the orce is worked out
using Newtons second law:
It is important to remember that even
though the speed o the object is constant,
its direction is changing all the time.
v m s -1
vA + change = vB
v m s -1
v m s- 1
Centripetal orce (CPF) Fce n trip e ta l = m ace n trip e ta l
m v2 [in towards the centre o the circle]
= ____
r
For example, i a car o mass 1 500 kg is travelling at a constant speed o 20 m s - 1
around a circular track o radius 50 m, the resultant orce that must be acting on it
works out to be
s -1
vm
v m s -1
speed is constant but the direction is
constantly changing
Circular motion the direction o
motion is changing all the time
1 500(20 ) 2
F = __________ = 1 2 000 N
50
It is really important to understand that centripetal orce is NOT a new orce that
starts acting on something when it goes in a circle. It is a way o working out what
the total orce must have been. This total orce must result rom all the other orces
on the object. See the examples below or more details.
This constantly changing direction means
that the velocity o the object is constantly
changing. The word acceleration is used
whenever an objects velocity changes.
This means that an object in uniorm
circular motion MUST be accelerating
even i the speed is constant.
One fnal point to note is that the centripetal orce does NOT do any work. (Work
done = orce distance in the direction of the force.)
The acceleration o a particle travelling in
circular motion is called the centripetal
acceleration. The orce needed to cause
the centripetal acceleration is called the
centripetal force.
exaMples
Earth's gravitational
attraction on
Moon
Moon
Earth
R
R cos
F
T cos
T
friction forces
between
tyres and road
T sin
R sin
mg
W
A conical pendulum centripetal force At a particular speed, the horizontal component
provided by horizontal component
of the normal reaction can provide all the
of tension.
centripetal force (without needing friction) .
C i r C u l a r m o t i o n a n d g r a v i tat i o n
65
au u m
radians
angUlar velocity, , and tiMe period, T
Angles measure the raction o a complete circle that has
been achieved. They can, o course, be measured in degrees
(symbol: ) but in studying circular motion, the radian
(symbol: rad) is a more useul measure.
radius
An object travelling in circular motion must be constantly
changing direction. As a result its velocity is constantly
changing even i its speed is constant (uniorm circular motion) .
We defne the average angular velocity, symbol (omega) as:
angle turned
a v e ra ge = ____________ = ____
time taken
t
The units o angular velocity are radians per second (rad s 1 ) .
r
s distance along
circular arc
The instantaneous angular velocity is the rate o change o angle:
d
= rate o change o angle = ___
dt
1 . Link between and v
The raction o the circle that has been achieved is the ratio o
arc length s to the circumerence:
s
raction o circle = ____
2r
In degrees, the whole circle is divided up into 360 which
defnes the angle as:
s
(in degrees) = ____
360
2r
In radians, the whole circle is divided up into 2 radians
which defnes the angle as:
Angle/
Angle/radian
s
(in radians) = ____
2 = __rs
2r
0
0.00
For small angles (less than
5
0.09
about 0.1 rad or 5) , the arc and
45
0.74
= __
the two radii orm a shape that
4
60
1 .05
approximates to a triangle. Since
radians are just a ratio, the
90
1 .57 = __
2
ollowing relationship applies i
1 80
3.1 4 =
working in radians:
3
270
4.71 = ___
2
sin tan
360
6.28 = 2
y
In a time t, the object rotates
an angle
= __rs s = r
r
s
v = ___
=
= r
t
t
v = r
_
2.
v
x
Link between and time period T
The time period T is the time taken to complete one ull circle.
In this time, the total angle turned is 2 radians, so:
2 or T = ___
2
= ___
T
3. Circular motion equations
Substitution o the above equations into the ormulae or
centripetal orce and centripetal acceleration (page 65) provide
versions that are sometime more useul:
v 2 = r 2 = _____
4 2 r
centripetal acceleration, a = ___
r
T2
m v2
4 2 mr
2
_______
centripetal orce, F = ____
r = mr = T2
circUlar Motion in a vertical plane
1.
Uniorm circular motion o a mass on the end o a string in
a horizontal plane requires a constant centripetal orce to act
and the magnitude o the tension in the string will not change.
Circular motion in the vertical plane is more complicated as the
weight o the object always acts in the same vertical direction.
The object will speed up and slow down during its motion due
to the component o its weight that acts along the tangent to
the circle. The maximum speed will be when the object is at
the bottom and the minimum speed will occur at the top. The
tension in the string will also change during one revolution.
The tension in the string, T, and the weight, mg, are in the same
direction and add together to provide the CPF:
mvto p 2
Tto p + mg = ______
r
To remain in the vertical circle, the object must be moving with
a certain minimum speed. At this minimum top speed, vto p m in ,
the tension is zero and the centripetal orce is provided by the
objects weight:
In a vertical circle, the tension o the string will always act at 90
to the objects velocity so this orce does no work in speeding it
up or slowing it down. The conservation o energy means that:
1 mv2 = constant
mgy + __
2
a) SITUATION DIAGRAM
H
r
small mass
m
s t ri
ng
O
path
y
At the top o the circle:,
m( vto p m in ) 2
mg = __________
r
vto p m in = rg
2.
At the bottom o the circle:,
The tension in the string, T, and the weight, mg, are in opposite
directions and the resultant orce provides the CPF:
mvb o tto m 2
Tb o tto m -mg = ________
r
In order to complete the vertical circle, the KE at the bottom o
the circle must be large enough or the object to arrive at the
top o the circle with sufcient speed( vto p m in = rg
) to complete
the circle. Energetically the object gains PE (= mg 2r) so it
must lose the same amount o KE:
1 m( v
1 m( v
1 mrg
__
) 2 - mg2r = __
) 2 = __
b o tto m m in
to p m in
2
2
2
L
b) FREE-BODY DIAGRAM
( vb o tto m m in ) 2 -4gr = rg
vb o tto m m in = 5rg
F
mg
66
instantaneous
acceleration
The mathematics in the above example (a mass on the end o a
string) can also apply or any vehicle that is looping the loop.
In place o T, the tension in the rope, there is N, the normal
reaction rom the surace.
C i r C u l a r m o t i o n a n d g r a v i tat i o n
n
newtons law of Universal gravitation
Universal gravitational constant G = 6.67 1 0 - 1 1 N m2 kg- 2
I you trip over, you will all down towards the ground.
The ollowing points should be noticed:
Newtons theory o universal gravitation explains what is
going on. It is called universal gravitation because at the core
o this theory is the statement that every mass in the Universe
attracts all the other masses in the Universe. The value o the
attraction between two point masses is given by an equation.
There is a orce acting on each o the masses. These orces are
EQUAL and OPPOSITE (even i the masses are not equal) .
m1
The law only deals with point masses.
The orces are always attractive.
m2
r
F
m1m2
F=
r2
The interaction between two spherical masses turns out to be
the same as i the masses were concentrated at the centres o
the spheres.
Gm 1 m 2
r2
gravitational field strength
The table below should be compared with the one on page 61 .
Gravitational feld strength
In the example on the let the numerical value or the
gravitational feld can be calculated using Newtons law:
GMm
F = _____
r2
g
Caused by
Masses
Aects
Masses
One type o
Mass
Simple orce rule:
All masses attract
orce
Acceleration = _____
mass (rom F = ma)
For the Earth
The gravitational feld is thereore defned as the orce
F
per unit mass. g = __
m m = small point test mass
F
M = 6.0 1 0 2 4 kg
r = 6.4 1 0 6 m
test
mass m 2
value of g =
F
m2
mass M1 producing
gravitational eld g
The SI units or g are N kg- 1 . These are the same as m s - 2 .
Field strength is a vector quantity and can be represented by
the use o feld lines.
gravitational
eld lines
6.67 1 0 - 1 1 6.0 1 0 2 4 = 9.8 m s - 2
g = ________________________
(6.4 1 0 6 ) 2
exaMple
In order to calculate the overall gravitational feld strength
at any point we must use vector addition. The overall
gravitational feld strength at any point between the Earth and
the Moon must be a result o both pulls.
There will be a single point somewhere between the Earth and
the Moon where the total gravitational feld due to these two
masses is zero. Up to this point the overall pull is back to the
Earth, ater this point the overall pull is towards the Moon.
Earth
sphere
GM
g = ____
r2
The gravitational feld strength at the surace o a planet must
be the same as the acceleration due to gravity on the surace.
orce
Field strength is defned to be _____
mass
Symbol
F
Gravitation orces act between ALL objects in the Universe.
The orces only become signifcant i one (or both) o the
objects involved are massive, but they are there nonetheless.
up to X overall pull
is back to Earth
beyond X overall
pull is towards
Moon
Moon
point mass
r1
Field strength around masses (sphere and point)
gravitational
eld
X r2
distance between Earth and Moon = (r1 + r2 )
I resultant gravitational feld at X = zero,
EARTH
Gravitational feld near surace o the Earth
GME a rth GM
M oon
_______
= _______
r1 2
r2 2
C i r C u l a r m o t i o n a n d g r a v i tat i o n
67
iB Questons crcular moton and gravtaton
1.
A ball is tied to a string and rotated at a uniorm speed in
a vertical plane. The diagram shows the ball at its lowest
position. Which arrow shows the direction o the net orce
acting on the ball?
[1 ]
a) (i) On the diagram above, draw and label arrows
to represent the orces on the ball in the
position shown.
A
b) Determine the speed o rotation o the ball.
5.
A particle o mass m is moving with constant speed v in
uniorm circular motion. What is the total work done by
the centripetal orce during one revolution?
mv2
A. Zero
B. ____
2
C. mv2
D. 2mv2
Which diagram correctly shows the direction o the
velocity v and acceleration a o the particle P in the
position shown?
B.
a
v
6.
D.
v
a
[1 ]
a
P
P
C.
4.
[1 ]
A particle P is moving anti-clockwise with constant
speed in a horizontal circle.
A.
P
(i) Derive an expression or the gravitational
feld strength at the surace o a planet in terms
o its mass M, its radius R and the gravitational
constant G.
[2]
(ii) Use your expression in (b) (i) above to estimate
the mass o Jupiter.
[2]
Gravitational felds and potential
a) Derive an expression or the gravitational feld
strength as a unction o distance away rom a
point mass M.
[3]
b) The radius o the Earth is 6400 km and the
gravitational feld strength at its surace is 9.8 N kg- 1 .
Calculate a value or the mass o the Earth.
[2]
c) On the diagram below draw lines to represent the
gravitational feld outside the Earth.
[2]
d) A satellite that orbits the Earth is in the gravitational
feld o the Earth. Discuss why an astronaut inside
the satellite eels weightless.
[3]
v
a and v
P
This question is about circular motion.
A ball o mass 0.25 kg is attached to a string and is made to
rotate with constant speed v along a horizontal circle o radius
r = 0.33 m. The string is attached to the ceiling and makes an
angle o 30 with the vertical.
30
vertical
r = 0.33 m
68
[2]
b) The gravitational feld strength at the surace o
Jupiter is 25 N kg- 1 and the radius o Jupiter is
7.1 1 0 7 m.
D
3.
[3]
This question is about gravitational felds.
a) Defne gravitational feld strength.
B
C
2.
[2]
(ii) State and explain whether the ball is in equilibrium. [2]
i B Q u e s t i o n s C i r C u l a r m o t i o n a n d g r a v i tat i o n
7 ato m i c , n u clE ar an d par ti clE ph ys i cs
E e
Emission spEctra and absorption spEctra
Explanation of atomic spEctra
When an element is given enough energy it emits light. This
light can be analysed by splitting it into its various colours
(or requencies) using a prism or a diraction grating. I all
possible requencies o light were present, this would be
called a continuous spectrum. The light an element emits,
its emission spectrum, is not continuous, but contains
only a ew characteristic colours. The requencies emitted
are particular to the element in question. For example, the
yellow-orange light rom a street lamp is oten a sign that
the element sodium is present in the lamp. Exactly the same
particular requencies are absent i a continuous spectrum o
light is shone through an element when it is in gaseous orm.
This is called an absorption spectrum.
In an atom, electrons are bound to the nucleus. See page 77,
the atomic model. This means that they cannot escape without
the input o energy. I enough energy is put in, an electron can
leave the atom. I this happens, the atom is now positive overall
and is said to be ionized. Electrons can only occupy given energy
levels the energy o the electron is said to be quantized. These
energy levels are fxed or particular elements and correspond
to allowed obitals. The reason why only these energies are
allowed orms a signifcant part o quantum theory (see HL
topic 1 2).
spectra: emission set-up
prism
(or diraction
grating)
slit
sample
of gas
spectral
lines
light emitted
from gas
spectra: absorption set-up
prism
( or diraction
grating)
slit
light
source
spectral
lines
all frequencies sample of gas
(continuous spectrum)
330
mercury 313
helium
415 420
334
365366
398405
389 403
361 371 382 396 412
319
384 397
410
380 389
hydrogen
546 579
439 447 471 492502
434
energy in joules
E = hf
Plancks constant
6.63 1 0 - 3 4 J s
requency o
light in Hz
Speed o light in m s - 1
Since c = f
hc
= ___
E
Wavelength in m
Thus the requency o the light, emitted or absorbed, is fxed
by the energy dierence between the levels. Since the energy
levels are unique to a given element, this means that the
emission (and the absorption) spectrum will also be unique.
569 590 615
436
The energy o a photon is given by
energy
sodium
When an electron moves between energy levels it must emit
or absorb energy. The energy emitted or absorbed corresponds
to the dierence between the two allowed energy levels.
This energy is emitted or absorbed as packets o light called
photons. A higher energy photon corresponds to a higher
requency (shorter wavelength) o light.
588
668
486
allowed
energy
levels
656
wavelength, / nm
300
310
320
330
340
350 360 370 380 390 400
450
500
550
600 650 700
yellow
invisible (UV)
approximate colour
violet indigo
blue
green
red
orange invisible
(IR)
Emission spectra
m e rcu ry
33 0
334
3 13
h e liu m
415 4 2 0
3 65 3 6 6
39 8 405
389 403
3 61 3 71 3 8 2 3 9 6 412
3 19
384 397
380 389
410
h y d ro ge n
photon of pa rticular
frequency absorbed
5 69 5 9 0 615
43 6
energy
s o d iu m
electron promoted from low
energy level to higher energy level
5 4 6 57 9
43 9 4 47
471 49 2 5 0 2
43 4
588
486
668
allowed
energy
levels
65 6
wa ve le n gth , / n m
300
3 10
320
330
340
35 0 3 60 370 3 8 0 39 0 40 0
45 0
50 0
550
60 0
65 0
70 0
y e l lo w
a p p ro xim a te c o lo u r
i n vi s i b le ( U V)
vio le t in d igo
blue
gre e n
re d
o ra n ge i n vi s i b le
( IR )
Absorption spectra
electron falls from high
energy level to lower energy level
photon of pa rticular
frequency emitted
Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
69
ne
isotopEs
nuclEar stability
When a chemical reaction takes place, it involves the outer
electrons o the atoms concerned. Dierent elements have
dierent chemical properties because the arrangement o
outer electrons varies rom element to element. The chemical
properties o a particular element are fxed by the amount o
positive charge that exists in the nucleus in other words, the
number o protons. In general, dierent nuclear structures
will imply dierent chemical properties. A nuclide is the
name given to a particular species o atom (one whose nucleus
contains a specifed number o protons and a specifed number
o neutrons) . Some nuclides are the same element they have
the same chemical properties and contain the same number o
protons. These nuclides are called isotopes they contain the
same number o protons but dierent numbers o neutrons.
Many atomic nuclei are unstable. The process by which they
decay is called radioactive decay (see page 72) . It involves
emission o alpha () , beta () or gamma () radiation.
The stability o a particular nuclide depends greatly on the
numbers o neutrons present. The graph below shows the
stable nuclides that exist.
For small nuclei, the number o neutrons tends to equal
the number o protons.
For large nuclei there are more neutrons than protons.
Nuclides above the band o stability have too many
neutrons and will tend to decay with either alpha or beta
decay (see page 72) .
neutron number, N
Nuclides below the band o stability have too ew
neutrons and will tend to emit positrons (see page 73) .
notation
A
mass number equal to number
of nucleons
chemical symbol
Z
160
150
140
130
120
110
atomic number equal to number of
protons in the nucleus
100
Nuclide notation
90
80
ExamplEs
Notation
70
Description
Comment
1
12
6
C
carbon-1 2
isotope o 2
2
13
6
C
carbon-1 3
isotope o 1
3
238
92
U
uranium-238
4
1 98
78
Pt
platinum-1 98
same mass number
as 5
5
1 98
80
Hg
mercury-1 98
same mass number
as 4
60
50
40
30
20
10
Each element has a unique chemical symbol and its own
atomic number. No.1 and No.2 are examples o two isotopes,
whereas No.4 and No.5 are not.
In general, when physicists use this notation they are
concerned with the nucleus rather than the whole atom.
Chemists use the same notation but tend to include the
overall charge on the atom. Thus 1 26 C can represent the
carbon nucleus to a physicist or the carbon atom to a chemist
depending on the context. I the charge is present the
situation becomes unambiguous. 31 57 C l must reer to a chlorine
ion an atom that has gained one extra electron.
70
0
10 20 30 40 50 60 70 80 90 100
atomic number, Z
Key
N number o neutrons
Z number o protons
naturally occurring stable nuclide
naturally occurring -emitting nuclide
artifcially produced -emitting nuclide
naturally occurring - -emitting nuclide
artifcially produced + -emitting nuclide
artifcially produced - -emitting nuclide
artifcially produced electron-capturing nuclide
artifcial nuclide decaying by spontaneous fssion
At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
fe e
strong nuclEar forcE
WEak nuclEar forcE
The protons in a nucleus are all positive. Since like charges
repel, they must be repelling one another all the time. This
means there must be another orce keeping the nucleus
together. Without it the nucleus would fy apart. We know a
ew things about this orce.
The strong nuclear orce (see box let) explains why nuclei
do not fy apart and thus why they are stable. Most nuclei,
however, are unstable. Mechanisms to explain alpha and
gamma emission (see page 72) can be identied but another
nuclear orce must be involved i we wish to be able to explain
all aspects o the nucleus including beta emission. We know a
ew things about this orce:
It must be strong. I the proton repulsions are calculated it is
clear that the gravitational attraction between the nucleons
is ar too small to be able to keep the nucleus together.
It must be very short-ranged as we do not observe this
orce anywhere other than inside the nucleus.
It must be weak. Many nuclei are stable and beta emission
does not always occur.
It must be very short-ranged as we do not observe this
orce anywhere other than inside the nucleus.
It is likely to involve the neutrons as well. Small nuclei
tend to have equal numbers o protons and neutrons.
Large nuclei need proportionately more neutrons in order
to keep the nucleus together.
Unlike the strong nuclear orce, it involves the lighter
particles (e.g. electrons, positrons and neutrinos) as well as
the heavier ones (e.g. protons and neutrons) .
The name given to this orce is the strong nuclear force.
The name given to this orce is the weak nuclear force.
othEr fundamEntal forcEs/intEractions
Electromagnetic
The standard model o particle physics is based around the orces
that we observe on a daily basis along with the two new orces
that have been identied as being involved in nuclear stability
(above) . As a result in the standard model, there are only our
undament orces (or interactions) that are known to exist.
These are Gravity, Electromagnetic, Strong and Weak. More
detail about all these orces is discussed on page 78. Outline
inormation about two everyday interactions is listed below:
This single orce includes all the orces that we normally
categorize as either electrostatic or magnetic.
Gravity
Gravity is the orce o attraction between all objects that
have mass.
Gravity is always attractive masses are pulled together.
The range o the gravity orce is innite.
Despite the above, the gravity orce is relatively quite weak.
At least one o the masses involved needs to be large or the
eects to be noticeable. For example, the gravitational orce
o attraction between you and this book is negligible, but the
orce between this book and the Earth is easily demonstrable
(drop it) .
Newtons law o gravitation describes the mathematics
governing this orce.
Electromagnetic orces involve charged matter.
Electromagnetic orces can be attractive or repulsive.
The range o the electromagnetic orce is innite.
The electromagnetic orce is relatively strong tiny
imbalances o charges on an atomic level give rise to
signicant orces on the laboratory scale.
At the end o the 1 9th century, Maxwell showed that the
electrostatic orce and the magnetic orce were just two
dierent aspects o the more undamental electromagnetic
orce.
The mathematics o the electromagnetic orce is described by
Maxwells equations.
Friction (and many other everyday orces) is simply the
result o the orce between atoms and this is governed by the
electromagnetic interaction.
The electromagnetic orce and the weak nuclear orce are now
considered to be aspects o the single electroweak orce.
particlEs that ExpEriEncE and mEdiatE thE fundamEntal forcEs.
See page 78 onwards or more details about the standard model or the undamental structure o matter. The ollowing table
summarizes which particles experience these orces and how they are mediated.
Particles experience
Particles mediate
Gravitational
Weak
Electromagnetic
Strong
All
Quark, Gluon
Charged
Quark, Gluon
Graviton
W+ , W- , Z 0
Gluon
Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
71
rv 1
ioniZing propErtiEs
EffEcts of radiation
Many atomic nuclei are unstable. The
process by which they decay is called
radioactive decay. Every decay
involves the emission o one o three
dierent possible radiations rom the
nucleus: alpha () , beta () or
gamma () .
At the molecular level, an ionization could cause damage directly to a biologically
important molecule such as DNA or RNA. This could cause it to cease unctioning.
Alternatively, an ionization in the surrounding medium is enough to interere with
the complex chemical reactions (called metabolic pathways) taking place.
Molecular damage can result in a disruption to the unctions that are taking place
within the cells that make up the organism. As well as potentially causing the cell
to die, this could just prevent cells rom dividing and multiplying. On top o this, it
could be the cause o the transormation o the cell into a malignant orm.
As all body tissues are built up o cells, damage to these can result in damage to the body
systems that have been aected. The non-unctioning o these systems can result in death
or the animal. I malignant cells continue to grow then this is called cancer.
radiation safEty
Alpha, beta and gamma all come rom
the nucleus
All three radiations are ionizing. This
means that as they go through a
substance, collisions occur which cause
electrons to be removed rom atoms.
Atoms that have lost or gained electrons
are called ions. This ionizing property
allows the radiations to be detected. It
also explains their dangerous nature.
When ionizations occur in biologically
important molecules, such as DNA,
unction can be aected.
There is no such thing as a sae dose o ionizing radiation. Any hospital procedures
that result in a patient receiving an extra dose (or example having an X-ray scan)
should be justiable in terms o the inormation received or the benet it gives.
There are three main ways o protecting onesel rom too large a dose. These can be
summarized as ollows:
Run away!
The simplest method o reducing the dose received is to increase the distance
between you and the source. Only electromagnetic radiation can travel large
distances and this ollows an inverse square relationship with distance.
Dont waste time!
I you have to receive a dose, then it is important to keep the time o this
exposure to a minimum.
I you cant run away, hide behind something!
Shielding can always be used to reduce the dose received. Lead-lined aprons can
also be used to limit the exposure or both patient and operator.
propErtiEs of alpha, bEta and gamma radiations
Property
Alpha,
Beta,
Gamma,
Eect on photographic lm
Yes
Yes
Yes
Approximate number o
ion pairs produced in air
1 0 4 per mm travelled
1 0 2 per mm travelled
1 per mm travelled
Typical material needed to
absorb it
1 0 - 2 mm aluminium; piece o paper
A ew mm aluminium
1 0 cm lead
Penetration ability
Low
Medium
High
Typical path length in air
A ew cm
Less than one m
Eectively innite
Defection by E and B elds
Behaves like a positive charge
Behaves like a negative charge
Not defected
Speed
About 1 0 7 m s - 1
About 1 0 8 m s - 1 , very variable
3 1 08 m s- 1
naturE of alpha, bEta and
gamma dEcay
When a nucleus decays the mass
numbers and the atomic numbers must
balance on each side o the nuclear
equation.
Alpha particles are helium nuclei, 42
or 42 He 2 + . In alpha decay, a chunk
o the nucleus is emitted. The portion
that remains will be a dierent
nuclide.
A
Z
X
e.g.
72
( A - 4)
(Z - 2)
2 41
95
4
2
Y+
Am
237
93
4
2
Np +
The atomic numbers and the mass
numbers balance on each side o the
equation.
(95 = 93 + 2 and 241 = 237 + 4)
Beta particles are electrons, -01
or -01 e - , emitted rom the nucleus.
The explanation is that the electron is
ormed when a neutron decays. At the
same time, another particle is emitted
called an antineutrino.
0
1
n 11 p +
0
-1
_
+
Since an antineutrino has no charge and
virtually no mass it does not aect the
equation.
At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
A
Z
X ( Z + A1 ) Y +
e.g.
90
38
Sr
90
39
0
-1
_
+
Y+
0
-1
_
+
Gamma rays are unlike the other two
radiations in that they are part o the
electromagnetic spectrum. Ater their
emission, the nucleus has less energy
but its mass number and its atomic
number have not changed. It is said to
have changed rom an excited state
to a lower energy state.
A
Z
X* AZ X + 00
Excited
state
Lower energy
state
rv 2
antimattEr
The nuclear model given on page 77 is somewhat simplifed.
One important thing that is not mentioned there is the
existence o antimatter. Every orm o matter has its
equivalent orm o antimatter. I matter and antimatter came
together they would annihilate each other. Not surprisingly,
antimatter is rare but it does exist. For example, another
orm o radioactive decay that can take place is beta plus or
positron decay. In this decay a proton decays into a neutron,
and the antimatter version o an electron, a positron, is
emitted.
background radiation
Some cosmic gamma rays will be responsible, but there will also be
, and radiation received as a result o radioactive decays that
are taking place in the surrounding materials. The pie chart below
identifes typical sources o background radiation, but the actual
value varies rom country to country and rom place to place.
Radioactive decay is a natural phenomenon and is going on
around you all the time. The activity o any given source is
measured in terms o the number o individual nuclear decays
that take place in a unit o time. This inormation is quoted in
becquerels (Bq) with 1 Bq = 1 nuclear decay per second.
Experimentally this would be measured using a Geiger
counter, which detects and counts the number o ionizations
taking place inside the GM tube. A working Geiger counter
will always detect some radioactive ionizations taking place
even when there is no identifed radioactive source: there is a
background count as a result o the background radiation.
A reading o 30 counts per minute, which corresponds to the
detector registering 30 ionizing events, would not be unusual.
To analyse the activity o a given radioactive source, it is
necessary to correct or the background radiation taking place. It
would be necessary to record the background count without the
radioactive source present and this value can then be subtracted
rom all readings with the source present.
random dEcay
Radioactive decay is a random process and is not aected by
external conditions. For example, increasing the temperature o
a sample o radioactive material does not aect the rate o decay.
This means that is there no way o knowing whether or not a
particular nucleus is going to decay within a certain period o
time. All we know is the chances o a decay happening in that time.
Although the process is random, the large numbers o atoms
involved allows us to make some accurate predictions. I we
1
1
p 10 n +
19
10
Ne
19
9
0
+1
+ +
F+
0
+1
+ +
The positron, , emission is accompanied by a neutrino.
+
The antineutrino is the antimatter orm o the neutrino.
For more details see page 78.
foo d
ra d on
m ed icin e
n u clea r in d u stry
bu ild ings/soil
cosm ic
m ed icin e 14%
n u clea r in d u stry 1 %
bu ild in gs/soil 18 %
cosm ic 14%
ra d on 42 %
foo d /
d rin king water 11 %
n a tu ra l
ra d ia tion
85 %
start with a given number o atoms then we can expect a certain
number to decay in the next minute. I there were more atoms in
the sample, we would expect the number decaying to be larger.
On average the rate o decay o a sample is proportional to the
number o atoms in the sample. This proportionality means that
radioactive decay is an exponential process. The number o
atoms o a certain element, N, decreases exponentially over time.
Mathematically this is expressed as:
dN
___
-N
dt
Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
73
h-e
amount o su stance
ExamplE
In simple situations, working out how much radioactive
material remains is a matter o applying the hal-lie property
several times. A common mistake is to think that i the hallie o a radioactive material is 3 days then it will all decay in
six days. In reality, ater six days (two hal-lives) a hal o a
hal will remain, i.e. a quarter.
number of nuclides available to decay
half-lifE
There is a temptation to think that every quantity that
decreases with time is an exponential decrease, but
exponential curves have a particular mathematical property.
In the graph shown below, the time taken or hal the number
o nuclides to decay is always the same, whatever starting
value we choose. This allows us to express the chances o
decay happening in a property called the half-life, T__1 . The
2
hal-lie o a nuclide is the time taken or hal the number
o nuclides present in a sample to decay. An equivalent
statement is that the hal-lie is the time taken or the rate o
decay (or activity) o a particular sample o nuclides to halve.
A substance with a large hal-lie takes a long time to decay. A
substance with a short hal-lie will decay quickly. Hal-lives
can vary rom ractions o a second to millions o years.
T1
2
(x)
decay of radioactive
parent nuclei
increase of stable
daughter nuclei
after 2 half-lives
3 of the nuclei are daughter nuclei
4
after 2 half-lives
1 of the original parent nuclei will remain
4
1
2
number = x
N0
1
The time taken to halve from
any point is always T1 .
2
number = x
2
N1/2
( 2x )
N 1/4
3
6
9
12 time / days
The decay o parent into daughter
e.g. The hal-lie o
14
6
C is 5570 years.
Approximately how long is needed beore less than 1 % o a
sample o 1 64 C remains?
N 1/8
N1/16
T1
2
T1
T1
2
2
T1
2
time
half-life
Hal-lie o an exponential decay
Time
Fraction left
T__1
50%
2
2 T__1
25%
2
3 T__1
1 2.5%
2
4T__1
invEstigating
half-lifE ExpErimEntally
When measuring the activity o a source, the background rate
should be subtracted.
I the hal-lie is short, then readings can be taken o
activity against time.
A simple graph o activity against time would produce
the normal exponential shape. Several values o hal-lie
could be read rom the graph and then averaged. This
method is simple and quick but not the most accurate.
A graph o ln (activity) against time could be produced.
This should give a straight line and the decay constant can
be calculated rom the gradient. See page 21 7.
I the hal-lie is long, then the activity will eectively be
constant over a period o time. In this case one needs to fnd
a way to calculate the number o nuclei present, N, and then
use
5 T__1
~ 3.1 %
2
6 T__1
~ 1 .6%
2
7 T__1
~ 0.8%
2
6 hal lives
= 33420 years
7 hal lives
= 38990 years
approximately 37000 years needed
simulation
The result o the throw o a die is a random process and can
be used to simulate radioactive decay. The dice represent
nuclei available to decay. Each throw represents a unit o
time. Every six represents a nucleus decaying meaning this die
is no longer available.
dN
___
= - N.
dt
74
~ 6.3%
2
At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
ne e
artificial transmutations
units
There is nothing that we can do to change the likelihood
o a certain radioactive decay happening, but under certain
conditions we can make nuclear reactions happen. This can
be done by bombarding a nucleus with a nucleon, an alpha
particle or another small nucleus. Such reactions are called
artifcial transmutations. In general, the target nucleus
frst captures the incoming object and then an emission
takes place. The frst ever artifcial transmutation was carried
out by Rutherord in 1 91 9. Nitrogen was bombarded by
alpha particles and the presence o oxygen was detected
spectroscopically.
Using Einsteins equation, 1 kg o mass is equivalent to
9 1 0 1 6 J o energy. This is a huge amount o energy. At
the atomic scale other units o energy tend to be more
useul. The electronvolt (see page 53) , or more usually, the
megaelectronvolt are oten used.
4
2
He 2 + +
14
7
17
8
N
O + 11 p
The mass numbers (4 + 1 4 = 1 7 + 1 ) and the atomic numbers
(2 + 7 = 8 + 1 ) on both sides o the equation must balance.
1 eV = 1 .6 1 0
-19
1 MeV = 1 .6 1 0
J
-13
J
1 u o mass converts into 931 .5 MeV
Since mass and energy are equivalent it is sometimes useul to
work in units that avoid having to do repeated multiplications
by the (speed o light) 2 . A new possible unit or mass is thus
MeV c- 2 . It works like this:
I 1 MeV c- 2 worth o mass is converted you get 1 MeV worth
o energy.
unifiEd mass units
The individual masses involved in nuclear reactions are
tiny. In order to compare atomic masses physicists oten use
unifed mass units, u. These are defned in terms o the most
common isotope o carbon, carbon-1 2. There are 1 2 nucleons
in the carbon-1 2 atom (6 protons and 6 neutons) and one
unifed mass unit is defned as exactly one twelth the mass
o a carbon-1 2 atom. Essentially, the mass o a proton and the
mass o a neutron are both 1 u as shown in the table below.
1 mass o a (carbon-1 2) atom = 1 .66 1 0 - 2 7 kg
1 u = ___
12
mass* o 1 proton = 1 .007 276 u
mass* o 1 neutron = 1 .008 665 u
WorkEd ExamplEs
Question:
How much energy would be released i 1 4 g o carbon-1 4
decayed as shown in the equation below?
14
6
C
14
7
N+
0
-1
_
+
Answer:
Inormation given
atomic mass o carbon-1 4 = 1 4.003242 u;
atomic mass o nitrogen-1 4 = 1 4.003074 u;
mass o electron = 0.000549 u
mass o let-hand side = nuclear mass o
mass* o 1 electron = 0.000 549 u
= 1 3.999948 u
nuclear mass o
mass dEfEct and binding EnErgy
14
7
N = 1 4.003074 - 7(0.000549) u
= 1 3.999231 u
The table above shows the masses o neutrons and protons.
It should be obvious that i we add together the masses o
6 protons, 6 neutrons and 6 electrons we will get a number
bigger than 1 2 u, the mass o a carbon-1 2 atom. What has
gone wrong? The answer becomes clear when we investigate
what keeps the nucleus bound together.
The dierence between the mass o a nucleus and the masses
o its component nucleons is called the mass deect. I one
imagined assembling a nucleus, the protons and neutrons
would initially need to be brought together. Doing this takes
work because the protons repel one another. Creating the bonds
between the protons and neutrons releases a greater amount
o energy than the work done in bringing them together. This
energy released must come rom somewhere. The answer lies in
Einsteins amous massenergy equivalence relationship.
E = mc2
mass in kg
C
= 1 4.003242 - 6(0.000549) u
* = Technically these are all rest masses see option A
energy in joules
14
6
mass o right-hand side = 1 3.999231 + 0.000549 u
= 1 3.999780 u
mass dierence = LHS - RHS
= 0.0001 68 u
energy released per decay = 0.0001 68 931 .5 MeV
= 0.1 56492 MeV
1 4g o C-1 4 is 1 mol
Total number o decays = NA = 6.022 1 0 2 3
Total energy release = 6.022 1 0 2 3 0.1 56492 MeV
= 9.424 1 0 2 2 MeV
= 9.424 1 0 2 2 1 .6 1 0 - 1 3 J
= 1 .51 1 0 1 0 J
1 5 GJ
speed o light in m s - 1
In Einsteins equation, mass is another orm o energy and it
is possible to convert mass directly into energy and vice versa.
The binding energy is the amount o energy that is released
when a nucleus is assembled rom its component nucleons.
It comes rom a decrease in mass. The binding energy would
also be the energy that needs to be added in order to separate
a nucleus into its individual nucleons. The mass deect is thus
a measure o the binding energy.
NB Many examination calculations avoid the need to consider
the masses o the electrons by providing you with the nuclear
mass as opposed to the atomic mass.
Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
75
f a
fission
Fission is the name given to the nuclear reaction whereby large
nuclei are induced to break up into smaller nuclei and release
energy in the process. It is the reaction that is used in nuclear
reactors and atomic bombs. A typical single reaction might
involve bombarding a uranium nucleus with a neutron. This
can cause the uranium nucleus to break up into two smaller
nuclei. A typical reaction might be:
1
0
n+
235
92
U
1 41
56
Ba +
92
36
Since the one original neutron causing the reaction has resulted
in the production o three neutrons, there is the possibility o a
chain reaction occurring. It is technically quite difcult to get
the neutrons to lose enough energy to go on and initiate urther
reactions, but it is achievable.
Kr + 3 10 n + energy
Ba-141
n
U-235
Kr-92
A fssion reaction
A chain reaction
fusion
binding EnErgy pEr nuclEon
Fusion is the name given to the
nuclear reaction whereby small nuclei
are induced to join together into
larger nuclei and release energy in the
process. It is the reaction that uels
all stars including the Sun. A typical
reaction that is taking place in the Sun
is the usion o two dierent isotopes
o hydrogen to produce helium.
Whenever a nuclear reaction (fssion or usion) releases energy, the products o the
reaction are in a lower energy state than the reactants. Mass loss must be the source
o this energy. In order to compare the energy states o dierent nuclei, physicists
calculate the binding energy per nucleon. This is the total binding energy or the
nucleus divided by the total number o nucleons. One o the nuclei with the largest
binding energy per nucleon is iron-56, 52 66 Fe.
H + 31 H 42 He + 10 n + energy
hydrogen-2
helium-4
fusion
hydrogen-3
neutron
One o the usion reactions happening
in the Sun
binding energy per nucleon / MeV
2
1
A reaction is energetically easible i the products o the reaction have a greater
binding energy per nucleon when compared with the reactants.
iron-56
10
8
6
4
2
fusion energetically
possible
20
40
60
80
Graph o binding energy per nucleon
76
ssion energetically
possible
At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
100 120 140 160 180 200
nucleon
number
A Head f e
se
introduction
atomic modEl
All matter that surrounds us, living
or otherwise, is made up o dierent
combinations o atoms. There are only a
hundred, or so, dierent types o atoms
present in nature. Atoms o a single
type orm an element. Each o these
elements has a name and a chemical
symbol; e.g. hydrogen, the simplest o all
the elements, has the chemical symbol
H. Oxygen has the chemical symbol O.
The combination o two hydrogen atoms
with one oxygen atom is called a water
molecule - H2 O. The ull list o elements
is shown in a periodic table. Atoms
consist o a combination o three things:
protons, neutrons and electrons.
The basic atomic model, known as the nuclear model, was developed during the
last century and describes a very small central nucleus surrounded by electrons
arranged in dierent energy levels. The nucleus itsel contains protons and neutrons
(collectively called nucleons) . All o the positive charge and almost all the mass o
the atom is in the nucleus. The electrons provide only a tiny bit o the mass but all o
the negative charge. Overall an atom is neutral. The vast majority o the volume is
nothing at all a vacuum. The nuclear model o the atom seems so strange that there
must be good evidence to support it.
Protons
Relative mass
Charge
1
+1
nucleus
Neutrons
Electrons
1
Negligible
Neutral
-1
Electron clouds. The positions of the 6 electrons
are not exactly known but they are most likely to
be found in these orbitals. The dierent orbitals
correspond to dierent energy levels.
10 -10 m
cell
10 -14 m
DNA
protons
neutron
nucleus
Atomic model of carbon
atom
This simple model has limitations. Accelerated charges are known to radiate energy
so orbital electrons should constantly lose energy (the changing direction means the
electrons are accelerating) .
In the basic atomic model, we are made
up of protons, neutrons, and electrons
nothing more.
EvidEncE
One o the most convincing pieces o evidence or the nuclear model o the atom comes rom the RutherordGeigerMarsden
experiment. Positive alpha particles were red at a thin gold lea. The relative size and velocity o the alpha particles meant that
most o them were expected to travel straight through the gold lea. The idea behind this experiment was to see i there was any
detectable structure within the gold atoms. The amazing discovery was that some o the alpha particles were defected through
huge angles. The mathematics o the experiment showed that numbers being defected at any given angle agreed with an inverse
square law o repulsion rom the nucleus. Evidence or electron energy levels comes rom emission and absorption spectra. The
existence o isotopes provides evidence or neutrons.
positive nucleus
vacuum
gold foil target
about 10 -8 m thick
screens
source of
-particles
beam of
-particles
most pass
straight
through
detector
about 1 in 8000
is repelled back
some are deviated
through a large
angle
RutherordGeigerMarsden experiment
1 in 8000
particles
rebound
from the foil.
gold
atom
stream of
-particles
positive
-particle
deected by
nucleus
N B no t to sca le. On ly a m in u te percentage
of -p a rticles a re scattered or rebou nd .
Atomic explanation o RutherordGeigerMarsden
experiment
Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
77
de f e
classiication o particlEs
consErvation laWs
Particle accelerator experiments identiy
many, many new particles. Two original
classes o particles were identifed the
leptons (= light) and the hadrons (=
heavy) . Protons and neutrons are hadrons
whereas electrons are leptons. The hadrons
were subdivided into mesons and baryons.
Protons and neutrons are baryons. Another
class o particles is involved in the mediation
o the interactions between the particles.
These were called gauge bosons or
exchange bosons.
Not all reactions between particles are possible. The study o the reactions
that did take place gave rise to some experimental conservation laws that
applied to particle physics. Some o these laws were simply confrmation o
conservation laws that were already known to physicists charge, momentum
(linear and angular) and mass-energy. On top o these undamental laws there
appeared to be other rules that were never broken e.g. the law o conservation
o baryon number. I all baryons were assigned a baryon number o 1 (and
all antibaryons were assigned a baryon number o -1 ) then the total number
o baryons beore and ater a collision was always the same. A similar law o
conservation o lepton number applies.
Particles are called elementary i they have
no internal structure, that is, they are not
made out o smaller constituents. The classes
o elementary particles are quarks, leptons
and the exchange particles. Another particle,
the Higgs boson, is also an elementry particle.
Combinations o elementary particles are
called composite particles. All hadrons are
composed o combinations o quarks. Inside
all baryons there are three quarks (or three
antiquarks) ; inside all mesons there is one
quark and one antiquark.
Other reactions suggested new and dierent particle properties that were oten,
but not always, conserved in reactions. Strangeness and charm are examples
o two such properties. Strangeness is conserved in all electromagnetic and
strong interactions, but not always in weak interactions.
All particles, whether they are elementary or composite, can be specifed in
terms o their mass and the various quantum numbers that are related to the
conservation laws that have been discovered. The quantum numbers that are
used to identiy particles include:
electric charge, strangeness, charm, lepton number, baryon number and colour
(this property is not the same as an objects actual colour see page 79).
Every particle has its own antiparticle. An antiparticle has the same mass as
its particle but all its quantum numbers (including charge, etc.) are opposite.
There are some particles (e.g. the photon) that are their own antiparticle.
thE standard modEl lEptons
There are six dierent leptons and six dierent antileptons. The
six leptons are considered to be in three dierent generations or
amilies in exactly the same way that there are considered to be
three dierent generations o quarks (see page 79) .
Similar principles are used to assign lepton numbers o +1 or
-1 to the muon and the tau amily members.
Lepton family number is also conserved in all reactions.
For example, whenever a muon is created, an antimuon or
an antimuon neutrino must also be created so that the total
number o leptons in the muon amily is always conserved.
ExchangE particlEs
There are only our undamental interactions that exist: Gravity,
Electromagnetic, Strong and Weak.
All our interactions can be thought o as being mediated by an
exchange o particles. Each interaction has its own exchange
particle or particles. The bigger the mass o the exchange boson,
the smaller the range o the orce concerned.
Generation
1
2
3
0
e
( electronneutrino)
M = 0 or
almost 0
( muonneutrino)
M = 0 or
almost 0
( tau- neutrino)
M = 0 or
almost 0
-1
e
( electron)
M = 0.51 1
MeV c- 2
( muon)
M = 1 05
MeV c- 2
( tau)
M = 1 784 MeV
MeV c- 2
Lepton
The electron and the electron neutrino have a lepton (electron
amily) number o +1 . The antielectron and the antielectron
neutrino have a lepton (electron amily) number o -1 .
E lectric
charge
The greater the mass o the exchange particle, the smaller the time
or which it can exist. The range o the weak interaction is small
as the masses o its exchange particles (W+ , W- and Z0 ) are large.
In particle physics, all real particles can be thought o as being
surrounded by a cloud o virtual particles that appear and
disappear out o the surrounding vacuum. The lietime o these
particles is inversely proportional to their mass. The interaction
between two particles takes place when one or more o the
virtual particles in one cloud is absorbed by the other particle.
Interaction
Relative Range Exchange Particles
strength (m)
particle
experience
Strong
1
Electromagnetic 1 0 - 2
~1 0 - 1 5
infnite photon
The exchange results in repulsion between the two particles
From the point o view o quantum mechanics, the energy
needed to create these virtual particles, E is available so long
as the energy o the particle does not exist or a longer time t
than is proscribed by the uncertainty principle (see page 1 26) .
78
8 dierent
gluons
W ,W ,Z
+
-
Weak
1 0- 1 3
~1 0 - 1 8
Gravity
1 0- 3 9
infnite graviton
Quarks,
gluons
Charged
0
Quarks,
lepton
All
Leptons and bosons are unaected by the strong orce.
At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
Q
standard modEl Quarks
Isolated quarks cannot exist. They can exist only in twos or
threes. Mesons are made rom two quarks (a quark and an
antiquark) whereas baryons are made up o a combination o
three quarks (either all quarks or all antiquarks) .
The standard model o particle physics is the theory that says
that all matter is considered to be composed o combinations o
six types o quark and six types o lepton. This is the currently
accepted theory. Each o these particles is considered to be
undamental, which means they do not have any deeper
structure. Gravity is not explained by the standard model.
All hadrons are made up rom dierent combinations o
undamental particles called quarks. There are six dierent types
o quark and six types o antiquark. This very neatly matches the
six leptons that are also known to exist. Quarks are aected by
the strong orce (see below) , whereas leptons are not. The weak
interaction can change one type o quark into another.
Quarks
E lectric Generation
charge 1
u
2
+ __ e ( up)
3
M = 5 MeV c- 2
2
c
( charm)
M = 1 500 MeV c- 2
Name o particle
Quark structure
Baryons
proton (p)
neutron (n)
lambda
_
antiproton ( p )
u
u
u
__
u
Mesons
- (pi-minus)
+ (pi-plus)
K 0 (Kz e ro )
u
d __
u __d
d s
u
d
d
__
u
d
d
s_
d
__
The orce between quarks is still the strong interaction but
the ull description o this interaction is termed QCD theory
quantum chromodynamics. The quantum dierence between
the quarks is a property called colour. All quarks can be red (r) ,
_
_
green (g) or blue
(b) . Antiquarks can be antired ( r) , antigreen ( g)
__
or antiblue ( b ) . The two up quarks in a proton are not identical
because they have dierent colours.
3
t
( top)
M = 1 74 MeV c- 2
d
s
b
( down)
( strange)
( bottom)
M = 1 0 MeV c- 2 M = 200 MeV c- 2 M = 4700 MeV c- 2
1
All quarks have a baryon number o + __,
3
1
All antiquarks have a baryon number o - __
3
All quarks have a strangeness number o 0 except the
s quark that has a strangeness number o - 1 .
1
- __ e
3
Only white (colour neutral) combinations are possible.
_ _ __
Baryons must contain r, g and b quarks (or r, g, b ) whereas
_
mesons
contain a colour and the anticolour (e.g. r and r or b
__
and b , etc.) The orce between quarks is sometimes called the
colour orce. Eight dierent types o gluon mediate it.
The details o QCD do not need to be recalled.
The c quark is the only quark with a charm number = +1 ,
all other quarks have charm number o 0.
Quantum chromodynamics (Qcd)
increases. Isolated quarks and gluons cannot be observed.
I sufcient energy is supplied to a hadron in order to attempt to
isolate a quark, then more hadrons or mesons will be produced
rather than isolated quarks. This is known as quark confnement.
The interaction between objects with colour is called the
colour interaction and is explained by a theory called quantum
chromodynamics. The orce-carrying particle is called the gluon.
There are eight dierent types o gluon each with zero mass. Each
gluon carries a combination o colour and anticolour and their
emission and absorption by dierent quarks causes the colour orce.
The six colour-changing gluons are: Gr _b , Gr _g , Gb _g , Gb _r , Gg _b , Gg _r .
For example when a blue up quark emits the gluon Gb _r it loses
its blue colour and becomes a red up quark (the gluon contains
antired, so red colour must be let behind) . A red down quark
absorbing this gluon will become a blue down quark.
As the gluons themselves are coloured, there will be a colour
interaction between gluons themselves as well as between quarks.
The overall eect is that they bind quarks together. The orce
between quarks increases as the separation between quarks
u
r
+
b
G rb
g
G gb
g
G0
b
G bg
u
p
d
r
b
g
g
There are two additional colour-neutral gluons: G0 and G 0 ,
making a total o eight gluons.
u
r
b
b
G rb
g
r
G gb
g
G rg
b
d
g
b
r
r
G gr
g
G rb
r
b
strong intEraction
higgs boson
The colour interaction and the strong interaction are essentially
the same thing. Properly, the colour interaction is the
undamental orce that binds quarks together into baryons
and mesons. It is mediated by gluons. The residual strong
interaction is the orce that binds colour-neutral particles (such
as the proton and neutron) together in a nucleus. The overall
eect o the interactions between all the quarks in the nucleons
is a short-range interaction between colour-neutral nucleons.
In addition to the three generations o leptons and quarks
in the standard model there are the our classes o gauge
boson and an additional highly massive boson, the Higgs
boson. This was proposed in 1 964 to explain the process by
which particles can acquire mass. In 201 3 scientists working
with the Large Hadron Collider announced the experimental
detection o a particle that that matched the standard models
predictions or the Higgs boson.
The particles mediating the strong interaction can be
considered to involve the exchange o composite particles
( mesons: + , - or ) whereas the undamental colour
interaction is always seen as the exchange o gluons.
Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
79
fe
rulEs for draWing fEynman diagrams
Some simple rules help in the construction o correct diagrams:
Feynman diagrams can be used to represent possible particle
interactions. The diagrams are used to calculate the overall
probability o an interaction taking place. In quantum
mechanics, in order to nd out the overall probability o an
interaction, it is necessary to add together all the possible ways
in which an interaction can take place. Used properly they are a
mathematical tool or calculations but, at this level, they can be
seen as a simple pictorial model o possible interactions.
Each junction in the diagram (vertex) has an arrow going
in and one going out. These will represent a leptonlepton
transition or a quarkquark transition.
Quarks or leptons are solid straight lines.
Exchange particles are either wavy or broken (photons,
W or Z) or curly (gluons) .
Time fows rom let to right. Arrows rom let to right
represent particles travelling orward in time. Arrows rom
right to let represent antiparticles travelling orward in time.
The labels or the dierent particles are shown at the end o
the line.
The junctions will be linked by a line representing the
exchange particle involved.
In the Feynman diagrams below the x-axis represents time
going rom let to right and the y-axis represents space (some
books reverse these two axes) . To view them in the alternative
way, turn the page anti-clockwise by 90.
ExamplEs
e-
An electron emits a photon.
e-
e-
An electron absorbs a photon.
e-
A positron emits a photon.
e+
e+
e+
A positron absorbs a photon.
e+
e-
A photon produces an electron and a
positron (an electron- positron pair) .
e+
u
d
before
W-
after
e-
Beta decay. A down quark
changes into an up quark
with the emission o a Wparticle. This decays into an
e
electron and an antineutrino.
The top vertex involves
quarks,
the bottom vertex
einvolves leptons.
An electron and positron annihilate
to produce two photons.
An electron and a positron meet and
annihilate
(disappear) , producing a
photon.
ee+
d
W+
+
+
u
Pion decay. The quark and antiquark annihilate to produce
a W+ particle. This decays into an antimuon and a muon
neutrino.
u
u
g
n
p
Simple diagrams can also be
drawn with exchanges between
e - hadrons.
W-
p
An up quark (in a proton) emits
a gluon which in turn transorms
+
into a down/antidown quark pair.
This reaction could take place as a
result o a protonproton collision:
p + p p + n + + .
A mediates the strong nuclear
orce between a proton and a
neutron in a nucleus.
p
Beta decay (hadron version)
e
usEs of fEynman diagrams
Once a possible interaction has been identied with a Feynman
diagram, it is possible to use it to calculate the probabilities
or certain undamental processes to take place. Each line and
vertex corresponds to a mathematical term. By adding together
all the terms, the probability o the interaction can be calculated
using the diagram.
More complicated diagrams with the same overall outcome
need to be considered in order to calculate the overall
80
d
d
e+
n
n
probability o a chosen outcome. The more diagrams that are
included in the calculation, the more accurate the answer.
In a Feynman diagram, lines entering or leaving the diagram
represent real particles and must obey mass, energy and
momentum relationships. Lines in intermediate stages in the
diagram represent virtual particles and do not have to obey
energy conservation providing they exist or a short enough
time or the uncertainty relationship to apply. Such virtual
particles cannot be detected.
At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
ib Questons atomc, nuclear and partcle physcs
1.
A sample o radioactive material contains the element Ra 226.
The hal-lie o Ra 226 can be dened as the time it takes or
(ii) Explain why it is necessary to give the deuterons
a certain minimum kinetic energy beore they
can react with the magnesium nuclei.
A. the mass o the sample to all to hal its original value.
B. hal the number o atoms o Ra 226 in the sample to decay.
C. hal the number o atoms in the sample to decay.
D. the volume o the sample to all to hal its original value.
2.
Oxygen-1 5 decays to nitrogen-1 5 with a hal-lie o
approximately 2 minutes. A pure sample o oxygen-1 5, with a
mass o 1 00 g, is placed in an airtight container. Ater 4 minutes,
the masses o oxygen and nitrogen in the container will be
Mass of oxygen
3.
4.
Mass of nitrogen
A. 0 g
1 00 g
B. 25 g
25 g
C. 50 g
50 g
D. 25 g
75 g
B. , ,
C. ,
D. , ,
a) Carbon-1 4 decays by beta-minus emission to nitrogen-1 4.
Write the equation or this decay.
[2]
(i) Explain why the beta activity rom the bowl
diminishes with time, even though the probability o
decay o any individual carbon-1 4 nucleus
is constant.
[3]
[3]
1 0. This question is about a nuclear ssion reactor or providing
electrical power.
In the Rutherord scattering experiment, a stream o
particles is red at a thin gold oil. Most o the particles
In a nuclear reactor, power is to be generated by the ssion
o uranium-235. The absorption o a neutron by 2 3 5 U results
in the splitting o the nucleus into two smaller nuclei plus a
number o neutrons and the release o energy. The splitting
can occur in many ways; or example
B. rebound.
C. are scattered uniormly.
D. go through the oil.
n+
A piece o radioactive material now has about 1 /1 6 o its
previous activity. I the hal-lie is 4 hours the dierence in
time between measurements is approximately
235
92
U
90
38
Sr +
1 43
54
Xe + neutrons + energy
a) The nuclear fssion reaction
(i) How many neutrons are produced in this
reaction?
A. 8 hours.
[1 ]
(ii) Explain why the release o several neutrons in
each reaction is crucial or the operation o a ssion
reactor.
[2]
B. 1 6 hours.
C. 32 hours.
D. 60 hours.
6.
The carbon in trees is mostly carbon-1 2, which is
stable, but there is also a small proportion o carbon-1 4,
which is radioactive. When a tree is cut down, the carbon-1 4
present in the wood at that time decays with a hal-lie o
5,800 years.
(ii) Calculate the approximate age o the
wooden bowl.
A. are scattered randomly.
5.
[2]
Radioactive carbon dating
b) For an old wooden bowl rom an archaeological site, the
average count-rate o beta particles detected per kg o
carbon is 1 3 counts per minute. The corresponding count
rate rom newly cut wood is 52 counts per minute.
A radioactive nuclide Z X undergoes a sequence o radioactive
decays to orm a new nuclide Z + 2 Y. The sequence o emitted
radiations could be
A. ,
9.
(iii) The sum o the rest masses o the uranium plus
neutron beore the reaction is 0.22 u greater than
the sum o the rest masses o the ssion products.
What becomes o this missing mass?
[1 ]
a) Use the standard model to describe, in terms o
undamental particles, the internal structure o:
(i) A proton
(iv) Show that the energy released in the above
ssion reaction is about 200 MeV.
(ii) An electron
(iii) Baryons
[2]
b) A nuclear fssion power station
(iv) Mesons
(i)
b) Draw Feynman diagram or + decay.
7.
A proton undergoes a strong interaction with a - particle
(quark content:
ud) to produce a neutron and another
particle. Use conservation laws to deduce the structure o the
particle produced in this reaction.
8.
a) Two properties o the isotope o uranium,
(i) it decays radioactively (to
234
90
238
92
U are:
Th)
(ii) it reacts chemically (e.g. with fuorine to orm UF 6 ) .
What eatures o the structure o uranium atoms are
responsible or these two widely dierent properties?
Suppose a nuclear ssion power station generates
electrical power at 550 MW. Estimate the minimum
number o ssion reactions occurring each second in
the reactor, stating any assumption you have made
about eciency.
[4]
1 1 . Which o the ollowing is a correct list o particles upon which
the strong nuclear orce may act?
A. protons and neutrons
B. protons and electrons
C. neutrons and electrons
D. protons, neutrons and electrons
[2]
2
1
b) A beam o deuterons (deuterium nuclei, H) are
accelerated through a potential dierence and are then
incident on a magnesium target ( 21 62 Mg) . A nuclear reaction
occurs resulting in the production o a sodium nucleus and
an alpha particle.
(i) Write a balanced nuclear equation or
this reaction.
[2]
i B Q u e s t i o n s Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s
81
8 EN ERGY PRODUCTI ON
Energy nd poer genertion snkey digrm
ENERGY CONvERsIONs
ElECTRICal POwER PRODUCTION
The production o electrical power around the world is
achieved using a variety o dierent systems, oten starting
with the release o thermal energy rom a uel. In principle,
thermal energy can be completely converted to work in a
single process, but the continuous conversion o this energy
into work implies the use o machines that are continuously
repeating their actions in a xed cycle. Any cyclical process
must involve the transer o some energy rom the system
to the surroundings that is no longer available to perorm
useul work. This unavailable energy is known as degraded
energy, in accordance with the principle o the second law o
thermodynamics (see page 1 62) .
In all electrical power stations the process is essentially the
same. A uel is used to release thermal energy. This thermal
energy is used to boil water to make steam. The steam is used
to turn turbines and the motion o the turbines is used to
generate electrical energy. Transormers alter the potential
dierence (see page 1 1 4) .
useful
electrical output
energy in from fuel
Energy conversions are represented using Sankey diagrams.
An arrow (drawn rom let to right) represents the energy
changes taking place. The width o the arrow represents the
power or energy involved at a given stage. Created or degraded
energy is shown with an arrow up or down.
heating and
sound in
transformers
friction and heating
losses
cooling tower
losses (condenser)
Note that Sankey diagrams are to scale. The width o the
useul electrical output in the diagram on the right is 2.0 mm
compared with 1 2.0 mm or the width o the total energy rom
the uel. This represents an overall eciency o 1 6.7%.
Sankey diagram representing the energy fow in a typical
power station
steam
fuel (coal)
turbine
to transformer
generator
boiler
water
condenser
Electrical energy generation
POwER
Power is dened as the rate at which energy is converted. The units o power are J S - 1 or W.
energy
Power = _
time
82
EN ERG Y PRO D U CTI O N
Priry energy ource
RENEwablE / NON-RENEwablE ENERGY sOURCEs
The law o conservation o energy states that energy is neither
created nor destroyed, it just changes orm. As ar as human
societies are concerned, i we wish to use devices that require
the input o energy, we need to identiy sources o energy.
Renewable sources o energy are those that cannot be used up,
whereas non-renewable sources o energy can be used up and
eventually run out.
Renewable sources
Non-renewable sources
hydroelectric
coal
photovoltaic cells
oil
active solar heaters
natural gas
wind
nuclear
It is possible or a uel to be managed in a renewable or
a non-renewable way. For example, i trees are cut
down as a source o wood to burn then this is clearly
non-renewable. It is, however, possible to replant trees at the
same rate as they are cut down. I this is properly managed,
it could be a renewable source o energy.
O course these possible sources must have got their energy
rom somewhere in the rst place. Most o the energy used by
humans can be traced back to energy radiated rom the Sun,
but not quite all o it. Possible sources are:
the Suns radiated energy
gravitational energy o the Sun and the Moon
biouels
nuclear energy stored within atoms
Sometimes the sources are hard to classiy so care needs to be
taken when deciding whether a source is renewable or not.
One point that sometimes worries students is that the Sun
will eventually run out as a source o energy or the Earth, so
no source is perectly renewable! This is true, but all o these
sources are considered rom the point o view o lie on Earth.
When the Sun runs out, then so will lie on Earth. Other things
to keep in mind include:
Nuclear sources (both ssion and usion) consume a material
as their source so they must be non-renewable.
sPECIfIC ENERGY aND ENERGY
DENsITY
On the other hand, the supply available can make the source
eectively renewable (usion) .
the Earths internal heat energy.
Although you might think that there are other sources o
energy, the above list is complete. Many everyday sources o
energy (such as coal or oil) can be shown to have derived their
energy rom the Suns radiated energy. On the industrial scale,
electrical energy needs to be generated rom another source.
When you plug anything electrical into the mains electricity you
have to pay the electricity-generating company or the energy
you use. In order to provide you with this energy, the company
must be using one (or more) o the original list o sources.
COmPaRIsON Of ENERGY sOURCEs
Fuel
Renewable?
Two quantities are useul to consider
when making comparisons between
dierent energy sources the specifc
energy and the energy density.
Specic energy provides a useul
comparison between uels and is dened
as the energy liberated per unit mass
o uel consumed. Specic energy is
measured in J kg- 1
specic energy
energy released rom uel
= ___
mass o uel consumed
Fuel choice can be particularly
infuenced by specic energy when
the uel needs to be transported: the
greater the mass o uel that needs to
be transported, the greater the cost.
Energy density is dened as the energy
liberated per unit volume o uel
consumed. The unit is J m- 3
energy density
CO 2
emission
Specifc
energy(MJ kg - 1 )
(values vary
depending on
type)
Energy
density
(MJ m - 3 )
Coal
No
Yes
2233
23,000
Oil
No
Yes
42
36,500
Gas
No
Yes
54
37
Nuclear
(uranium)
No
No
8.3 1 0 7
1 .5 1 0 1 2
Waste
No
Yes
10
variable
n/a
Solar
Yes
No
n/a
Wind
Yes
No
n/a
n/a
Hydro water
stored in dams
Yes
No
n/a
n/a
Tidal
Yes
No
n/a
n/a
Pumped storage
n/a
No
n/a
n/a
Wave
Yes
No
n/a
n/a
Geothermal
Yes
No
n/a
n/a
Biouels e.g.
ethanol
Some types
Yes
30
21 ,000
energy release rom uel
= ___
volume o uel consumed
EN ERGY PRO D U CTI O N
83
foi ue poer production
ORIGIN Of fOssIl fUEl
Coal, oil and natural gas are known as fossil fuels. These
uels have been produced over a timescale that involves tens
or hundreds o millions o years rom accumulations o dead
matter. This matter has been converted into ossil uels by
exposure to the very high temperatures and pressure that exist
beneath the Earths surace.
Coal is ormed rom the dead plant matter that used to grow
in swamps. Layer upon layer o decaying matter decomposed.
As it was buried by more plant matter and other substances,
the material became more compressed. Over the geological
timescale this turned into coal.
Oil is ormed in a similar manner rom the remains o
microscopic marine lie. The compression took place under the
sea. Natural gas, as well as occurring in underground pockets,
can be obtained as a by-product during the production o oil. It is
also possible to manuacture gas rom coal.
ENERGY TRaNsfORmaTIONs
Fossil uel power stations release energy in uel by burning it. The thermal energy is then used to convert water into steam that
once again can be used to turn turbines. Since all ossil uels were originally living matter, the original source o this energy was the
Sun. For example, millions o years ago energy radiated rom the Sun was converted (by photosynthesis) into living plant matter.
Some o this matter has eventually been converted into coal.
solar energy
photosynthesis
chemical
energy
in plants
chemical
energy in
fossil fuels
compression
Energy storage in ossil uels
ExamPlE
EffICIENCY Of fOssIl fUEl POwER sTaTIONs
Use the data on this page and the previous page to calculate
the typical rate (in tonnes per hour) at which coal must be
supplied to a 500 MW coal fred power station.
The efciency o dierent power stations depends on the
design. At the time o publishing, the ollowing fgures apply.
Answer
Electrical power supply
= 500 MW = 5 1 0 J s
Power released rom uel
= 5 1 0 8 / efciency
= 5 1 0 8 / 0.35
= 1 .43 1 0 9 J s - 1
8
Rate o consumption o coal = 1 .43 1 0 / 3.3 1 0 kg s
9
Fossil uel
Typical efciency
Current
maximum
efciency
Coal
35%
42%
-1
7
-1
= 43.3 kg s - 1
= 43.3 60 60 kg hr- 1
Natural gas
45%
52%
Oil
38%
45%
Note that thermodynamic considerations limit the maximum
achievable efciency (see page 1 63) .
= 1 .56 1 0 5 kg hr- 1
1 60 tonnes hr- 1
aDvaNTaGEs aND DIsaDvaNTaGEs
Advantages
Disadvantages
Very high specifc energy and energy density a great deal
o energy is released rom a small mass o ossil uel.
Combustion products can produce pollution, notably
acid rain.
Fossil uels are relatively easy to transport.
Combustion products contain greenhouse gases.
Still cheap when compared to other sources o energy.
Extraction o ossil uels can damage the environment.
Power stations can be built anywhere with good transport
links and water availability.
Non-renewable.
Can be used directly in the home to provide heating.
84
EN ERG Y PRO D U CTI O N
Coal-fred power stations need large amounts o uel.
Nucer power proce
PRINCIPlEs Of ENERGY PRODUCTION
Many nuclear power stations use uranium-235 as the uel.
This uel is not burned the release o energy is achieved using
a fssion reaction. An overview o this process is described on
page 76. In each individual reaction, an incoming neutron
causes a uranium nucleus to split apart. The ragments are
moving ast. In other words the temperature is very high.
Among the ragments are more neutrons. I these neutrons go
on to initiate urther reactions then a chain reaction is created.
The design o a nuclear reactor needs to ensure that, on
average, only one neutron rom each reaction goes on to initiate
a urther reaction. I more reactions took place then the number
o reactions would increase all the time and the chain reaction
would run out o control. I ewer reactions took place, then the
number o reactions would be decreasing and the fssion process
would soon stop.
The chance that a given neutron goes on to cause a fssion
reaction depends on several actors. Two important ones are:
the number o potential nuclei in the way
the speed (or the energy) o the neutrons.
As a general trend, as the size o a block o uel increases so do the
chances o a neutron causing a urther reaction (beore it is lost rom
the surace o the block). As the uel is assembled together a stage
is reached when a chain reaction can occur. This happens when a
so-called critical mass o uel has been assembled. The exact value o
the critical mass depends on the exact nature o the uel being used
and the shape o the assembly.
There are particular neutron energies that make them more
likely to cause nuclear fssion. In general, the neutrons created
by the fssion process are moving too ast to make reactions
likely. Beore they can cause urther reactions the neutrons
have to be slowed down.
mODERaTOR, CONTROl RODs aND hEaT
ExChaNGER
Three important components in the design o all
nuclear reactors are the moderator, the control
rods and the heat exchanger.
core
coolant
steam turbine
kinetic energy
nuclear
energy
electrical
energy
thermal
energy
distributors
to electricity
consumers
thermal
energy
losses
Collisions between the neutrons and the nuclei o
the moderator slow them down and allow urther
reactions to take place.
The control rods are movable rods that readily
absorb neutrons. They can be introduced or
removed rom the reaction chamber in order to
control the chain reaction.
The heat exchanger allows the nuclear reactions
to occur in a place that is sealed o rom the rest
o the environment. The reactions increase the
temperature in the core. This thermal energy is
transerred to heat water and the steam that is
produced turns the turbines.
generators
concrete shields
control rods
(moveable)
pressurizer
moderator
A general design or one type o nuclear reactor
(PWR or pressurized water reactor) is shown here.
It uses water as the moderator and as a coolant.
HOT
WATER
aDvaNTaGEs aND DIsaDvaNTaGEs
steam to
turbines
secondary
coolant
circuit
Advantages
Extremely high specifc energy a great deal
o energy is released rom a very small mass
o uranium.
Reserves o uranium large compared to oil.
heat
exchange
steel pressure vessel
pump
fuel rods
Disadvantages
Process produces radioactive nuclear waste
that is currently just stored.
pump
primary
coolant
circuit
Larger possible risk i anything should go
wrong.
Non-renewable (but should last a long time) .
Pressurized water nuclear reactor (PWR)
EN ERGY PRO D U CTI O N
85
Nucer poer ety nd ri
ENRIChmENT aND REPROCEssING
NUClEaR wEaPONs
Naturally occurring uranium contains less than 1 % o
uranium-235. Enrichment is the process by which this
percentage composition is increased to make nuclear fssion
more likely.
A nuclear power station involves controlled nuclear fssion
whereas an uncontrolled nuclear fssion produces the huge
amount o energy released in nuclear weapons. Weapons have
been designed using both uranium and plutonium as the uel.
Issues associated with nuclear weapons include:
In addition to uranium-235, plutonium-239 is also capable
o sustaining fssion reactions. This nuclide is ormed as a
by-product o a conventional nuclear reactor. A uranium-238
nucleus can capture ast-moving neutrons to orm
uranium-239. This undergoes -decay to neptunium-239
which undergoes urther -decay to plutonium-239:
238
92
239
92
U + 10 n
U
239
93
239
93
Np
239
92
Np +
239
94
U
0
-1
Pu +
_
+
0
-1
_
+
Reprocessing involves treating used uel waste rom nuclear
reactors to recover uranium and plutonium and to deal with
other waste products. A ast breeder reactor is one design that
utilizes plutonium-239.
hEalTh, safETY aND RIsk
Issues associated with the use o nuclear power stations or
generation o electrical energy include:
I the control rods were all removed, the reaction would
rapidly increase its rate o production. Completely
uncontrolled nuclear fssion would cause an explosion and
thermal meltdown o the core. The radioactive material
in the reactor could be distributed around the surrounding
area causing many atalities. Some argue that the terrible
scale o such a disaster means that the use o nuclear
energy is a risk not worth taking. Nuclear power stations
could be targets or terrorist attacks.
The reaction produces radioactive nuclear waste.
While much o this waste is o a low level risk and will
radioactively decay within decades, a signifcant amount
o material is produced which will remain dangerously
radioactive or millions o years. The current solution is to
bury this waste in geologically secure sites.
The uranium uel is mined rom underground and any
mining operation involves signifcant risk. The ore is also
radioactive so extra precautions are necessary to protect
the workers involved in uranium mines.
The transportation o the uranium rom the mine to a
power station and o the waste rom the nuclear power
station to the reprocessing plant needs to be secure
and sae.
By-products o the civilian use o nuclear power can be
used to produce nuclear weapons.
86
EN ERG Y PRO D U CTI O N
Moral issues associated with any weapon o aggression that
is associated with warare. Nuclear weapons have such
destructive capability that since the Second World War the
threat o their deployment has been used as a deterrent to
prevent non-nuclear aggressive acts against the possessors
o nuclear capability.
The unimaginable consequences o a nuclear war have
orced many countries to agree to non-prolieration
treaties, which attempt to limit nuclear power technologies
to a small number o nations.
A by-product o the peaceul use o uranium or energy
production is the creation o plutonium-239 which
could be used or the production o nuclear weapons. Is
it right or the small number o countries that already
have nuclear capability to prevent other countries rom
acquiring that knowledge?
fUsION REaCTORs
Fusion reactors oer the theoretical potential o signifcant
power generation without many o the problems associated
with current nuclear fssion reactors. The uel used, hydrogen,
is in plentiul supply and the reaction (i it could be sustained)
would not produce signifcant amounts o radioactive waste.
The reaction is the same as takes place in the Sun (as outlined
on page 76) and requires creating temperatures high enough
to ionize atomic hydrogen into a plasma state (this is the
ourth state o matter, in which electrons and protons are
not bound in atoms but move independently) . Currently the
principal design challenges are associated with maintaining
and confning the plasma at sufciently high temperature and
density or usion to take place.
sor poer nd ydroeectric poer
sOlaR POwER (TwO TYPEs)
solar radiation
There are two ways o harnessing the radiated energy that
arrives at the Earths surace rom the Sun.
A photovoltaic cell (otherwise known as a solar cell or
photocell) converts a portion o the radiated energy directly
into a potential dierence (voltage) . It uses a piece o
semiconductor to do this. Unortunately, a typical photovoltaic
cell produces a very small voltage and it is not able to provide
much current. They are used to run electrical devices that do
not require a great deal o energy. Using them in series would
generate higher voltages and several in parallel can provide a
higher current.
glass/plastic
cover
warmer
water out
cold water in
solar radiation
slices of
semiconductor
copper pipe (black)
solar energy
metal layer
output voltage
reective insulator
behind pipe
active solar heater
thermal energy
in water
aDvaNTaGEs aND DIsaDvaNTaGEs
Advantages
Very clean production no harmul chemical by-products.
solar energy
photocell
Renewable source o energy.
electrical energy
Source o energy is ree.
Disadvantages
Can only be utilized during the day.
An active solar heater (otherwise known as a solar panel) is
designed to capture as much thermal energy as possible. The
hot water that it typically produces can be used domestically
and would save on the use o electrical energy.
A very large area would be needed or a signicant amount
o energy.
Source o energy is unreliable could be a cloudy day.
hYDROElECTRIC POwER
aDvaNTaGEs aND DIsaDvaNTaGEs
The source o energy in a hydroelectric power station is the
gravitational potential energy o water. I water is allowed to move
downhill, the fowing water can be used to generate electrical energy.
Advantages
The water can gain its gravitational potential energy in several ways.
Renewable source o energy.
As part o the water cycle, water can all as rain. It can be
stored in large reservoirs as high up as is easible.
Source o energy is ree.
Tidal power schemes trap water at high tides and release it
during a low tide.
Water can be pumped rom a low reservoir to a high
reservoir. Although the energy used to do this pumping must
be more than the energy regained when the water fows
back down hill, this pumped storage system provides one
o the ew large-scale methods o storing energy.
Very clean production no harmul chemical
by-products.
Disadvantages
Can only be utilized in particular areas.
Construction o dams will involve land being submerged
under water.
energy lost due to friction throughout
gravitational
PE of water
KE of water
+
KE of turbines
electrical energy
EN ERGY PRO D U CTI O N
87
wind poer nd oter tecnoogie
ENERGY TRaNsfORmaTIONs
maThEmaTICs
density of air
There is a great deal o kinetic energy involved in the
winds that blow around the Earth. The original source
o this energy is, o course, the Sun. Dierent parts o
the atmosphere are heated to dierent temperatures.
The temperature dierences cause pressure dierences,
due to hot air rising or cold air sinking, and thus air
fows as a result.
r
wind speed
blades turn
wind
The area swept out by the blades o the turbine = A = r 2
In one second the volume o air that passes the turbine = v A
heating
Earth
solar energy
KE of wind
energy lost
due to friction electric energy
throughout
KE of turbine
So mass o air that passes the turbine in one second = v A
1 mv2
Kinetic energy m available per second = _
2
1
_
= (vA) v2
2
1
_
= Av3
2
1
_
3
In other words, power available = Av
2
In practice, the kinetic energy o the incoming wind is easy to
calculate, but it cannot all be harnessed as the air must continue to
move in other words the wind turbine cannot be one hundred per
cent ecient. A doubling o the wind speed would mean that the
available power would increase by a actor o eight.
aDvaNTaGEs aND DIsaDvaNTaGEs Disadvantages
Advantages
Very clean production no harmul
chemical by-products.
Renewable source o energy.
Some consider large wind generators
to spoil the countryside.
Source o energy is unreliable could
be a day without wind.
Can be noisy.
A very large area would need be covered
or a signicant amount o energy.
Best positions or wind generators are
oten ar rom centres o population.
Source o energy is ree.
sECONDaRY ENERGY sOURCEs
By ar the most common primary energy sources in use worldwide
are the three main ossil uels: oil, coal and natural gas. With the
inclusion o uranium, at the time o writing this guide, this accounts
or 90% o the worlds energy consumption. Other primary uels
include the renewables: solar, wind, tidal, biomass and geothermal.
With global energy demand expected to rise in the uture, the
hope is that developments with renewable energy can help to
reduce the dependence on ossil uels.
Primary energy sources are not convenient or individual users
and typically a conversion process takes place that results in a
NEw aND DEvElOPING TEChNOlOGIEs
It is impossible to predict technological developments that are
going to take place over the coming years. Current models,
however, predict a continuing dependence on the use o ossil
uels or many years to come. The hope is that we will be able
to decrease this dependency over time. It is important to be
88
EN ERG Y PRO D U CTI O N
secondary energy source that can be widely used in society.
The most common secondary sources are electrical energy (a
very versatile secondary source) or rened uels (e.g. petrol) .
The storage o electrical energy is a challenge, with everyday
devices (e.g. batteries or capacitors) having a very limited
capability when compared with typical everyday demands.
Power companies need to vary the generation o electrical
energy to match consumer demand. Currently pumped storage
hydroelectric systems are the only viable large-scale method
o storing spare electrical energy capacity or uture use. The
eciency o a typical system is approximately 75% meaning
that one quarter o the energy supplied is wasted.
aware o the development o new technologies particularly
those associated with:
renewable energy sources
improving the eciency o our energy conversion process.
Ter energy trner
PROCEssEs Of ThERmal ENERGY TRaNsfER
CONvECTION
There are several processes by which the transer o thermal
energy rom a hot object to a cold object can be achieved. Three
very important processes are called conduction, convection
and radiation. Any given practical situation probably involves
more than one o these processes happening at the same time.
There is a ourth process called evaporation. This involves the
aster moving molecules leaving the surace o a liquid that is
below its boiling point. Evaporation causes cooling.
In convection, thermal energy moves between two points because
o a bulk movement o matter. This can only take place in a fuid
(a liquid or a gas). When part o the fuid is heated it tends to
expand and thus its density is reduced. The colder fuid sinks and
the hotter fuid rises up. Central heating causes a room to warm
up because a convection current is set up as shown below.
Cool air is denser and
sinks downwards.
Hot air is less dense
and is forced upwards.
CONDUCTION
In thermal conduction, thermal energy is transerred along
a substance without any bulk (overall) movement o the
substance. For example, one end o a metal spoon soon eels
hot i the other end is placed in a hot cup o tea.
Conduction is the process by which kinetic energy is passed
rom molecule to molecule.
macroscopic view
Convection in a room
HOT
COLD
thermal energy
thermal energy
RESERVOIR
RESERVOIR
Thermal energy ows along the material as a result
of the temperature dierence across its ends.
microscopic view
HOT
COLD
The faster-moving molecules at the hot end pass
on their kinetic energy to the slower-moving
molecules as a result of intermolecular collisions.
Points to note:
Poor conductors are called thermal insulators.
Metals tend to be very good thermal conductors. This is
because a dierent mechanism (involving the electrons)
allows quick transer o thermal energy.
All gases (and most liquids) tend to be poor conductors.
Examples:
Most clothes keep us warm by trapping layers o air a
poor conductor.
I one walks around a house in bare eet, the foors that are
better conductors (e.g. tiles) will eel colder than the foors
that are good insulators (e.g. carpets) even i they are at
the same temperature. (For the same reason, on a cold day
a piece o metal eels colder than a piece o wood.)
When used or cooking ood, saucepans conduct thermal
energy rom the source o heat to the ood.
ExamPlE
cork a poor conductor
hot liquid
surfaces silvered
so as to reduce
radiation
air gap
(poor conductor)
A thermos fask prevents heat loss
The ow of air around a room Air is warmed
is called a convection current. by the heater.
outer plastic cover
partial vacuum between
glass walls to prevent
convection and
conduction
insulating space
Points to note:
Convection cannot take place in a solid.
Examples:
The pilots o gliders (and many birds) use naturally occurring
convection currents in order to stay above the ground.
Sea breezes (winds) are oten due to convection. During
the day the land is hotter than the sea. This means hot air
will rise rom above the land and there will be a breeze
onto the shore. During the night, the situation is reversed.
Lighting a re in a chimney will mean that a breeze fows
in the room towards the re.
RaDIaTION
Matter is not involved in the transer o thermal energy by
radiation. All objects (that have a temperature above zero
kelvin) radiate electromagnetic waves. I you hold your
hand up to a re to eel the heat, your hands are receiving
the radiation.
For most everyday
objects this radiation is
in the inra-red part o
the electromagnetic
spectrum. For
more details o the
electromagnetic spectrum,
see page 37.
HOT
OBJECT
Electromagnetic
radiation is
given o from
all surfaces.
Points to note:
An object at room temperature absorbs and radiates
energy. I it is at constant temperature (and not changing
state) then the rates are the same.
A surace that is a good radiator is also a good absorber.
Suraces that are light in colour and smooth (shiny) are
poor radiators (and poor absorbers) .
Suraces that are dark and rough are good radiators (and
good absorbers) .
I the temperature o an object is increased then the
requency o the radiation increases. The total rate at
which energy is radiated will also increase.
Radiation can travel through a vacuum (space) .
Examples:
The Sun warms the Earths surace by radiation.
Clothes in summer tend to be white so as not to absorb
the radiation rom the Sun.
EN ERGY PRO D U CTI O N
89
Rdition: wien nd the stenbotnn
blaCk-bODY RaDIaTION: sTEfaN-bOlTzmaNN law
wIENs law
In general, the radiation given out rom a hot object depends
on many things. It is possible to come up with a theoretical
model or the perect emitter o radiation. The perect
emitter will also be a perect absorber o radiation a black
object absorbs all o the light energy alling on it. For this
reason the radiation rom a theoretical perect emitter is
known as black-body radiation.
Wiens displacement law relates the wavelength at which
the intensity o the radiation is a maximum m a x to the
temperature o the black body T. This states that
intensity / arbitrary units
5
2.90 1 0
_
-3
m a x (metres) =
intensity
Black-body radiation does not depend on the nature o the
emitting surace, but it does depend upon its temperature.
At any given temperature there will be a range o dierent
wavelengths (and hence requencies) o radiation that are
emitted. Some wavelengths will be more intense than others.
This variation is shown in the graph below.
m a x T = constant
The value o the constant can be ound by experiment. It is
2.9 1 0 3 m K. It should be noted that in order to use this
constant, the wavelength should be substituted into the
equation in metres and the temperature in kelvin.
T(kelvin)
The peak wavelength rom the Sun
is approximately 500 nm.
m a x = 500 nm
= 5 1 0- 7 m
6000 K
2.9 1 0
_
K
-3
4
3
so T =
visible
region
5 1 0- 7
= 5800 K
5000 K
2
wavelength / nm
1
max = 5 0 0 n m
4000 K
3000 K
0
1200
800
violet
blue
green
yellow
orange
red
400
1600
wavelength / nm
To be absolutely precise, it is not correct to label the y-axis on
the above graph as the intensity, but this is oten done. It is
actually something that could be called the intensity unction.
This is dened so that the area under the graph (between two
wavelengths) gives the intensity emitted in that wavelength
range. The total area under the graph is thus a measure o the
total power radiated. The power radiated by a Black-body (See
page 1 95) is given by:
Surface area in m 2
absolute temperature in kelvins
We can analyse light rom a star and calculate a value or
its surace temperature. This will be much less than the
temperature in the core. Hot stars will give out all requencies
o visible light and so will tend to appear white in colour.
Cooler stars might well only give out the higher wavelengths
(lower requencies) o visible light they will appear red.
Radiation emitted rom planets will peak in the inra-red.
INTENsITY, I
The intensity o radiation is the power per unit area that is
received by the object. The unit is W m- 2 .
Power
I = _.
A
P = AT4
Total power radiated in W
Stefan-Boltzmann constant
Although stars and planets are not perect emitters, their radiation
spectrum is approximately the same as black-body radiation.
EqUIlIbRIUm aND EmIssIvITY
I the temperature o a planet is constant, then the power being
absorbed by the planet must equal the rate at which energy is
being radiated into space. The planet is in thermal equilibrium.
I it absorbs more energy than it radiates, then the temperature
must go up and i the rate o loss o energy is greater than its rate
o absorption then its temperature must go down.
In order to estimate the power absorbed or emitted, the
ollowing concepts are useul.
Emissivity
The Earth and its atmosphere are not a perect black body.
Emissivity, e, is dened as the ratio o power radiated per unit
area by an object to the power radiated per unit area by a black
body at the same temperature. It is a ratio and so has no units.
90
EN ERG Y PRO D U CTI O N
power radiated by object per unit area
e = ____________________________________________
power radiated per unit area by black body at same temperature
thus
p = e A T 4
albEDO
Some o the radiation received by a planet is refected straight
back into space. The raction that is refected back is called the
albedo, .
The Earths albedo varies daily and is dependent on season
(cloud ormations) and latitude. Oceans have a low value but
snow has a high value. The global annual mean albedo is 0.3
(30% ) on Earth.
total scattered power
albedo = __
total incident power
sor power
sOlaR CONsTaNT
The amount o power that arrives rom the Sun is measured by the solar constant. It is properly defned as the amount o solar
energy that alls per second on an area o 1 m2 above the Earths atmosphere that is at right angles to the Suns rays. Its average
value is about 1 400 W m- 2 .
This is not the same as the power that arrives on 1 m2 o the Earths surace. Scattering and absorption in the atmosphere means
that oten less than hal o this arrives at the Earths surace. The amount that arrives depends greatly on the weather conditions.
1% absorbed
in stratosphere
stratosphere
15 km
incoming solar
radiation
100%
NB These gures are only guidelines
because gures vary with cloud cover,
water vapour, etc.
tropopause
troposphere
clouds reect 23%
24% absorbed
in troposphere
clouds absorb 3%
4% reected from
the Earths surface
24% direct
surface of
the Earth
21% diuse
45% reaches Earths surface
Fate o incoming radiation
Dierent parts o the Earths surace (regions at dierent latitudes) will receive dierent amounts o solar radiation. The amount
received will also vary with the seasons since this will aect how spread out the rays have become.
atmosphere is a near-uniform
thickness all around the Earth
re
ce
he
rfa
Tro p
' s su
o sp
M
North Pole
Ea rth
a tm
ed ge o f
23.5
ic of
Eq u a
Tro
pi c
of C
MN > PQ
RS > TU
R
N
S
60
C anc
er
30
to r
a p ri c
incoming solar
radiation
travelling in
near parallel
lines
0 TP U
Q
o rn
South Pole
30
Radiation has to travel through a
greater depth of atmosphere ( RS as
compared with TU) in high latitudes.
When it reaches the surface the radiation
is also spread out over a greater area (MN as
compared with PQ) than in lower latitudes.
60
90
The eect o latitude on incoming solar radiation
Tropic of
Cancer
SUN
Summer
in northern hemisphere
Tropic of
Capricorn
Summer
in southern hemisphere
The Earths orbit and the seasons
EN ERGY PRO D U CTI O N
91
Te greenoue efect
PhYsICal PROCEssEs
A
Short wavelength radiation is received rom
the Sun and causes the surace o the Earth
to warm up. The Earth will emit inra-red
radiation (longer wavelengths than the
radiation coming rom the Sun) because
the Earth is cooler than the Sun. Some o
this inra-red radiation is absorbed by gases
in the atmosphere and re-radiated in all
directions.
This is known as the greenhouse eect
and the gases in the atmosphere that absorb
inra-red radiation are called greenhouse
gases. The net eect is that the upper
atmosphere and the surace o the Earth are
warmed. The name is potentially conusing,
as real greenhouses are warm as a result o
a dierent mechanism.
The temperature o the Earths surace will
be constant i the rate at which it radiates
energy equals the rate at which it absorbs
energy. The greenhouse eect is a natural
process and without it the temperature
o the Earth would be much lower; the
average temperature o the Moon is more
than 30 C colder than the Earth.
SUN
S
P
H
E
R
E
Some solar radiation is
Some of the infra-red
reected by the atmosphere radiation passes through
and Earths surface
the atmosphere and is
lost in space
Outgoing solar radiation:
Solar radiation passes 103 W m - 2
through the clear atmosphere.
Net outgoing infra-red
Incoming solar radiation:
radiation: 240 W m - 2
-2
343 W m
Net incoming
solar radiation:
240 W m - 2
Solar energy absorbed
by atmosphere:
72 W m - 2
M
O
G R E E N H O U S E
G A S E S
Some of the infra-red radiation is
absorbed and re-emitted by the
greenhouse gas molecules. The
direct eect is the warming of the
Earths surface and the troposphere.
Surface gains more heat and
infra-red radiation is emitted again
Solar energy is absorbed by the
Earths surface and warms it...
168 W m - 2
...and is converted into heat causing
the emission of longwave (infra-red)
rediation back to the atmosphere
E
A
R
T
H
Sources: Okanagan University College in Canada; Department of Geography, University of Oxford;
United States Environmental Protection Agency (EPA) , Washington; Climate change 1995, The
science of climate change, contribution of working group 1 to the second assessment report of the
Intergovernmental Panel on Climate Change, UNEP and WMO, Cambridge Press, 1996
GREENhOUsE GasEs
Chlorofuorocarbons (CFCs) . Used as rerigerants,
propellants and cleaning solvents. They also have the eect
o depleting the ozone layer.
The main greenhouse gases are naturally occurring but the
balance in the atmosphere can be altered as a result o their
release due to industry and technology. They are:
Methane, CH4 . This is the principal component o
natural gas and the product o decay, decomposition or
ermentation. Livestock and plants produce signifcant
amounts o methane.
Carbon dioxide, CO 2 . Combustion releases
carbon dioxide into the atmosphere which can
signifcantly increase the greenhouse eect.
Overall, plants (providing they are growing)
remove carbon dioxide rom the atmosphere
during photosynthesis. This is known as
carbon xation.
Nitrous oxide, N2 O. Livestock and industries
(e.g. the production o Nylon) are major sources
o nitrous oxide. Its eect is signifcant as it can
remain in the upper atmosphere or long periods.
In addition the ollowing gases also contribute to the
greenhouse eect:
Absorptivity
Water, H2 O. The small amounts o water vapour in the upper
atmosphere (as opposed to clouds which are condensed water
vapour) have a signifcant eect. The average water vapour
levels in the atmosphere do not appear to alter
greatly as a result o industry, but local levels
can vary.
1
Ozone, O 3 . The ozone layer is an important
region o the atmosphere that absorbs high
energy UV photons which would otherwise be
harmul to living organisms. Ozone also adds to
the greenhouse eect.
T
Each o these gases absorbs inra-red radiation as a result o
resonance (see page 168). The natural requency o oscillation o
the bonds within the molecules o the gas is in the inra-red region.
I the driving requency (rom the radiation emitted rom the Earth)
is equal to the natural requency o the molecule, resonance will
occur. The amplitude o the molecules vibrations increases and the
temperature will increase. The absorption will take place at specifc
requencies depending on the molecular energy levels.
Absorption spectra for major natural
greenhouse gases in the Earths atmosphere
Methane
CH 4
0
1
Nitrous oxide
N 2O
0
1
Oxygen, O 2
& Ozone, O 3
0
1
Carbon
dioxide
CO 2
0
1
Water vapour
H 2O
0
1
0
0.1
Total
atmosphere
0.2 0.30.4 0.60.8 1 1.5 2
3 4 5 6 8 10
20 30
Wavelength (m)
[After J.N. Howard, 1959: Proc. I. R.E . 47, 1459: and R.M. Goody
and G.D. Robinson, 1951: Quart. J. Roy Meteorol. Soc. 77, 153]
92
EN ERG Y PRO D U CTI O N
Go ring
POssIblE CaUsEs Of GlObal waRmING
Records show that the mean temperature o the Earth has been
increasing in recent years.
annual mean
5-year mean
In 201 3, the IPCC (Intergovernmental Panel on Climate
Change) report stated that It is extremely likely that human
infuence has been the dominant cause o the observed
warming since the mid20th century.
0.2
0
-0.2
-0.4
1880
Cyclical changes in the Earths orbit and volcanic activity.
The rst suggestion could be caused by natural eects or could
be caused by human activities (e.g. the increased burning o
ossil uels) . An enhanced greenhouse effect is an increase
in the greenhouse eect caused by human activities.
0.6
0.4
Changes in the intensity o the radiation emitted by the Sun
linked to, or example, increased solar fare activity.
1900
1920
1940
1960
1980
2000
Although it is still being debated, the generally accepted view
is that that the increased combustion o ossil uels has released
extra carbon dioxide into the atmosphere, which has enhanced
the greenhouse eect.
All atmospheric models are highly complicated. Some possible
suggestions or this increase include.
Changes in the composition o greenhouse gases in the
atmosphere.
EvIDENCE fOR GlObal waRmING
One piece o evidence that links global warming to increased levels o greenhouse gases comes rom ice core data. The ice core has
been drilled in the Russian Antarctic base at Vostok. Each years new snow all adds another layer to the ice.
Antarctic Ice Core
Temperature Variation
CO 2 Concentration
380
340
300
260
220
180 40 0,0 00 350,0 00
3 00 ,00 0 250,0 00 200 ,0 00 150 ,0 00
ppmv = parts per million by volume
mEChaNIsms
Predicting the uture eects o global warming involves a great
deal o uncertainty, as the interactions between dierent systems
in the Earth and its atmosphere are extremely complex.
There are many mechanisms that may increase the rate o
global warming.
Global warming reduces ice/snow cover, which in turn
reduces the albedo. This will result in an increase in the
overall rate o heat absorption.
Temperature increase reduces the solubility o CO 2 in the sea
and thus increases atmospheric concentrations.
4
2
0
-2
-4
-6
-8
-10
10 0,00 0 50,00 0
0
Years before present
C
CO 2 / ppmv
Isotopic analysis allows the temperature to be estimated and air bubbles trapped in the ice cores can be used to measure the
atmospheric concentrations o greenhouse gases. The record provides data rom over 400,000 years ago to the present. The
variations o temperature and carbon dioxide are very closely correlated.
Regions with rozen subsoil exist (called tundra) that support
simple vegetation. An increase in temperature may cause a
signicant release o trapped CO 2 .
Not only does deorestation result in the release o urther
CO 2 into the atmosphere, the reduction in number o trees
reduces carbon xation.
The rst our mechanisms are examples o processes whereby a
small initial temperature increase has gone on to cause a urther
increase in temperature. This process is known as positive
feedback. Some people have suggested that the current
temperature increases may be corrected by a process which
involves negative eedback, and temperatures may all in the uture.
Continued global warming will increase both evaporation
and the atmospheres ability to hold water vapour. Water
vapour is a greenhouse gas.
EN ERGY PRO D U CTI O N
93
Ib questions energy production
1.
A wind generator converts wind energy into electric energy.
The source o this wind energy can be traced back to solar
energy arriving at the Earths surace.
a) Outline the energy transormations involved as
solar energy converts into wind energy.
[2]
b) List one advantage and one disadvantage o the
use o wind generators.
[2]
Calculate
a) the energy per second carried away by the water
in the cooling tower;
5.
The expression or the maximum theoretical power, P,
available rom a wind generator is
1 Av3
P= _
2
where
is the density o air and
v is the wind speed.
[2]
d) In practice, under these conditions, the generator
only provides 3 MW o electrical power.
(i)
Calculate the eciency o this generator.
(ii) Give two reasons explaining why the actual
power output is less than the maximum
theoretical power output.
2.
[2]
6.
4.
[2]
[2]
d) the mass o coal burnt each second.
[1 ]
This question is about tidal power systems.
a) Describe the principle o operation o such a system.
[2]
b) Outline one advantage and one disadvantage o
using such a system.
[2]
Height between high tide and
low tide
4m
Trapped water would cover an
area o
1 .0 1 0 6 m2
Density o water
1 .0 1 0 3 kg m3
Number o tides per day
2
[2]
[4]
Solar power and climate models.
a) Distinguish, in terms o the energy changes involved,
between a solar heating panel and a photovoltaic cell.
[2]
b) State an appropriate domestic use or a
This question is about energy sources.
(i)
a) Give one example o a renewable energy source and
one example o a non-renewable energy source and
explain why they are classied as such.
(ii) photovoltaic cell.
[4]
b) A wind arm produces 35,000 MWh o energy in a
year. I there are ten wind turbines on the arm
show that the average power output o one
turbine is about 400 kW.
[3]
c) State two disadvantages o using wind power to generate
electrical power.
[2]
3.
b) the energy per second produced by burning the coal;
c) the overall eciency o the power station;
c) A small tidal power system is proposed. Use the data in
the table below to calculate the total energy available and
hence estimate the useul output power o this system.
A is the area swept out by the blades,
c) Calculate the maximum theoretical power, P, or
a wind generator whose blades are 30 m long
when a 20 m s - 1 wind blows. The density o air
is 1 .3 kg m- 3 .
[2]
solar heating panel.
[1 ]
[1 ]
c) The radiant power o the Sun is 3.90 1 0 2 6 W. The
average radius o the Earths orbit about the Sun is
1 .50 1 0 1 1 m. The albedo o the atmosphere is 0.300
and it may be assumed that no energy is absorbed by
the atmosphere.
Show that the intensity incident on a solar heating panel
at the Earths surace when the Sun is directly overhead
[3]
is 966 W m2 .
Wind power can be used to generate electrical energy.
d) Show, using your answer to (c) , that the average intensity
incident on the Earths surace is 242 Wm2 .
[3]
Construct an energy fow diagram which shows the energy
transormations, starting with solar energy and ending with
electrical energy, generated by windmills. Your diagram
should indicate where energy is degraded.
[7]
e) Assuming that the Earths surace behaves as a black-body
and that no energy is absorbed by the atmosphere, use
your answer to (d) to show that the average temperature
o the Earths surace is predicted to be 256 K.
[2]
This question is about a coal-red power station which is
water cooled.
) Outline, with reerence to the greenhouse eect, why
the average surace temperature o the Earth is higher
than 256 K.
[4]
This question is about energy transormations.
Data:
Electrical power output rom
the station
Temperature at which water
enters cooling tower
= 200 MW
= 288 K
Temperature at which water
leaves cooling tower
= 348 K
Rate o water fow through tower
= 4000 kg s - 1
Energy content o coal
= 2.8 1 0 7 J kg- 1
Specic heat o water
= 4200 J kg1 K- 1
94
I B Q U EsTI O N s EN ERG Y PRO D U CTI O N
9 W av e p h e n o m e n a
hL Sil i i
SImpLe harmonIc motIon (Shm) equatIon
tWo exampLeS of Shm
SHM occurs when the orces on an object are such that
the resultant acceleration, a, is directed towards, and is
proportional to, its displacement, x, rom a fxed point.
a - x or a = - (constant) x
The mathematics o SHM is simplifed i the constant o
proportionality between a and x is identifed as the square
o another constant which is called the angular requency.
Thus the general orm or the equation that defnes SHM is:
a = - 2 x
The solutions or this equation ollow below. The angular
requency has the units o rad s - 1 and is related to the time
period, T, o the oscillation by the ollowing equation.
2
=_
T
Two common situations that approximate to SHM are:
1.
Provided that:
the mass o the spring is negligible compared to the
mass o the load
riction (air riction) is negligible
the spring obeys Hookes law with spring constant, k at
all times (i.e. elastic limit is not exceeded)
the gravitational feld strength g is constant
the fxed end o the spring cannot move.
Then it can be shown that:
k
2 = _
m
__
m
Or T = 2 _
k
The simple pendulum o length l and mass m
IdentIfIcatIon of Shm
In order to analyse a situation to decide i SHM is taking place,
the ollowing procedure should be ollowed.
2.
Provided that:
Identiy all the orces acting on an object when it is
displaced an arbitrary distance x rom its rest position using
a ree-body diagram.
the mass o the string is negligible compared with the
mass o the load
riction (air riction) is negligible
Calculate the resultant orce using Newtons second law. I
this orce is proportional to the displacement and always
points back towards the mean position (i.e. F - x) then
the motion o the object must be SHM.
the maximum angle o swing is small ( 5 or 0.1 rad)
the gravitational feld strength g is constant
the length o the pendulum is constant.
Once SHM has been identifed, the equation o motion
must be in the ollowing orm:
restoring orce per unit displacement, k
a = - ____ x
oscillating mass, m
(
Mass, m, on a vertical spring
Then it can be shown that:
g
2 = _
l
__
l
Or T = 2 _
g
Note that the mass o the pendulum bob, m, is not in this
equation and thus does not aect the time period o the
pendulum, T.
)
( )
k
This identifes the angular requency as 2 = _
m or
k
_
=
m . Identifcation o allows quantitative
equations to be applied.
( )
exampLe
acceLeratIon, veLocIty and dISpLacement
durIng Shm
A 600 g mass is attached to a light spring with spring constant
30 N m1 .
The variation with time o the acceleration, a, velocity, v,
and displacement, x, o an object doing SHM depends on the
angular requency .
(b) Calculate the requency o its oscillation.
The precise ormat o the relationships depends on where the
object is when the clock is started (time t = zero) . The let
hand set o equations correspond to an oscillation when the
object is in the mean position when t = 0. The right hand set
o equations correspond to an oscillation when the object is at
maximum displacement when t = 0.
x = x0 sin t
x = x0 cos t
v = x0 cos t
v = -x0 sin t
a = - 2 x0 sin t
a = - 2 x0 cos t
The frst two equations can be rearranged to produce the
ollowing relationship:
(a) Show that the mass does SHM.
(a) Weight o mass = mg = 6.0 N
Additional displacement x down means that resultant orce
on mass = k x upwards. Since F - x, the mass will oscillate
with SHM.
_____
____
m = 2 _
0.6 = 0.889s
(b) Since SHM, T = 2 _
30
k
( )
( )
1 = _
1
f= _
= 1 .1 Hz
T
0.889
displacement
x0
x0
_______
velocity
v = (x0 - x2 )
x0 is the amplitude o the oscillation measured in m
t is the time taken measured in s
is the angular requency measured in rad s 1
t is an angle that increases with time measured in radians. A
ull oscillation is completed when ( t) = 2 rad.
The angular requency is related to the time period T by the
ollowing equation.
___
2
m
_
T= _
= 2 k
2 x0
T
4
T
2
3T
4
T
time
acceleration
acceleration leads velocity by 90
velocity leads displacement by 90
acceleration and displacement are 1 80 out o phase
displacement lags velocity by 90
velocity lags acceleration by 90
W av e p h e n o m e n a
95
enrgy cangs during simpl armonic motion
hL
During SHM, energy is interchanged between KE and PE.
Providing there are no resistive forces which dissipate this
energy, the total energy must remain constant. The oscillation is
said to be undamped.
The kinetic energy can be calculated from
The total energy is
1 m 2 (x - x2 ) + _
1 m 2 x2 = _
1 m 2 x
E = Ek + Ep = _
0
0
2
2
2
Energy in SHM is proportional to:
the mass m
1 m 2 (x - x2 )
1 mv2 = _
Ek = _
0
2
2
the (amplitude) 2
the (frequency) 2
The potential energy can be calculated from
1 m 2 x2
Ep = _
2
E
total
p
Graph showing the
variation with distance, x
of the energy, during SHM
k
x0
-x0
x
total
k
Graph showing the
variation with time, t
of the energy,
during SHM
p
t
T
4
96
W av e p h e n o m e n a
T
2
3T
4
T
2
hL
difi
BaSIc oBServatIonS
The intensity plot or a single slit is:
Diraction is a wave eect. The objects involved
(slits, apertures, etc.) have a size
that is o the same order o magnitude
as the wavelength o visible light.
nu
bsl
gmil
sw
intensity
There is a central maximum intensity.
Other
maxima occur roughly halfway
10
between the minima.
difi
As the angle increases,
the intensity of the
maxima decreases.
() straight
edge
1.1
0.4
b = slit width
(b) single
long slit
b ~ 3
angle
1st minimum 2
3
=
b
= = b
b
The angle of the rst minimum is given by sin = b .
For small angles, this can be simplied to = b .
() circular
aperture
() single
long slit
b ~ 5
expLanatIon
The shape o the relative intensity versus angle plot can be
derived by applying an idea called Huygens principle. We
can treat the slit as a series o secondary wave sources. In the
orward direction ( = zero) these are all in phase so they add
up to give a maximum intensity. At any other angle, there is a
path dierence between the rays that depends on the angle.
The overall result is the addition o all the sources. The condition
or the frst minimum is that the angle must make all o the
sources across the slit cancel out.
The condition or the frst maximum out rom the centre is
3
. At this
when the path dierence across the whole slit is ___
2
angle the slit can be analysed as being three equivalent sections
each having a path dierence o __
across its length. Together,
2
two o these sections will destructively interere leaving the
resulting amplitude to be __13 o the maximum. Since intensity
(amplitude) 2 , the frst maximum intensity out rom the
centre will be __19 o the central maximum intensity. By a similar
argument, the second maximum intensity out rom the centre will
1
o the central
have __15 o the maximum amplitude and thus be __
25
maximum intensity.
For 1 st m in im u m :
b sin =
sin =
b
b
Sin ce a n gle is sm a ll,
sin
=
b
for 1 st m in im u m
path dierence across slit = b sin
SIngLe-SLIt dIractIon WIth WhIte LIght
When a single slit is illuminated with white light, each
component colour has a specifc wavelength and so the
associated maxima and minima or each wavelength will be
located at a dierent angle. For a given slit width, colours
with longer wavelengths (red, orange, etc.) will diract more
than colours with short wavelengths (blue, violet, etc.) . The
maxima or the resulting diraction pattern will show all the
colours o the rainbow with blue and violet nearer to the
central position and red appearing at greater angles.
incident white light
Red
rst order
Violet
zero order
Violet
rst order
Red
W av e p h e n o m e n a
97
hL
tw-s i ws: ys blsli i
douBLe-SLIt Interference
The double-slit intererence pattern shown on page 47 was derived assuming that each slit was behaving like a perect point source.
This can only take place i the slits are infnitely small. In practice they have a fnite width. The diraction pattern o each slit needs
to be taken into account when working out the overall double slit intererence pattern as shown below.
Decreasing the slit width will mean that the observed pattern becomes more and more idealized. Unortunately, it will also mean
that the total intensity o light will be decreased. The intererence pattern will become harder to observe.
(a) Youngs fringes for innitely narrow slits
relative intensity
(c) Youngs fringes for slits of nite width
intensity
angle
bright
fringes
(b) diraction pattern for a nite-width slit
intensity
angle
s = D still applies but dierent fringes
d
will have dierent intensities with
it being possible for some fringes
to be missing.
angle
InveStIgatIng youngS douBLe-SLIt experImentaLLy
Possible set-ups or the double-slit experiment are shown on
page 47.
Set-up 1
region in which
superposition occurs
separation
monochromatic of slits
light source
S0
In the simplifed version (set-up 2) o the experiment, ringes
can still be bright enough to be viewed several metres away
rom the slits and thus they can be projected onto an opaque
screen (it is dangerous to look into a laser beam) . Their
separation can be then be directly measured with a ruler.
S1
S2
source
twin source
slit
slits (less than 5 mm)
0.1 m
1m
possible
screen
positions
In the original set-up (set-up 1 ) light rom the monochromatic
source is diracted at S 0 so as to ensure that S 1 and S 2 are
receiving coherent light. Diraction takes place providing S 1
and S 2 are narrow enough. The slit separations need to be
approximately 1 mm (or less) thus the slit widths are o the
order o 0.1 mm (or less) . This would provide ringes that
were separated by approximately 0.5mm on a screen (semitransparent or translucent) situated 1 m away. The laboratory
will need to be darkened to allow the ringes to be visible and
they can be viewed using a microscope.
98
W av e p h e n o m e n a
The most accurate measurements or slit separation and ringe
width are achieved using a travelling microscope. This is a
microscope that is mounted on a rame so that it can be moved
perpendicular to the direction in which it is pointing. The
microscope is moved across ten or more ringes and the distance
moved by the microscope can be read o rom the scale. The
precision o this measurement is oten improved by utilizing a
vernier scale.
Set-up 2
laser
double
slit
screen
mlipl-sli ifi
hL
the dIractIon gratIng
(a) 2 slits
The diraction that takes place at an individual slit aects
the overall appearance o the ringes in Youngs doubleslit experiment (see page 98 or more details) . This section
considers the eect on the fnal intererence pattern o adding
urther slits. A series o parallel slits (at a regular separation) is
called a diffraction grating.
Additional slits at the same separation will not aect the
condition or constructive intererence. In other words, the
angle at which the light rom slits adds constructively will
be unaected by the number o slits. The situation is
shown below.
path dierence
between slits = d sin
(b) 4 slits
(c) 50 slits
Grating patterns
uSeS
d
For constructive
interference:
path dierence = n
between slits [, 2, 3]
n = d sin
This ormula also applies to the Youngs double-slit
arrangement. The dierence between the patterns is most
noticeable at the angles where perect constructive intererence
does not take place. I there are only two slits, the maxima will
have a signifcant angular width. Two sources that are just out
o phase interere to give a resultant that is nearly the same
amplitude as two sources that are exactly in phase.
resultant interference
pattern
source A
time
source B
The addition o more slits will mean that each new slit is
just out o phase with its neighbour. The overall intererence
pattern will be totally destructive.
overall interference pattern is totally destructive
One o the main uses o a diraction grating is the accurate
experimental measurement o the dierent wavelengths o
light contained in a given spectrum. I white light is incident
on a diraction grating, the angle at which constructive
intererence takes place depends on wavelength. Dierent
wavelengths can thus be observed at dierent angles. The
accurate measurement o the angle provides the experimenter
with an accurate measurement o the exact wavelength
(and thus requency) o the colour o light that is being
considered. The apparatus that is used to achieve this accurate
measurement is called a spectrometer.
third (and part of the fourth)
order spectrum not shown
R
2nd order
V
R
1st order
V
white
light
white central
maximum
V
diraction
grating
R
V
R
time
The addition o urther slits at the same slit separation has the
ollowing eects:
the principal maxima maintain the same separation
the principal maxima become much sharper
the overall amount o light being let through is increased,
so the pattern increases in intensity.
W av e p h e n o m e n a
99
ti lll fls
hL
phaSe changeS
condItIonS or Intererence patternS
There are many situations when intererence can take place
that also involve the refection o light. When analysing
in detail the conditions or constructive or destructive
intererence, one needs to take any phase changes into
consideration. A phase change is the inversion o the wave
that can take place at a refection interace, but it does not
always happen. It depends on the two media involved.
A parallel-sided lm can produce intererence as a result o the
refections that are taking place at both suraces o the lm.
E
D
A
C
The technical term or the inversion o a wave is that it has
undergone a phase change o .
When light is refected back rom an optically denser
medium there is a phase change o .
thickness d
transmitted wave
(no phase change)
n1 < n2
lm
(refractive index = n)
When light is refected back rom an optically less dense
medium there is no phase change.
air
B
air
= ze ro
wh e n vie we d
a lo n g th e n o rm a l
n1
n2
F
reected wave
(no phase change)
incident wave
n1 < n2
incident wave
reected wave
( phase change)
1. along path AE in air
2. along ABCD in the lm of thickness d
These rays then interfere and we need to calculate
the optical path difference.
The path AE in air is equivalent to CD in the lm
So path difference = (AB + BC) in the lm.
n1
In addition, the phase change at A is equivalent to path
2
difference.
n2
transmitted wave
(no phase change)
exampLe
The equations in the box on the right work out the angles or
which constructive and destructive intererence take place or a
given wavelength. I the source o light is an extended source, the
eye receives rays leaving the lm over a range o values or .
I white light is used then the situation becomes more
complex. Provided the thickness o the lm is small, then one
or two colours may reinorce along a direction in which others
cancel. The appearance o the lm will be bright colours, such
as can be seen when looking at
an oil lm on the surace o water or
rays from an
extended source
W av e p h e n o m e n a
So total path difference = (AB + BC) in lm +
2
= n(AB + BC) +
2
By geometry:
(AB + BC) = FC
= 2 d cos
path difference = 2dn cos +
2
if 2dn cos = m : destructive
or when = 0, 2dn = m: destructive
if 2dn cos = m + : constructive
2
or when = 0, 2dn = m: constructive
m = 0,1 ,2,3,4
appLIcatIonS
soap bubbles.
100
From point A, there are two possible paths:
eye focused
at innity
Applications o parallel thin lms include:
The design o non-refecting radar coatings or military
aircrat. I the thickness o the extra coating is designed so
that radar signals destructively interere when they refect
rom both suraces, then no signal will be refected and an
aircrat could go undetected.
Measurements o thickness o oil slicks caused by spillage.
Measurements o the wavelengths o electromagnetic signals
that give constructive and destructive intererence (at known
angles) allow the thickness o the oil to be calculated.
Design o non-refecting suraces or lenses (blooming), solar
panels and solar cells. A strong refection at any o these suraces
would reduce the amount o energy being useully transmitted.
A thin surace lm can be added so that destructive intererence
takes place or a typical wavelength and thus maximum
transmittance takes place at this wavelength.
hL
rsli
These examples look at the situation o a line source o light
and the diraction that takes place at a slit. A more common
situation would be a point source o light, and the diraction
that takes place at a circular aperture. The situation is exactly
the same, but diraction takes place all the way around the
aperture. As seen on page 97, the diraction pattern o the
point source is thus concentric circles around the central
position. The geometry o the situation results in a slightly
dierent value or the frst minimum o the diraction
pattern.
relative intensity
dIffractIon and reSoLutIon
(a) resolved
I two sources o light are very close in angle to one
another, then they are seen as one single source o light. I
the eye can tell the two sources apart, then the sources are
said to be resolved. The diraction pattern that takes place
at apertures aects the eyes ability to resolve sources. The
examples to the right show how the appearance o two line
sources will depend on the diraction that takes place at
a slit. The resulting appearance is the addition o the two
overlapping diraction patterns. The graph o the resultant
relative intensity o light at dierent angles is also shown.
angle
appearance
two sources clearly separate
resultant intensity
diraction pattern
of source B
(b) just resolved
diraction pattern
of source A
angle
slightly dimmer
appearance
For a slit, the frst minimum was at the angle
=_
b
two maxima visible
(c) not resolved
resultant intensity
For a circular aperture, the frst minimum is at the angle
1 .22
=_
b
diraction pattern
of source B
diraction pattern
of source A
I two sources are just resolved, then the frst minimum o
one diraction pattern is located on top o the maximum o
the other diraction pattern. This is known as the Rayleigh
criterion.
angle
appearance
appears as one source
exampLe
Late one night, a student was observing a
car approaching rom a long distance away.
She noticed that when she frst observed the
headlights o the car, they appeared to be
one point o light. Later, when the car was
closer, she became able to see two separate
points o light. I the wavelength o the light
can be taken as 500 nm and the diameter o
her pupil is approximately 4 mm, calculate
how ar away the car was when she could
frst distinguish two points o light. Take the
distance between the headlights to be 1 .8 m.
Since small
1 .8
=_
x [x is distance to car]
When just resolved
1 .8
x = __
1 .525 1 0 - 4
1 .22
=_
b
= 1 1 .803
1 2 km
1 .22 5 1 0 - 7
= __
0.004
= 1 .525 1 0 - 4
reSoLvance of dIffractIon gratIngS
Example:
As a result o Rayleighs criterion, there is a limit placed
on a gratings ability to resolve dierent wavelengths. The
resolvance, R, o a diraction grating is defned as the ratio
between a wavelength being investigated, , and the smallest
possible resolvable wavelength dierence, .
In the sodium emission spectrum there are two wavelengths
that are close to one another (the Na D-lines) . These are
589.00 nm and 589.59 nm. In order or these to be resolved by
a diraction grating, the resolvance must be
R= _
For any given grating, R is dependent on the diraction order,
m, being observed (frst order: m = 1 ; second order: m = 2, etc.)
and the total number o slits, N, on the grating that are being
illuminated.
= _
589.00 = 1 000
R= _
0.59
In the frst order spectrum, at least 1 000 slits must be
illuminated whereas in the second order spectrum, the
requirement drops to only 500 slits.
= mN
R= _
W av e p h e n o m e n a
101
hL
t dl f
doppLer eect
The Doppler eect is the name given to the change o requency
o a wave as a result o the movement o the source or the
movement o the observer.
When a source o sound is moving:
Sound waves are emitted at a particular requency rom the
source.
The speed o the sound wave in air does not change, but
the motion o the source means that the wave ronts are all
bunched up ahead o the source.
This means that the stationary observer receives sound
waves o reduced wavelength.
Reduced wavelength corresponds to an increased requency
o sound.
The overall eect is that the observer will hear sound at a
higher requency than it was emitted by the source. This applies
when the source is moving towards the observer. A similar
analysis quickly shows that i the source is moving away rom
the observer, sound o a lower requency will be received.
A change o requency can also be detected i the source is
stationary, but the observer is moving.
When a police car or ambulance passes you on the road, you can
hear the pitch o the sound change rom high to low requency. It
is high when it is approaching and low when it is going away.
Radar detectors can be used to measure the speed o a
moving object. They do this by measuring the change in the
requency o the refected wave.
For the Doppler eect to be noticeable with light waves, the
source (or the observer) needs to be moving at high speed.
I a source o light o a particular requency is moving away
rom an observer, the observer will receive light o a lower
requency. Since the red part o the spectrum has lower
requency than all the other colours, this is called a red shift.
I the source o light is moving towards the observer, there
will be a blue shift.
movIng Source
mathematIcS o the doppLer eect
Source moves rom A to D with velocity, u s , speed o waves is v.
Mathematical equations that apply to sound are
stated on this page.
Unortunately the same analysis does not apply to light
the velocities can not be worked out relative to the
medium. It is, however, possible to derive an equation
or light that turns out to be in exactly the same orm as
the equation or sound as long as two conditions are met:
o
moving source
Q
A BC D
stationary
observer
receives sound
at lower frequency
f' = f
v
v us
P
u s t
the relative velocity o source and detector is used
in the equations.
stationary
observer
receives sound
at higher frequency
Received
frequency
at P
Received
frequency
at Q
f' = f
v
v - us
f' = f
v
v + us
this relative velocity is a lot less than the speed o light.
Providing v << c
change in
change wavelength
relative speed o
requency due to relative motion source and observer
f
_
v
_
= _
c
requency o f
wavelength when no speed o light
source
relative motion
movIng oBServer
exampLe
The requency o a cars horn is
measured by a stationary observer as
200 Hz when the car is at rest. What
requency will be heard i the car is
approaching the observer at
30 m s - 1 ? (Speed o sound in air is
330 m s - 1 .)
uo
S
O
f = 2 00 Hz
f = ?
us = 3 0 m s- 1
in a time t, observer has moved u o t
v = 3 3 0 m s- 1
300
f = 200 _
300 - 30
(
= 200 1 .1
f' = f v u o
v
102
If observer is moving away from source:
f' = f v - u o
v
If observer is moving towards source:
f' = f v + u o
v
W av e p h e n o m e n a
= 220 Hz
)
hL
1.
exampls an applications o t dopplr fct
Train going through a station
4.
The sound emitted by a moving trains whistle is o constant
requency, but the sound received by a passenger standing
on the platorm will change. At any instant o time, it is
the resolved component o the trains velocity towards the
passenger that is used to calculate the requency received.
The light shows a characteristic absorption spectrum.
The measured wavelengths are not the same as those
associated with particular elements as measured in the
laboratory.
received frequency
train passing
through station
2.
Receding galaxies red shit
The relative intensities o the dierent wavelengths o
light received rom the stars in distant galaxies can be
analysed.
For the vast majority o stars, all the received requencies
have been shited towards the red end o the visible
spectrum (i.e. to lower requencies) . The light shows a
red shift (see page 202) .
time
The magnitude o the red shit is used to calculate the
recessional velocity and provides evidence or the Big
Bang model o the creation o the Universe.
Radars speed measurement
In many countries the police use radar to measure speed o
vehicles to see i they are breaking the speed limit.
reected
wave
RADAR
transmitter
stationary
5.
The rotation o luminous objects (e.g. the Sun) can be
measured by looking or a dierent Doppler shit on one
side o the object compared with the other.
transmitted
RADAR wave
'
Rotating object
light red shifted
moving car
V
rotating star
Police
light blue shifted
Pulse o microwave radiation o known requency
emitted.
Pulse is refected o moving car and received back at
source.
Dierence in emitted and received requencies is used to
calculate speed o car.
Double Doppler eect taking place:
3.
view above pole
6.
Broadening o spectral lines
Absorption and emission spectra provide evidence or
discrete atomic energy levels (see page 69) .
Moving car receives a requency that is higher than
emitted as it is a moving observer.
Precise measurements show that each individual level
is actually equivalent to a small but dened wavelength
range.
Moving car acts as a moving source when sending
signal back.
The gas molecules are moving so light rom molecules
will be subjected to Doppler shit.
Medical physics blood fow measurements
Doctors can use a pulse o ultrasound to measure the speed o
red blood cells in an analogous way that a pulse o microwaves
is used to measure the speed o a moving car (above).
transmitter receiver
Dierent molecules have a range o speeds so there
will be a general Doppler broadening o the discrete
wavelengths.
A higher temperature means a wider distribution o
kinetic energies and hence more broadening to the
spectral line.
skin
incident
sound
reected
sound
s
red blood cell
W av e p h e n o m e n a
103
IB Questions wave phenomena
HL
1.
When a train travels towards you sounding its whistle, the
pitch o the sound you hear is dierent rom when the train
is at rest. This is because
much less massive than the carbon atom such that any
displacement o the carbon atom may be ignored.
The graph below shows the variation with time t o
the displacement x rom its equilibrium position o a
hydrogen atom in a molecule o methane.
A. the sound waves are travelling aster toward you.
B. the wave ronts o the sound reaching you are spaced
closer together.
x / 10 -10 m
C. the wave ronts o the sound reaching you are spaced
urther apart.
D. the sound requency emitted by the whistle changes with
the speed o the train.
2.
3.
A car is travelling at constant speed towards a stationary
observer whilst its horn is sounded. The requency o the note
emitted by the horn is 660 Hz. The observer, however, hears a
note o requency 720 Hz.
a) With the aid o a diagram, explain why a higher
requency is heard.
[2]
b) I the speed o sound is 330 m s - 1 , calculate the
speed o the car.
[2]
This question is about using a diraction grating to view
the emission spectrum o sodium.
a) Determine the angular separation o the two lines when
viewed in the second order spectrum.
[4]
4.
0.05
[1 ]
0.10
0.15
0.20
0.25
0.30
t / 10 - 13 s
The mass o hydrogen atom is 1 .7 1 0 2 7 kg. Use data
rom the graph above
(i)
Light rom a sodium discharge tube is incident normally
upon a diraction grating having 8.00 1 0 5 lines per metre.
The spectrum contains a double yellow line o wavelengths
589 nm and 590 nm.
b) State why it is more dicult to observe the double
yellow line when viewed in the rst order spectrum.
2.0
1.5
1.0
0.5
0.0
0.5
1.0
1.5
2.0
to determine its amplitude o oscillation.
[1 ]
(ii) to show that the requency o its oscillation is
9.1 1 0 1 3 Hz.
[2]
(iii) to show that the maximum kinetic energy o the
hydrogen atom is 6.2 1 0 1 8 J.
[2]
c) Sketch a graph to show the variation with time t o
the velocity v o the hydrogen atom or one period o
oscillation starting at t = 0. (There is no need to add
values to the velocity axis.)
[3]
d) Assuming that the motion o the hydrogen atom is
simple harmonic, its requency o oscillation f is given
by the expression
This question is about thin lm intererence.
A transparent thin lm is sometimes used to coat spectacle
lenses as shown in the diagram below.
___
1 _
k ,
f= _
2 m p
air, refractive coating, refractive glass lens, refractive
index = 1.00
index = 1.30
index = 1.53
where k is the orce per unit displacement between a
hydrogen atom and the carbon atom and m p is the mass
o a proton.
incoming
light
(i) Show that the value o k is approximately
560 N m1 .
[1 ]
(ii) Estimate, using your answer to (d) (i) , the
maximum acceleration o the hydrogen atom.
[2]
e) Methane is classied as a greenhouse gas.
boundary A boundary B
a) State the phase change which occurs to light that
(i)
is transmitted at boundary A into the lm.
[1 ]
(ii) is refected at boundary B.
[1 ]
(iii) is transmitted at boundary A rom the lm into
the air.
[1 ]
b) Light o wavelength 570 nm in air is incident on the
coating. Determine the smallest thickness o the coating
required so that the refection is minimized or normal
incidence.
[2]
5.
Simple harmonic motion and the greenhouse eect
a) A body is displaced rom equilibrium. State the two
conditions necessary or the body to execute simple
harmonic motion.
[2]
b) In a simple model o a methane molecule, a hydrogen
atom and the carbon atom can be regarded as two
masses attached by a spring. A hydrogen atom is
104
I B Q u e s t I o n s w av e p h e n o m e n a
(i) Describe what is meant by a greenhouse gas.
[2]
(ii) Electromagnetic radiation o requency
9.1 1 0 1 3 Hz is in the inrared region o the
electromagnetic spectrum. Suggest, based on
the inormation given in (b) (ii) , why methane is
classied as a greenhouse gas.
[2]
10 f i e l d s
Hl P ( )
describing fields: e
The concept o eld lines can be used to visually represent:
The direction o the orce is represented by the direction o
the eld lines.
the gravitational eld, g, around a mass (or collection o
masses)
This means that, or both gravitational and electric elds, as a
test object is moved:
the electric eld, E, around a charge (or collection o
charges) .
along a eld line, work will be done (orce and distance
moved are in the same direction)
Magnetic elds can also be represented using eld lines (see
page 61 ) . In all cases the eld is the orce per unit test point
object placed at a particular point in the eld with:
at right angles to a eld line, no work will be done (orce
and distance moved are perpendicular) .
gravitational eld = orce per unit test point mass
( units: N kg - 1 )
electric eld = orce per unit test point positive charge
(units: N C - 1 )
An alternative method o mapping the elds around an object
is to consider the energy needed to move between points in the
eld. This denes the new concepts o electric potential and
gravitational potential (see below) .
Forces are vectors and eld lines represent both the magnitude
and the direction o the orce that would be elt by a test object.
The magnitude o the orce is represented by how close
the eld lines are to one another ( or an example o a
more precise denition, see denition o magnetic fux on
page 1 1 2 ) .
Potential, V (gravitational
or electric)
The feld (gravitational or electric) is
dened as the orce per unit test point
object placed at a particular point in
the eld. In an analogous denition,
the potential (gravitational, Vg , or
electric, Ve ) is dened as the energy per
unit test point object that the object
has as a result o the eld. The ull
mathematical relationships are shown
on page 1 1 0.
energy
Gravitational potential, Vg = _
mass
Units o Vg = J kg- 1
energy
Electric potential, Ve = _
charge
Units o Ve = J C
-1
(or volts)
Potential difference V (electric and gravitational)
Potential is the energy per unit test object. In general, moving a mass between two
points, A and B, in a gravitational eld (or moving a charge between two points,
A and B, in an electric eld) means that work is done. When work is done, the
potential at A and the potential at B will be dierent. Between the points A and B,
there will be a potential dierence, V.
I positive work is done on a test object as it moves between two points then the
potential between the two points must increase.
I work is done by the test object as it moves between the two points then the
potential between the two points must decrease.
Gravitational potential dierence between two points,
work done moving a test mass
Vg = ___
test mass
Units o Vg = J kg1
Electric potential dierence between two points,
work done moving a test charge
Ve = ___
test charge
Units o Ve = J C - 1 or V (volts)
Thus to calculate the work done, W, in moving a charge q or a mass m between two
points in a eld we have:
W = qVe
W = mVg
fi eld s
105
Hl
ep
equiPotential surfaces
examPles of equiPotentials
The best way o representing how the electric potential varies
around a charged object is to identiy the regions where the
potential is the same. These are called equipotential suraces.
In two dimensions they would be represented as lines o
equipotential. A good way o visualizing these lines is to start
with the contour lines on a map.
The diagrams below show equipotential lines or various
situations.
-25
+
+30 V
+20 V
+10 V
contour lines
3-d surface
positively all -ve with units
charged
of J kg-1
sphere
+40 V
shallow
-50
-100
-200
mass
steep
contour lines
close
contour lines
further apart
The contour diagram on the right represents the changing
heights o the landscape on the let. Each line joins up points
that are at the same height. Points that are high up represent a
high value o gravitational potential and points that are low
down represent a low gravitational potential. Contour lines are
lines o equipotential in a gravitational feld.
The same can be done
with an electric feld.
Lines are drawn joining
up points that have the
same electric potential.
The situation right shows
the equipotentials or an
isolated positive point
charge.
point
charge
30 V
20 V
40 V
Equipotentials outside a charge-conducting sphere and a
point mass.
150 V
100 V
50 V
150 V
10 V
-200 J kg- 1
-150 J kg-1
-100 J kg- 1
Equipotentials or two point charges (same charge) and two
point masses.
relationsHiP to field lines
There is a simple relationship between electric feld lines and
lines o equipotential they are always at right angles to one
another. Imagine the contour lines. I we move along a contour
line, we stay at the same height in the gravitational feld. This
does not require work because we are moving at right angles
to the gravitational orce. Whenever we move along an electric
equipotential line, we are moving between points that have the
same electric potential in other words, no work is being done.
Moving at right angles to the electric feld is the only way to
avoid doing work in an electric feld. Thus equipotential lines
must be at right angles to feld lines as shown below.
+
80 V
60 V
40 V
20 V
Equipotentials or two point charges (equal and opposite
charges) .
d
eld line
0V
-80 V
-60 V
-40 V
-20 V
+V = 70 V
60 V
50 V
40 V
30 V zero potential often
20 V taken to be negative
10 V terminal of battery
0V
Equipotential lines between charged parallel plates.
equipotential
lines
always at 90
Field lines and equipotentials are at right angles
106
fi eld s
It should be noted that although the correct defnition o
zero potential is at infnity, most o the time we are not
really interested in the actual value o potential, we are only
interested in the value o the dierence in potential. This means
that in some situations (such as the parallel conducting plates) it
is easier to imagine the zero at a dierent point. This is just like
setting sea level as the zero or gravitational contour lines rather
than correctly using infnity or the zero.
Hl
g p p
gravitational Potential energy
It is easy to work out the dierence in gravitational energy
when a mass moves between two dierent heights near the
Earths surace.
The dierence in energies = m g(h 2 - h 1 )
There are two important points to note:
this derivation has assumed that the gravitational feld
strength g is constant. However, Newtons theory o
universal gravitation states that the feld MUST CHANGE
with distance. This equation can only be used if the
vertical distance we move is not very large.
the equation assumes that the gravitational potential energy
gives zero PE at the surace o the Earth. This works or
everyday situations but it is not undamental.
I the potential energy o the mass, m, was zero at infnity, and
it lost potential energy moving in towards mass M, the potential
energy must be negative at a given point, P.
The value o gravitational potential energy o a mass at any
point in space is defned as the work done in moving it rom
infnity to that point. The mathematics needed to work this out
is not trivial since the orce changes with distance.
It turns out that
GMm
Gravitational potential energy o mass m = - _
r
(due to M)
This is a scalar quantity (measured in joules) and is independent
o the path taken rom infnity.
The true zero o gravitational potential energy is taken as infnity.
potential energy decreases as
gravitational force does work
M
F4
m
F3
zero of potential energy taken
to be at innity
m
F2 m
F1 m
as m moves towards M in the force on m increases
gravitational Potential
escaPe sPeed
We can defne the gravitational potential Vg that measures
the energy per unit test mass.
The escape speed o a rocket is the speed needed to be able to
escape the gravitational attraction o the planet. This means
getting to an infnite distance away.
W
Vg = _
m
(work done)
__
(test mass)
-1
The SI units o gravitational potential are J kg . It is a scalar
quantity.
Using Newtons law o universal gravitation, we can work out
the gravitational potential at a distance r rom any point mass.
Vg
r
We know that gravitational
GM .
potential at the surace o a planet = - _
Rp
(where Rp is the radius o the planet)
This means that or a rocket o
mass m, the dierence between
GMm
its energy at the surace and at infnity = _
Rp
GMm
Thereore the minimum kinetic energy needed = _
Rp
In other words,
Vg = - GMr
1 m (v ) 2 = _
GMm
_
esc
Rp
2
so
This ormula and the graph also works or spherical masses
(planets etc.) . The gravitational potential as a result o lots o
masses is just the addition o the individual potentials. This is
an easy sum since potential is a scalar quantity.
potential due to
m 1 = -40 J kg-1
A
potential due to
m 2 = -30 J kg-1
______
ve s c =
(
2GM
_
Rp
)
This derivation assumes the planet is isolated.
examPle
The escape speed rom an isolated planet like Earth (radius o
Earth RE = 6.37 1 0 6 m) is calculated as ollows:
__________________________
m1
overall potential
= (-40) + (-30) J kg-1
= -70 J kg-1
ve s c =
m2
Once you have the potential at one point and the potential at
another, the dierence between them is the energy you need
to move a unit mass between the two points. It is independent
o the path taken.
2 6.67 1 0 5.98 1 0
)
(___
6.37 1 0 m
-11
24
6
=
(1 .25 1 0 8 ) m s - 1
= 1 .1 2 1 0 4 m s - 1
1 1 km s - 1
The vast majority o rockets sent into space are destined to
orbit the Earth so they leave with a speed that is less than the
escape speed.
fi eld s
107
Hl
o m
gravitational Potential gradient
energy of an orbiting satellite
In the diagram below, a point test mass m moves in a
gravitational feld rom point A to point B.
GMm
We already know that the gravitational energy = - _
r
_____
1 m v2 but v =
The kinetic energy = _
2
The dierence in gravitational potential, Vg
average orce distance moved
= ______________________
= - g r
m
The negative sign is because g is directed towards M, but the orce
doing the work is directed away rom M and thus in the opposite
direction rom g. Since the gravitational orce is attractive, work
has to be done in going rom A to B, so the potential at A <
potential at B.
GM
( _
r )
(Circular motion)
1 m_
GM = _
GMm
1 _
kinetic energy = _
r
r
2
2
So total energy = KE + PE
1 _
GMm - _
GMm = - _
GMm
1 _
=_
r
r
r
2
2
Note that:
1 magnitude
In the orbit the magnitude o the KE = _
2
o the PE.
distance moved
r
The overall energy o the satellite is negative. (A satellite
must have a total energy less than zero otherwise it would
have enough energy to escape the Earths gravitional feld.)
mass M
distance r
A
B
average eld between A and B, g =
(towards M)
Faverage
m
In order to move rom a small radius orbit to a large radius
orbit, the total energy must increase. To be precise, an
increase in orbital radius makes the total energy go rom a
large negative number to a smaller negative number this
is an increase.
This can be summarized in graphical orm.
kinetic energy
in orbit
energy
Vg = -g r
V
g = -_
r
V
___
is called the potential gradient. It has units o J kg- 1 m- 1
r
(which are the same as N kg- 1 or m s - 2 ) .
total energy
The gravitational feld strength is equal to minus the potential
gradient. The equivalent relationship also applies or electric
felds (see page 1 09) .
orbital
radius
gravitational potential
energy in orbit
WeigHtlessness
One way o defning the weight o a person is to
say that it is the value o the orce recorded on a
supporting scale.
I the scales were set up in a lit, they would
record dierent values depending on the
acceleration o the lit.
An extreme version o these situations occurs i
the lit cable breaks and the lit (and passenger)
accelerates down at 1 0 m s 2 .
accelerating down at 10 m s -2
R = zero
a = 10 m s -2
W
resultant force
down = W
no weight will
be recorded
on scales
108
fi eld s
The person would appear to be weightless or the duration o the all. Given
the possible ambiguity o the term weight, it is better to call this situation
the apparent weightlessness o objects in ree-all together.
An astronaut in an orbiting space station would also appear weightless. The
space station and the astronaut are in ree-all together.
In the space station, the gravitational pull on the astronaut provides the
centripetal orce needed
to stay in the orbit. This
resultant orce causes the
centripetal acceleration.
orbital path
The same is true or the
is a circle
gravitational pull on the
satellite and the satellites
velocity
acceleration. There is no
contact orce between the
satellite and the astronaut
so, once again, we have
gravitational attraction on
apparent weightlessness.
astronaut provides centripetal
force needed to stay in orbit
Hl
e p y p
Potential and Potential difference
The concept o electrical potential dierence between two points was introduced
on page 1 05. As the name implies, potential dierence is just the dierence
between the potential at one point and the potential at another. Potential is
simply a measure o the total electrical energy per unit charge at a given point
in space. The defnition is very similar to that o gravitational potential.
+Q
potential increases
zero of potential
as charge is moved
in against repulsion
taken to be at
innity
q
q
F4
q
F3
Potential due to more tHan
one cHarge
I several charges all contribute to the total
potential at a point, it can be calculated by
adding up the individual potentials due to the
individual charges.
potential at point P
= (potential due to Q 1 ) Q2
+ (potential due to Q 2 )
+ (potential due to Q3 )
As q comes in the force on q increases.
potential V
I the total work done in bringing a positive test charge q rom infnity to a
point in an electric feld is W, then the electric potential at that point, V, is
defned to be
W
V= _
q
The units or potential are the same as the units or potential dierence:
J C - 1 or volts.
Q
V= _
4 o r
This equation only applies to
Q
a single point charge.
V = 4
or
Q1
The electric potential at any point outside a
charged conducting sphere is exactly the same
as i all the charge had been concentrated at
its centre.
Potential and field strengtH
potential dierence
= (VB - VA)
potential = VA
distance
+Q
Potential inside a cHarged sPHere
r2
r1
q F1
F2
Q3
r3
P
potential = VB
FB
B q
FA
q A
Charge will distribute itsel uniormly on the outside o a conducting sphere.
Outside the sphere, the feld lines and equipotential suraces are the same
as i all the charge was concentrated at a point at the centre o the sphere.
Inside the sphere, there is no net contribution rom the charges outside
the sphere and the electric feld is zero. The potential gradient is thus also
zero meaning that every point inside the sphere is at the same potential
the potential at the spheres surace.
distance d
Bringing a positive charge from A to B
means work needs to be done against
the electrostatic force.
The graphs below show how feld and potential vary or a sphere o radius a.
(a)
point charge
Q ( positive)
/V m - 1
point
charge
m a x
a
r/m
V/ V
point
charge
0
[
V since V = _
W
=-_
q
x
]
In words,
Vm a x
1 Q
Vm a x =
4 a
A
The work done W = - E q x [the negative sign
is because the direction o the orce needed to
do the work is opposite to the direction o E]
1 _
W
Thereore E = - _
q x
0
(b)
B
direction of force
applied by external
agent on test charge
q at A and B
slope falls
o as r- 2
1
Q
m a x = 4 2
a
direction of
at A and B
slope falls
o as r- 1
a
electric feld = - potential gradient
volt
Units = _
(V m- 1 ) , N C - 1
metre
r/m
fi eld s
109
e f p
Hl
comParison betWeen electric & gravitational ield
Electrostatics
Gravitational
Force can be attractive or repulsive
Force always attractive
Coulombs law or point charges
q1 q2
q1 q2
FE = _
=k_
4 o r2
r2
Newtons law or point masses
m1 m2
F= G _
r2
Electric eld
Gravitational eld
charge producing eld
mass producing eld
electric eld
gravitational eld
q
q1
= k 21
E= F=
q2 4o r2
r
g=
test charge
Gm 1
F
=
m2
r2
test mass
Electric potential due to a point charge
q1
q1
Ve = _
=k_
r
4 r
Gravitational potential due to a point mass, m 1
Electric potential gradient
Gravitational potential gradient
Gm 1
Vg = - _
r
o
Vg
g = - _
r
V
E = - _e
r
Electric potential energy
Gravitational potential energy
q1 q2
q1 q2
Ep = qVe = _
= k_
r
4 o r
GMm
Ep = mVg = - _
r
uniorm ields
Field strength is equal to minus the potential gradient.
A constant eld thus means:
A constant potential gradient i.e. a given increase in distance
will equate to a xed change in potential.
In 3D this means that equipotential suraces will be fat
planes that are equally spaced apart. In 2D equipotential
lines will be equally spaced.
Field lines (perpendicular to equipotential suraces) will be
equally spaced parallel lines.
1.
Constant gravitational eld
The gravitational eld near the surace o a planet is eectively
constant. At the surace o the Earth, the eld lines will be
perpendicular to the Earths surace. Since g = 9.81 m s - 2 , the
potential gradient must also be 9.81 J kg- 1 m- 1 . Equipotential
suraces that are 1 ,000 m apart represent changes o potential
approximately equal to 1 0 kJ kg- 1 .
PE u sing PE = m g h
zero a t su rfa ce
heigh t = 3 km
30 000 J
20 000 J
heigh t = 2 km
2.
Constant electrical eld
The electric eld in between charged parallel plates
(e.g. a capacitor see page 52) is eectively constant
in the middle section.
In the diagram below, the potential dierence across the
plates is V and the separation o the plates is d. Thus the
V
and the constant eld in the
electric potential gradient is __
d
V
centre o the plates, E = __
.
The
units V m- 1 and N C - 1 are
d
equivalent and can both be used or E.
Strictly, the electric eld between two charged parallel
plates cannot remain uniorm throughout the plates and
there will be an edge eect. It is straightorward to show
that at the edge, the eld must have dropped to hal the
value in the centre, but modelling the eld as constant
everywhere between the parallel plates with the edge eects
occurring beyond the limits o the plates can be acceptable.
PE from 1 st prin ciples
zero a t in n ity
- 6 .2575 10 7 J
- 6 .2584 10 7 J
d
+V = 70 V
60 V
50 V
40 V
30 V zero potential often
20 V taken to be negative
10 V terminal of battery
0V
Equipotentials lines between charged parallel plates.
10 0 0 0 J
heigh t = 1 km
- 6 .2593 10 7 J
PE d ieren ce, PE = 10 0 0 0 J
0J
110
su rface o f Ea rth
fi eld s
- 6 .2603 10 7 J
ib questons feds
Hl
1.
Which one o the ollowing graphs best represents the
variation o the kinetic energy, KE, and o the gravitational
potential energy, GPE, o an orbiting satellite at a distance r
rom the centre o the Earth?
A.
KE
0
KE
GPE
energy
energy
D.
KE
0
r
[2]
What orces, i any, act on the astronauts
inside the Space Shuttle whilst in orbit?
[1 ]
Explain why astronauts aboard the Space
Shuttle eel weightless.
[2]
c) Imagine an astronaut 2 m outside the exterior walls o the
Space Shuttle, and 1 0 m rom the centre o mass o the
Space Shuttle. By making appropriate assumptions and
approximations, calculate how long it would take or this
astronaut to be pulled back to the Space Shuttle by the
orce o gravity alone.
[7]
GPE
0
Calculate the energy needed to put the
Space Shuttle into orbit.
r
GPE
C.
[2]
(iii)
(ii)
0
r
Calculate the speed o the Space Shuttle
whilst in orbit.
b) (i)
energy
energy
B.
(ii)
r
KE
5.
a) The diagram below shows a planet o mass M and radius Rp .
GPE
2.
Rp
M
The diagram below illustrates some equipotential lines
between two charged parallel metal plates.
+
+
+
+
The gravitational potential V due to the planet at point X
distance R rom the centre o the planet is given by
80 V
GM
V= - _
R
60 V
where G is the universal gravitational constant.
0.1 m
40 V
Show that the gravitational potential V can be expressed as
20 V
-
-
-
-
-1
B. 8 NC - 1
3.
C. 600 NC
g0 R p 2
V= - _
R
where g0 is the acceleration o ree-all at the
surace o the planet.
The electric feld strength between the plates is
A. 6 NC
X
R
[3]
b) The graph below shows how the gravitational potential V
due to the planet varies with distance R rom the centre
o the planet or values o R greater than Rp , where
Rp = 2.5 1 0 6 m.
-1
D. 800 NC - 1
The diagram shows equipotential lines due to two objects
R/10 6 m
V/10 6 J kg-1
0
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
object 1
object 2
The two objects could be
A. electric charges o the same sign only.
B. masses only.
C. electric charges o opposite sign only.
D. masses or electric charges o any sign.
4.
[1 ]
The Space Shuttle orbits about 300 km above the surace o
the Earth. The shape o the orbit is circular, and the mass o
the Space Shuttle is 6.8 1 0 4 kg. The mass o the Earth is
6.0 1 0 2 4 kg, and radius o the Earth is 6.4 1 0 6 m.
a) (i)
Calculate the change in the Space Shuttles
gravitational potential energy between its
launch and its arrival in orbit.
5
10
15
Use the data rom the graph to
(i)
determine a value o g0 .
[2]
(ii)
show that the minimum energy required to
raise a satellite o mass 3000 kg to a height
3.0 1 0 6 m above the surface o the planet
is about 1 .7 1 0 1 0 J.
[3]
[3]
i B Q u es ti o n s fi eld s
111
11 E l E c t r o m a g n E t i c i n d u c t i o n
HL ie eee e (e)
inducEd Emf
When a conductor moves through a magnetic eld, an em is
induced. The em induced depends on:
The speed o the wire.
The strength o the magnetic eld.
The length o the wire in the magnetic eld.
We can calculate the magnitude o the induced em
by considering an electron at equilibrium in the middle
o the wire. The induced electric orce and the magnetic orce
are balanced.
magnetic eld lines
B
electric eld down wire
due to charge separation
-
charge q
Fe
B
B
Fm
An em is induced in a conductor whenever lines o magnetic
fux are cut. But fux is more than just a way o picturing the
situation; it has a mathematical denition.
I the magnetic eld is perpendicular to the surace, the
magnetic fux passing through the area A is dened in
terms o the magnetic eld strength B as ollows.
= B A, so B = _
A
In a uniorm eld, B = _
A
negative end
production of inducEd Emf by rElativE
motion
An alternative name or magnetic eld strength is fux density.
I the area is not perpendicular, but at an angle to the eld
lines, the equation becomes
potential
dierence V
= B A cos (units: T m2 )
length l
is the angle between B and the normal to the surace.
B
v
positive end
1 Wb = 1 T m2
V q
Fe = E q = _
l
Fm = B q v
( )
Electrical orce due to em,
Magnetic orce due to movement,
V q
Bqv = _
l
V= B l v
( )
So
Flux can also be measured in webers (Wb) , dened as ollows.
These relationships allow us to calculate the induced em in
a moving wave is terms o fux.
in a time t:
l
As no current is fowing, the em = potential dierence
N
S
= B lv
I the wire was part o a complete circuit (outside the magnetic
eld) , the em induced would cause a current to fow.
B
b
a
but l x = A, the area swept out by the conductor in a time
B A
t so = ____
t
c
external agent
exerts force F
l
but B A = so = ___
t
In words, the em induced is equal to the rate o cutting o
fux. I the conductor is kept stationary and the magnets are
moved, the same eect is produced.
velocity v
induced
current I
d
I this situation was repeated with a rectangular coil with N
turns, each section ab would generate an em equal to Bvl.
The total em generated will thus be
= B vlN
Note that in the situation above, a current only fows when
one side o the coil (ab) is moving through the magnetic eld
and the other side (cd) is outside the eld. I the whole coil
was inside the magnetic eld, each side would generate an
em. The two ems would oppose one another and no current
would fow.
112
area swept out A = lx
x then = _
B l x
= B l v since v = _
t
t
boundary
of B
rectangular
emf coil
x
ElEctro m ag n Eti c i n d u cti o n
ExamplE
An aeroplane fies at 200 m s - 1 . Estimate the maximum pd
that can be generated across its wings.
Vertical component
o Earths magnetic eld = 1 0 - 5 T (approximately)
Length across wings = 30 m (estimated)
em = 1 0 - 5 30 200
= 6 1 0- 2 V
= 0.06 V
HL
le' f'
lEnzs law
transformEr-inducEd Emf
Lenzs law states that
An em is also produced in a wire i the
magnetic eld changes with time.
The direction o the induced em is such that i an induced current were
able to fow, it would oppose the change which caused it.
(1)
(2)
I
I the amount o fux passing through one
turn o a coil is , then the total fux linkage
with all N turns o the coil is given by
Flux linkage = N
motion
N
motion
S
S
N
I
Current induced in this direction,
the force would be upwards
(left-hand rule)
original motion would
be opposed.
N
S
I
If current were induced this way,
the induced eld would repel
the magnet - opposing motion.
Lenzs law can be explained in terms o the conservation o energy. The electrical
energy generated within any system must result rom work being done on the
system. When a conductor is moved through a magnetic eld and an induced
current fows, an external orce is needed to keep the conductor moving (the
external orce balances the opposing orce that Lenzs law predicts) . The external
orce does work and this provides the energy or the current to fow.
Put another way, i the direction o an induced current did not oppose the
change that caused it, then it would be acting to support the change. I this was
the case, then a orce would be generated that urther accelerated the moving
object which would generate an even greater em electrical energy would be
generated without work being done.
The universal rule that applies to all
situations involving induced em can now be
stated as
The magnitude o an induced em is
proportional to the rate o change o fux
linkage.
This is known as Faradays law = N _
t
Faradays law and Lenzs law can be
combined together in the ollowing
mathematical statement or the em, ,
generated in a coil o N turns with a rate o
change o fux through the coil o ___
:
t
_
= -N
t
The dependence on the rate o change o
fux and the number o turns is Faradays
law and the negative sign (opposing the
change) is Lenzs law.
application of faradays law to moving and rotating coils
There are many situations involving magnetic elds with
moving or rotating coils. To decide whether or not an em
is generated and, i it is, to calculate its value, the ollowing
procedure can be used:
Example:
A physicist holds her hand so that the magnetic eld o the
Earth (50 T) passes through a ring on her hand.
Choose the period o time, t, over which the motion o the
coil is to be considered.
At the beginning o the period, work out the fux passing
through one turn o the coil, in itia l . Note that the shape o
the coil is not relevant just the cross-sectional area.
= BAcos.
At the end o the period, work out the fux passing through
one turn o the coil n a l using the equation above. Note that
the sense o the magnetic eld is important. I the magnitude
o the eld is the same but it is passing through the coil in
the opposite direction, then
n a l = - in itia l
B = 5 10 -5 T
Determine the change in fux, :
= n a l - in itia l
I there is no overall change o fux then, overall, no em will
be induced. I there is a change in fux then the em induced
in a coil o N turns will be:
= -N _
t
In 0.1 s, she quickly turns her hand through 90 so that the
magnetic eld o the Earth no longer goes through the ring.
Estimate the em generated in the ring.
Answer:
Estimate o cross-sectional area o ring, A 1 cm2 = 1 0 - 4 m2
in itia l = 5 1 0 - 5 1 0 - 4 cos(0) = 5 1 0 - 9 Wb
n a l = 0
= 5 1 0 - 9 Wb
5 1 0- 9 = 5 1 0- 8 V
magnitude o = N _ = _
1 0- 1
t
ElEctro m ag n Eti c i n d u cti o n
113
HL
ae e (1)
coil rotating in a magnEtic fiEld ac
gEnErator
The structure o a typical ac generator is shown below.
eld lines
B
C
A
N
coil (only one
turn shown)
D
S
carbon brush
slip rings (rotate with coil)
rms valuEs
I the output o an ac generator is connected to a resistor an
alternating current will fow. A sinusoidal potential dierence
means a sinusoidal current.
power,
P (= V I )
Po
average
power
P
= o
2
curve is a sin 2 curve (not a sin curve)
time
The graph shows that the average power dissipation is hal the
peak power dissipation or a sinusoidal current.
I0 2 R
_
Average power P = _
=
2
carbon
brush
output
ac generator
The coil o wire rotates in the magnetic eld due to an
external orce. As it rotates the fux linkage o the coil changes
with time and induces an em (Faradays law) causing a
current to fow. The sides AB and CD o the coil experience
a orce opposing the motion (Lenzs law) . The work done
rotating the coil generates electrical energy.
induced emf
A coil rotating at constant speed will produce a sinusoidal
induced em. Increasing the speed o rotation will reduce the
time period o the oscillation and increase the amplitude o the
induced em (as the rate o change o fux linkage is increased).
( )
I0
_
__
2
2
R
Thus the eective current through the resistor is
(mean value o I2 ) and it is called the root mean square
current or rms current, Irm s .
I0
__ (or sinusoidal currents)
Irm s = _
2
When ac values or voltage or current are quoted, it is the root
mean square value that is being used. In Europe this value is
230 V, whereas in the USA it is 1 20 V.
V0
__
Vrm s = _
2
_
1 IV
P = Vrm s Irm s = _
2 0 0
Pm a x = I0 V0
time
coil rotated
at double the
speed
V0
Vrm s
V= _
R=_
=_
I0
Irm s
I
constant speed of
rotation means induced
emf is sinusodial
transformEr opEration
An alternating potential dierence is put into the transormer,
and an alternating potential dierence is given out. The value
o the output potential dierence can be changed (increased
or decreased) by changing the turns ratio. A step-up
transormer increases the voltage, whereas a step-down
transormer decreases the voltage.
The ollowing sequence o calculations provides the correct
method or calculating all the relevant values.
The output voltage is xed by the input voltage and the
turns ratio.
The value o the load that you connect xes the output
current (using V = I R) .
iron core
secondary coil
primary coil
Ip
Is
input ac
voltage p
output ac
voltage s
number of turns Np
number of turns Ns
p
Np
Is
=
=
s
Ns
I
The value o the output power is xed by the values above
(P = V I) .
Transormer structure
The value o the input power is equal to the output power
or an ideal transormer.
The alternating pd across the primary creates an ac within the
coil and hence an alternating magnetic eld in the iron core.
The value o the input current can now be calculated (using
P = V I) .
This alternating magnetic eld links with the secondary
and induces an em. The value o the induced em depends
on the rate o change o fux linkage, which increases with
increased number o turns on the secondary. The input and
output voltages are related by the turns ratio.
So how does the transormer manage to alter the voltages in
this way?
114
ElEctro m ag n Eti c i n d u cti o n
HL
ae ue (2)
transmission of ElEctrical powEr
lossEs in thE transmission of powEr
Transormers play a very important role in the sae and
ecient transmission o electrical power over large distances.
I large amounts o power are being distributed, then the
currents used will be high. (Power = V I)
In addition to power losses associated with the resistance
o the power supply lines, which cause the power lines to
warm up, there are also losses associated with non-ideal
transormers:
The wires cannot have zero resistance. This means they
must dissipate some power
Resistance o the windings (joule heating) o a
transormer result in the transormer warming up.
Power dissipated is P = I2 R. I the current is large then the
(current) 2 will be very large.
Eddy currents are unwanted currents induced in the iron
core. The currents are reduced by laminating the core
into individually electrically insulated thin strips.
Over large distances, the power wasted would be very
signicant.
The solution is to choose to transmit the power at a very
high potential dierence.
Only a small current needs to fow.
Hysteresis losses cause the iron core to warm up as a
result o the continued cycle o changes to its magnetism.
Flux losses are caused by magnetic leakage. A transormer
is only 1 00% ecient i all o the magnetic fux that is
produced by the primary links with the secondary.
A very high potential dierence is much more ecient, but
very dangerous to the user.
diodE bridgEs
The ecient transmission o electrical power is best achieved
using alternating current (ac) and transormers can ensure
the appropriate Vrm s is supplied. Many electrical devices are,
however, designed to operate using direct current (dc) . The
conversion rom ac into dc is called rectifcation which relies
on diodes.
A diode is a two-terminal electrical device that has dierent
electrical characteristics depending on which way around it is
connected. An ideal diode allows current to fow in the orward
direction (negligible resistance with orward bias) but does not
allow current to fow in the reverse direction (innite resistance
with reverse bias) .
Symbol:
B
A
secondary
primary
Current is allowed to fow rom A to B (A is positive and B is
negative) but is prevented rom ollowing rom B to A (A is
negative and B is positive) .
Vd
diode current
Use step-up transormers to increase the voltage or the
transmission stage and then use step-down transormers
or the protection o the end user.
lamination
O
reverse bias
forward bias
diode voltage
typical diode characteristics
allowed current direction
ElEctro m ag n Eti c i n d u cti o n
115
ref
HL
rEctiication
smoothing circuits
1.
Diode-bridge circuits provide a current that fows in one
direction (dc) but still pulsates. In order to achieve a steady pd,
a smoothing device is required. One possibility is a capacitor
(see page 1 1 7 or more details) .
Hal-wave rectication
A single diode will convert ac into a pulsating dc:
+
+
AC
supply
load
output from
rectifying circuit
+
load
smoothed half-wave rectication
voltage
across
load
voltage
across
load
time
smoothed output
time
In hal-wave rectication, electrical energy that is
available in the negative cycle o the ac is not utilized.
2.
+
smoothed full-wave rectication
Full-wave rectication
A diode bridge (using our diodes) can utilize all the
electrical energy that is available during a complete cycle
as shown below.
voltage
across
load
time
A
+
Note that:
The output is still fuctuating slightly; this is known as the
output ripple.
load
AC
supply
unsmoothed output
B
C+
The capacitor is acting as a short-term store o electrical
energy.
The capacitor is constantly charging and discharging.
D
voltage
across
load
time
In order to ensure a slow discharge, the value o the
capacitor C needs to be chosen to ensure that the time
constant (see page 1 1 8) is suciently large.
invEstigating a diodE-bridgE rEctiication
circuit ExpErimEntally
In the positive hal o the cycle, current fows through the
diode bridge rom ACBD.
The display o the varying pd across the load is best achieved
using a cathode ray oscilloscope (CRO) .
In the negative hal o the cycle, current fows through
the diode bridge rom DCBA.
The y-input control, allows the sensitivity o the CRO to
appropriately display a changing pd on the y-axis. The timebase controls allows an appropriate calibration o the x-axis to
match the time period o the oscillations.
Note that:
Current always fows through the load resistor in the
same direction. (CB)
e bse
Diodes on parallel sides point in the same directions.
The ac signal is ed to the points where opposite ends
o two diodes join.
The positive output is taken rom the junction o the
negative side o two diodes.
The negative output is taken rom the junction o the
positive side o two diodes.
During each hal-cycle one set o parallel-side diodes
conducts.
ch 1
che 1 ch 2
che 2
sesvy p
sesvy p
time base set at 2.5 mS cm- 1
1 oscillation = 8 cm on screen = 20 mS
1 = 50 Hz
requency = _
0.02
116
ElEctro m ag n Eti c i n d u cti o n
HL
ce
capacitancE
Capacitors are devices that can store charge. The charge stored q
is proportional to the pd across the capacitor V and the constant
of proportionality is called the capacitance C.
The capacitance of a parallel plate capacitor depends on three
different factors:
The area of each plate, A. Each plate is assumed to have the
same area A and the plates overlap one another completely.
The separation of the plates, d
Symbol:
C
charge in coulombs
q
C= _
V
capacitance in farads
pd in volts
The farad (F) is a very large unit and practical capacitances are
measured in F, nF or pF.
1 F = 1 C V- 1
The material between the plates which is called the
dielectric material. Different materials will have different
values of a constant called its permittivity, . The
permittivity of air is effectively the same as the permittivity
of a vacuum (free space) , 0 = 8.85 1 0 - 1 2 C 2 N- 1 m- 2 .
The permittivity of all substances is greater than 0 .
The relationship is:
A
C=_
d
when a dielectric material is introduced, change separation
across the dielectric is induced. This increases the capacitance.
A measurement of the pd across a capacitance allows the charge
stored to be calculated.
capacitors in sEriEs and parallEl
2.
The effective total capacitance, Cto ta l, of the combination of
capacitors (C1 , C2 , C3 , etc.) in a circuit depends on whether
the capacitors are joined together in series or in parallel. The
capacitor equation can be used on individual capacitors or on
the combination.
q to ta l
q1
q
Cto ta l = _
and C1 = _
, C2 = _2 , etc.
Vto ta l
V1
V2
1.
In parallel
C1
+q 1
-q 1
pd V1
C2
+q 2
-q 2
In series
C1
+q
+
-
C2
-q
+q
V1
+
-
pd V2
C3
-q
+q
V2
+
-
-q
C3
+q 3
-q 3
V3
pd V3
Vtotal
The charge stored in each capacitor is the same, q and the
pds across the individual capacitors add together to give
the total pd
qto ta l = q 1 = q 2 = q 3 = q
Vto ta l = V1 + V2 + V3
pd Vtotal
The pd across each capacitor is the same, V and the charges
stored in each of the individual capacitors add together to
give the total charge stored.
q to ta l
q
q
q
_
= _1 + _2 + _3
Cto ta l
C1
C2
C3
Vto ta l = V1 = V2 = V3 = V
q
q
q
q
_ = _ + _ + _
Cs e rie s
C1
C2
C3
Cto ta l Vto ta l = C1 V1 + C2 V2 + C3 V3
1 = _
1 + _
1 +
_
Cs e rie s
C1
C2
Cp a ra lle l = C1 + C2 +
e.g. if three capacitors 5 F, 1 0 F and 20 F are added in
series, the combined capacitance is:
1 =_
1 + _
1 + _
1 =_
7 F - 1
_
5
Cs e rie s
10
20
20
qto ta l = q 1 + q 2 + q 3
Cp a ra lle l V = C1 V + C2 V + C3 V
e.g. if three capacitors 5 F, 1 0 F and 20 F are added in
parallel, the combined capacitance is:
Cp a ra lle l = 5 + 1 0 + 20 = 35 F
20 = 2.86 F
Cs e rie s = _
7
ElEctro m ag n Eti c i n d u cti o n
117
HL
c e
capacitor (rc) dischargE circuits
I the two ends o a charged capacitor are joined together with a
resistor, a current will fow until the capacitor is discharged.
The product o RC is called the time constant or the circuit
and is given the symbol (the Greek letter tau) .
= RC
The SI unit or will be seconds (NB: care needed with SI
multipliers) .
initial
pd V0
+q 0 -q 0
t
_
q = q 0 e-
Since the current I and the pd V are both proportional to the
charge, the ollowing equations also apply:
_t
when S is closed,
the current I will ow
in direction shown
switch S
I = I0 e-
t
_
V = V0 e-
Where
V
q0
I0 = _
= _0
R
RC
Example
R
A 1 0 F capacitor is discharged through a 20 k resistor.
Calculate (a) the time constant or the circuit and (b) the
raction o charge remaining ater one time constant
During the discharge process:
the value o the discharge current, I, drops rom an initial
maximum I0 down to zero
the value o the stored charge, q, drops rom an initial
maximum q0 down to zero
the value o the pd across the capacitor (which is also the
pd across the resistor) , V, drops rom an initial maximum V0
down to zero.
Applying Kirchos law around the loop gives
q
0 = IR + _
C
dq
Since I is the rate o fow o charge, _,
dt
q
dq
0 = R ___ + _
dt
C
dq
q
_ = - _
RC
dt
This has the rate o fow o charge proportional to the charge
stored. The solution is an exponential decrease o charge stored
given by:
time (s)
q = q0 e
charge
remaining
118
original
charge
t
-_
RC
capacitance (F)
resistance ()
ElEctro m ag n Eti c i n d u cti o n
a) = RC = 1 0 F 20 k = 200 ms
b) Ater one time constant,
q = q0 e- 1 = 0.37q 0
q/C
q = V0 C
q0
e
exponential
decay
O
t/ms
RC
time
charge
0
1 00%
1 RC
37%
2RC
1 4%
3RC
5%
4RC
2%
5RC
<1 %
Ater 5 time constents, the capacitor is eectively discharged
HL
c e
capacitor charging circuits
I the two ends o an uncharged capacitor are joined together with a resistor, a current will fow until the capacitor
is charged.
C
R
when S is closed, the current
I will ow in direction shown
until capacitor is charged
switch S
emf
During the charging process:,
the value o the charging current, I, drops rom an initial maximum I0 down to zero
the value o the stored charge, q, increases rom zero up to a nal maximum value, q 0
the value o the pd across the capacitor, V, increases rom zero up to a nal maximum value,
the value o the pd across the resistor drops rom an initial maximum down to zero.
q/C
q0
(1 - 1e ) q
I/mA
nal charge = 0c
exponential
growth
0
O
RC
initial current = R
I0
exponential
decay
I0
e
O
t/ms
RC
t/ms
The equation or the increase o charge on the capacitor (which does not need to be memorized) is:
_t
q = q 0 ( 1 - e- )
EnErgy storEd in a chargEd capacitor
A charged capacitor can provide a temporary store o electrical
energy when there is a potential dierence V across the
capacitor. The charge, q, that is stored is distributed with +q on
one plate and - q on the other plate as shown below. There is
an electric eld between the plates.
+q
V/V
V0
V0 (nal potential)
V0
2
1
2 V0 ( average potential)
-q
pd V
In the charging process, as more charge is added to the
capacitor, the pd across it also increases proportionally.
The graph (right) shows how the pd across the capacitor
varies with charge stored in the capacitor during the charging
process. The total energy stored, E, is represented by the area
under the graph.
area = 21 q 0 V0
O
q0
Q/C
q2
1 qV = _
1 _
1 CV2
E=_
=_
2
2 C
2
Note that both charging and discharging are exponential
processes. I a circuit is arranged in which a capacitor spends
equal time charging and discharging through the same value
resistor, then in one complete cycle, more charge will be added
to the capacitor during the charging time than it loses during
the discharging time. The result over several cycles will be or
the capacitor to charge up to the same pd as the power supply.
ElEctro m ag n Eti c i n d u cti o n
119
ib Questons electromagnetc nducton
HL
1.
The primary o an ideal transormer has 1 000 turns and the
secondary 1 00 turns. The current in the primary is 2 A and
the input power is 1 2 W.
5.
Which one o the ollowing about the secondary current
and the secondary power output is true?
secondary current
2.
A.
20 A
1 .2 W
0.2 A
12 W
C.
0.2 A
1 20 W
D.
20 A
12 W
1
I
A
1
t
t
2 (no current)
A
B
1
I
1
I
small coil
t
2
C
a) (i) State Faradays law o electromagnetic
induction.
(ii) Use the law to explain why, when the current
in the wire changes, an em is induced in
the coil.
[2]
[1 ]
The diagram shows a simple generator with the coil rotating
between magnetic poles. Electrical contact is maintained
through two brushes, each touching a slip ring.
S
t
D
2
A loop o wire o negligible resistance is rotated in a magnetic
eld. A 4 resistor is connected across its ends.
A cathode ray oscilloscope measures the varying induced
potential dierence across the resistor as shown below.
potential dierence / V
6.
N
~
2
A small coil is placed with its plane parallel to a long straight
current-carrying wire, as shown below.
3.
2
I
This question is about electromagnetic induction.
current-carrying
wire
1
The variation with time o the
current in loop 1 is shown as
line 1 in each o the graphs below. In which graph does line 2
best represent the current in loop 2?
secondary power output
B.
Two loops o wire are next to
each other as shown here.
There is a source o alternating
em connected to loop 1 and
an ammeter in loop 2.
2
1.5
1
0.5
0
0.05
0.5
1
1.5
2
0.15
0.25
0.35 time / s
a) I the coil is rotated at twice the speed, show on the axes
above how potential dierence would vary with time.
[2]
b) What is the rms value o the induced potential
dierence, Vrm s , at the original speed o rotation?
[1 ]
c) Draw a graph showing how the power dissipated in the
resistor varies with time, at the original speed o rotation. [3]
At the instant when the rotating coil is oriented as shown, the
voltage across the brushes
A. is zero.
B. has its maximum value.
C. has the same constant value as in all other orientations.
D. reverses direction.
4.
The rms current rating o an electric heater is 4A. What
direct current would produce the same power dissipation
in the electric heater?
[2]
4__ A
A. _
2
C.
__
4 2 A
120
B. 4A
D. 8A
7.
a) A 3F capacitor is charged to 240 V. Calculate the
charge stored.
[1 ]
b) Estimate the amount o time it would take or the charge
you have calculated in (a) to fow through a 60 W light
bulb connected to the 240 V mains electricity.
[2]
c) The charged capacitor in (a) is discharged through a
60 W 240 V light bulb.
(i)
Explain why the current during its discharge will
not be constant.
[2]
(ii) Estimate the time taken or the capacitor to
discharge through the light bulb.
[2]
(iii) Will the bulb light during discharge?
Explain your answer.
[2]
i B Q u Esti o n s ElEctro m ag n Eti c i n d u cti o n
12 Q u a n t u m a n d n u c l e a r P h yS i c S
HL P f
Photoelectric eect
einStein model
Under certain conditions, when light (ultra-violet) is shone onto a metal surace
(such as zinc) , electrons are emitted rom the surace.
Einstein introduced the idea o
thinking o light as being made up o
particles.
More detailed experiments (see below) showed that:
Below a certain threshold frequency, f0 , no photoelectrons are emitted, no
matter how long one waits.
Above the threshold requency, the maximum kinetic energy o these
electrons depends on the requency o the incident light.
The number o electrons emitted depends on the intensity o the light and does
not depend on the requency.
There is no noticeable delay between the arrival o the light and the emission o
electrons.
These observations cannot be reconciled with the view that light is a wave. A wave o
any requency should eventually bring enough energy to the metal plate.
StoPPing Potential exPeriment
UV
window to
transmit UV
(quartz)
The stopping potential depends on the
requency o UV light in the linear way
shown in the graph below.
stopping potential Vs
vacuum
anode
G microammeter cathode
V
variable power supply
(accelerating pd)
In the apparatus above, photoelectrons
are emitted by the cathode. They are
then accelerated across to the anode by
the potential dierence.
In this situation, the electrons are
decelerated. At a certain value o
potential, the stopping potential, Vs ,
no more photocurrent is observed. The
photoelectrons have been brought to rest
beore arriving at the anode.
photocurrent
The energy in each packet is fxed
by the requency o UV light that is
being used, whereas the number o
packets arriving per second is fxed
by the intensity o the source.
The energy carried by a photon is
given by
E = hf
energy in joules
Max KE o electrons = Vs e
and e = charge on an electron]
v=
I the energy o the photon is too
small, the electron will still gain this
amount o energy but it will soon
share it with other electrons.
2V e
_
m
s
Above the threshold requency,
incoming energy o photons = energy
needed to leave the surace + kinetic
energy.
high-intensity UV
In symbols,
Em a x = hf -
otential
hf = + Em a x or hf = + Vs e
This means that a graph o requency
against stopping potential should be a
straight line o gradient __he .
examPle
What is the maximum velocity o electrons
v=
emitted rom a zinc surace ( = 4.2 eV) when
illuminated by EM radiation o wavelength 200 nm?
=
= 4.2 eV = 4.2 1 .6 1 0 - 1 9 J = 6.72 1 0 - 1 9 J
c = ___
6.63 1 0 - 3 4 3 1 0 8
Energy o photon = h _
2 1 0- 7
Dierent electrons absorb dierent
photons. I the energy o the
photon is large enough, it gives the
electron enough energy to leave the
surace o the metal.
Any extra energy would be
retained by the electron as kinetic
energy.
energy
[since pd = _
charge
1 mv2 = V e
_
s
2
requency o
light in Hz
frequency
threshold frequency, f0
low-intensity UV
of same frequency
Vs
The UV light energy arrives in lots
o little packets o energy the
packets are called photons.
Plancks constant
6.63 1 0 3 4 J s
The stopping potential is a measure o
the maximum kinetic energy o the
electrons.
The potential between cathode and
anode can also be reversed.
His explanation was:
Electrons at the surace need a
certain minimum energy in order
to escape rom the surace. This
minimum energy is called the work
function o the metal and given
the symbol .
_____
2 KE
_
m
______________
2 3.225 1 0
__
9.1 1 0
-19
- 31
= 8.4 1 0 5 m s - 1
= 9.945 1 0 - 1 9 J
KE o electron = (9.945 - 6.72) 1 0 - 1 9 J
= 3.225 1 0 - 1 9 J
Q u a n t u m a n d n u c l e a r p h ys i c s
121
hl
m ws
WaveParticle duality
The photoelectric eect o light waves clearly demonstrates that
light can behave like particles, but its wave nature can also be
demonstrated it refects, reracts, diracts and intereres just
like all waves. So what exactly is it? It seems reasonable to ask
two questions.
o it as a wave and sometimes it helps to think o it as a particle,
but neither model is complete. Light is just light. This dual
nature o light is called waveparticle duality.
2.
If light waves can show particle properties, can particles such as
electrons show wave properties?
The correct answer to this question is yes! At the most
undamental and even philosophical level, light is just light.
Physics tries to understand and explain what it is. We do this by
imagining models o its behaviour. Sometimes it helps to think
Again the correct answer is yes. Most people imagine moving
electrons as little particles having a denite size, shape, position
and speed. This model does not explain why electrons can be
diracted through small gaps. In order to diract they must
have a wave nature. Once again they have a dual nature. See
the experiment below.
de Broglie hyPotheSiS
1.
I matter can have wave properties and waves can have matter
properties, there should be a link between the two models. The
de Broglie hypothesis is that all moving particles have a matter
wave associated with them. This matter wave can be thought o
as a probability unction associated with the moving particle. The
(amplitude) 2 o the wave at any given point is a measure o the
probability o nding the particle at that point. The wavelength o
this matter wave is given by the de Broglie equation:
In these situations, the rest energy o the particles can be
negligible compared with their energy o motion.
1.
Is light a wave or is it a particle?
At very high energies: pc = E
For example, the rest energy o an electron (0.51 1 MeV) is
negligible i it has been accelerated through an eective potential
dierence o 420 MV to have kinetic energy o 420 MeV. In
these circumstances the total energy o an electron is eectively
420 MeV. The de Broglie wavelength o 420 MeV electrons is:
6.6 1 0 - 3 4 3.0 1 0 8 = 2.9 1 0 - 1 5 m
= ___
420 1 0 6 1 .6 1 0 - 1 9
hc _
hc
= _
pc = E or photons
2.
is the wavelength in m
At low energies
In these situations the relationship can be restated in terms o
the momentum p o the particle measured in kg m s - 1 (in nonrelativistic mechanics, P = mass velocity) :
h is Planks constant = 6.63 1 0 - 3 4 J s
c is the speed o light = 3.0 1 0 8 m s - 1
p is the momentum o the particle
The higher the energy, the lower the de Broglie wavelength. This
equation was introduced on page 69 as the method o calculating
a photons wavelength rom its energy, E. In order or the wave
nature o particles to be observable in experiments, the particles
oten have very high velocities. In these situations the proper
calculations are relativistic but simplications are possible.
h
=_
p
For example, electrons accelerated through 1 kV would gain a KE
o 1 .6 10 - 1 6 J. Since KE and non-relativistic momentum are
p
related by EK = ___
, this gives p = 1 .7 1 0 - 2 3 kg m s - 1
2m
2
6.6 1 0 - 3 4 = 3.9 1 0 - 1 1 m
= __
1 .7 1 0 - 2 3
electron diffraction exPeriment
daviSSon and germer exPeriment (1927)
In order to show diraction, an electron wave must travel
through a gap o the same order as its wavelength. The atomic
spacing in crystal atoms provides such gaps. I a beam o
electrons impinges upon powdered carbon then the electrons
will be diracted according to the wavelength.
The diagram below shows the principle behind the Davisson
and Germer electron diraction experiment.
accelerating p.d.
~1000 V
movable
electron
detector
screen
+
electron
beam
~
heater
powdered
graphite
scattered electrons
vacuum
The circles correspond to the angles where constructive
intererence takes place. They are circles because the powdered
carbon provides every possible orientation o gap. A higher
accelerating potential or the electrons would result in a higher
momentum or each electron. According to the de Broglie
relationship, the wavelength o the electrons would thus decrease.
This would mean that the size o the gaps is now proportionally
bigger than the wavelength so there would be less diraction. The
circles would move in to smaller angles. The predicted angles o
constructive intererence are accurately veried experimentally.
122
lament
Q u a n t u m a n d n u c l e a r p h ys i c s
Target
A beam o electrons strikes a target nickel crystal. The electrons
are scattered rom the surace. The intensity o these scattered
electrons depends on the speed o the electrons (as determined
by their accelerating potential dierence) and the angle.
A maximum scattered intensity was recorded at an angle
that quantitatively agrees with the constructive intererence
condition rom adjacent atoms on the surace.
HL
a sp ss
introduction
Dierent atomic models have attempted to explain these energy
levels. The frst quantum model o matter was the Bohr model
or hydrogen: modern models describe the electrons by using
waveunctions (see page 1 25) .
As we have already seen, atomic spectra (emission and
absorption) provide evidence or the quantization o the
electron energy levels. See page 69 or the laboratory set-up.
hydrogen SPectrum
The emission spectrum o atomic hydrogen consists o particular
wavelengths. In 1 885 a Swiss schoolteacher called Johann Jakob
Balmer ound that the visible wavelengths ftted a mathematical
ormula.
These wavelengths, known as the Balmer series, were later
shown to be just one o several similar series o possible
wavelengths that all had similar ormulae. These can be
expressed in one overall ormula called the Rydberg formula.
(
)
1 =R _
1
1 _
_
H
n2 m2
the wavelength
m a whole number larger than 2 i.e. 3, 4, 5 etc
For the Lyman series o lines (in the ultra-violet range) n = 1 .
For the Balmer series n = 2. The other series are the Paschen
(n = 3) , Brackett (n = 4) , and the Pfund (n = 5) series. In
each case the constant RH , called the Rydberg constant, has
the one unique value, 1 .097 1 0 7 m1 .
examPle
The diagram below represents some o the electron energy levels in the hydrogen atom. Calculate the wavelength o the photon
emitted when an electron alls rom n = 3 to n = 2.
energy level / eV
0
-0.9
-1.5
-3.4
n=3
allowed energy
levels
n=2
ground state: n = 1
-13.6
Energy dierence in levels = 3.4 1 .5 = 1 .9 eV = 1 .9 1 .6 1 0 - 1 9 J = 3.04 1 0 - 1 9 J
E = __
3.04 1 0 - 1 9 = 4.59 1 0 1 4 Hz
Frequency o photon f = _
h
6.63 1 0 - 3 4
3.00 1 0 8 = 6.54 1 0 - 7 m = 654 nm
Wavelength o photon = _c = __
f
4.59 1 0 1 4
This is in the visible part o the electromagnetic spectrum and one wavelength in the Balmer series.
Pair Production and Pair annihilation
Matter and radiation interactions are not restricted to the
absorption or emission o radiation by matter (such as takes
place in absorption or emission spectra, above) . As introduced on
page 73, or every normal matter particle that exists, there will
be a corresponding antimatter particle which has the same mass
but every other property is opposite. For example:
The antiparticle o an electron, e - (or - ) is a positron,
e + (or + )
The antiparticle or a proton, p + is the antiproton, p _
The antiparticle or a neutrino, is an antineutrino,
When a particle and its corresponding antiparticle meet
they annihilate one another and the mass is converted into
radiation. As seen on page 78, these annihilations must obey
certain conservations and in particular the conservation o
energy, momentum and charge.
When an electron e - and a positron e + annihilate typically
they create two photons. Each photon has a momentum and
i combined momentum o the electronpositron pair was
initially zero, then the two photons will be travelling in opposite
directions. The reverse process is also possible photons o
sufcient energy can convert into a pair o particles (one matter
and one antimatter) . Much o the energy goes into the rest
masses o the particles with any excess going into the kinetic
energy o the particles that have been created. Typically or pair
production to take place, the photon needs to interact with a
nucleus. The nucleus is not changed in the interaction but is
involved in the overall conservation o momentum and energy
that must take place. Without its ability to absorb some o the
momentum, the interaction could not occur.
Q u a n t u m a n d n u c l e a r p h ys i c s
123
HL
B f
Bohr model
Niels Bohr took the standard planetary model o the hydrogen
atom and flled in the mathematical details. Unlike planetary
orbits, there are only a limited number o allowed orbits
or the electron. Bohr suggested that these orbits had fxed
multiples o angular momentum. The orbits were quantized
in terms o angular momentum. The energy levels predicted
by this quantization were in exact agreement with the
discrete wavelengths o the hydrogen spectrum. Although this
agreement with experiment is impressive, the model has some
problems associated with it.
The second postulate can be used (with the ull equation) to
predict the wavelength o radiation emitted when an electron
makes a transition between stable orbits.
hf = E2 - E1
m e e4 _
1 -_
1
= _
8 0 2 h 2 n 1 2
n2 2
(
)
but f = _c
m e e4 _
1 =_
1 -_
1
_
8 0 2 ch 3 n 1 2
n2 2
(
)
Bohr postulated that:
It should be noted that:
An electron does not radiate energy when in a stable orbit.
The only stable orbits possible or the electron are ones
where the angular momentum o the orbit is an integral
h
multiple o __
where h is a fxed number (6.6 1 0 3 4 J s)
2
called Plancks constant. Mathematically
this equation is o the same orm as the Rydberg ormula.
nh
me vr = _
2
[angular momentum is equal to m e vr]
the values predicted by this equation are in very good
agreement with experimental measurement.
the Rydberg constant can be calculated rom other (known)
constants. Again the agreement with experimental data is good.
The limitations to this model are:
When electrons move between stable orbits they radiate (or
absorb) energy.
i the same approach is used to predict the emission spectra
o other elements, it ails to predict the correct values or
atoms or ions with more than one electron.
the frst postulate (about angular momentum) has no
theoretical justifcation.
Fe le ctro s ta tic = centripetal orce
m e v2
e
_
=_
r
4 0 r2
nh
_
but v =
[rom 1 st postulate]
2m e r
2
0 n2 h2
r=_
(by substitution)
m e e2
theory predicts that electrons should, in act, not be stable
in circular orbits around a nucleus. Any accelerated electron
should radiate energy. An electron in a circular orbit is
accelerating so it should radiate energy and thus spiral in to
the nucleus.
Total energy o electron = KE + PE
it is unable to account or relative intensity o the dierent lines.
1 m v2 = _
e
1 _
where KE = _
2 e
2 (4 0 r)
e2 [electrostatic PE]
and PE = - _
4 0 r
m e e4
e2 = - _
1 _
so total energy En = - _
2 4 0 r
8 20 n 2 h 2
This fnal equation shows that:
2
it is unable to account or the fne structure o the spectral lines.
the electron is bound to (= trapped by) the proton because
overall it has negative energy.
1
the energy o an orbit is proportional to __
. In electronvolts
n
2
1 3.6
En = _
n2
nuclear radii and nuclear denSitieS
Not surprisingly, more massive nuclei have larger radii.
D etailed analysis o the data implies that the nuclei have a
spherical distribution o positive charge with an essentially
constant density. The results are consistent with a model
in which the protons and neutrons can be imagined to be
hard spheres that are bonded tightly together in a sphere
o constant density. A nucleus that is twice the size o a
smaller nucleus will have roughly 8 ( = 2 3 ) times the mass.
The nuclear radius R o element with atomic mass number A
can be modelled by the relationship:
1
_
R = R0 A 3
Where R0 is a constant roughly equal to 1 0 - 1 5 m (or 1 m) .
R0 = 1 .2 1 0 - 1 5 m = 1 .2 m.
e.g. The radius o a uranium-238 nucleus is predicted to be
1
_
R = 1 .2 1 0 - 1 5 (238) 3 m = 7.4 m
124
Q u a n t u m a n d n u c l e a r p h ys i c s
The volume o a nucleus, V, o radius, R is given by:
4 R3 = _
4 AR 3
V= _
0
3
3
Where the mass number A is equal to the number o nucleons
A= _
3A
The number o nucleons per unit volume = _
V 4AR0 3
3
=_
4R0 3
The mass o a nucleon is m ( 1 .7 1 0 - 2 7 kg) , so the nuclear
density is:
3m = __
3 1 .7 1 0 - 2 7 = 2 1 0 1 7 kg m- 3
=_
4R0 3
4(1 .2 1 0 - 1 5 ) 3
This is a vast density (a teaspoon o matter o this density has
a mass 1 0 1 2 kg) . The only macroscopic objects with the same
density as nuclei are neutron stars (see page 200) .
t S f a
Schrdinger model
Erwin Schrdinger (1 8871 961 ) built on the concept o matter
waves and proposed an alternative model o the hydrogen atom
using wave mechanics. The Copenhagen interpretation is a way
to give a physical meaning to the mathematics o wave mechanics.
The description o particles (matter and/or radiation) in
quantum mechanics is in terms o a wavefunction . This
waveunction has no physical meaning but the square o the
waveunction does.
is a complex number.
At any instant o time, the waveunction has dierent values
at dierent points in space.
The mathematics o how this waveunction develops with
time and interacts with other waveunctions is like the
mathematics o a travelling wave.
The probability o fnding the particle (electron or photon,
etc.) at any point in space within the atom is given by the
square o the amplitude o the waveunction at that point.
The square o the absolute value o ,|| 2 , is a real number
corresponding to the probability density o fnding the
particle in a given place.
When an observation is made the waveunction is said to
collapse, and the complete physical particle (electron or
photon, etc.) will be observed to be at one location.
The standing waves on a string have a fxed wavelength
but or energy reasons the same is not true or the electron
waveunctions. As an electron moves away rom the nucleus it
must lose kinetic energy because they have opposite charges.
Lower kinetic energy means that it would be travelling with
a lower momentum and the de Broglie relationship predicts a
longer wavelength. This means that the possible waveunctions
that ft the boundary conditions have particular shapes.
The electron in this orbital can be visualized as a cloud o
varying electron density. It is more likely to be in some places
than other places, but its actual position in space is undefned.
1s
Electron cloud or the 1 s orbital in hydrogen
There are other fxed total energies or the electron that result
in dierent possible orbitals. In general as the energy o the
electron is increased it is more likely to be ound at a urther
distance away rom the nucleus.
probability density
HL
= -2.18 10 -18 J
1s
10
5
10
5
10
5
10
5
10
The resulting orbital or the electron can be described in
terms o the probability o fnding the electron at a certain
distance away. The probability o fnding the electron at a
given distance away is shown in the graph below.
probability / %
10
5
di an
fr m nu l u / 1
- 10 m
15 20 25
15 20 25
Probability density unctions or some orbitals in the hydrogen
atom. The scale on the vertical axis is dierent rom graph to graph.
The waveunction is central to quantum mechanics and, in
principle, should be applied to all particles.
Example:
A particle is described as the ollowing waveunction:
B
A
x
probability of
detection outside
L
L is zero
mid-point
L
2
A
0
15 20 25
radius ( 10 - 10 m)
probability of detection
outside L is zero
In Schrdingers model there are dierent waveunctions depending
on the total energy o the electron. Only a ew particular energies
result in waveunctions that ft the boundary conditions electrons
can only have these particular energies within an atom. An electron
in the ground state has a total energy o 13.6 eV, but its position
at any given time is undefned in this model. The waveunction or
an electron o this energy can be used to calculate the probability o
fnding it at a given distance away rom the nucleus.
15 20 25
= -0.242 10 -18 J
3s
3d
5
The waveunction exists in all three dimensions, which makes
it hard to visualize. Oten the electron orbital is pictured as a
cloud. The exact position o the electron is not known but we
know where it is more likely to be.
15 20 25
3p
p(r) = | | 2 V
probability o detecting the electrons in a small volume o
space, V
15 20 25
= -0.545 10 - 18 J
2s
2p
The waveunction provides a way o working out the probability
o fnding an electron at that particular radius. || 2 at any given
point is a measure o the probability o fnding the electron at
that distance away rom the nucleus in any direction.
100
80
60
40
20
0
5
B
x
The particle will not be detected at the mid point and the
probability o detection at A = probability o detection at B.
Q u a n t u m a n d n u c l e a r p h ys i c s
125
hl
t hsb pp d ss
ds
heiSenBerg uncertainty PrinciPle
The Heisenberg uncertainty principle identifes a
undamental limit to the possible accuracy o any physical
measurement. This limit arises because o the nature o
quantum mechanics and not as a result o the ability (or
otherwise) o any given experimenter. He showed that it was
impossible to measure exactly the position and the momentum
o a particle simultaneously. The more precisely the position is
determined, the less precisely the momentum is known in this
instant, and vice versa. They are linked variables and are called
conj ugate quantities.
There is a mathematical relationship linking these uncertainties.
h
xp _
4
x
The uncertainty in the measurement o position
p
The uncertainty in the measurement o momentum
The implications o this lack o precision are proound. Beore
quantum theory was introduced, the physical world was best
described by deterministic theories e.g. Newtons laws. A
deterministic theory allows us (in principle) to make absolute
predictions about the uture.
Quantum mechanics is not deterministic. It cannot ever predict
exactly the results o a single experiment. It only gives us the
probabilities o the various possible outcomes. The uncertainty
principle takes this even urther. Since we cannot know the
precise position and momentum o a particle at any given time,
its uture can never be determined precisely. The best we can do
is to work out a range o possibilities or its uture.
It has been suggested that science would allow us to
calculate the uture so long as we know the present exactly.
As Heisenberg himsel said, it is not the conclusion o this
suggestion that is wrong but the premise.
Measurements o energy and time are also linked variables.
h
Et _
4
E
The uncertainty in the measurement o energy
t
The uncertainty in the measurement o time
eStimateS from the uncertainty PrinciPle
Example calculation: The position o a proton is measured
with an accuracy o 1 .0 1 0 - 1 1 m. What is the minimum
uncertainty in the protons position 1 .0 s later?
h
xp _
4
This calculation is a very rough estimate but correctly predicts
the right order o magnitude or the electrons kinetic energy
(ground state o electron in H atom is -1 3.6 eV) .
2.
h
x mv _
4
The above calculation can be repeated imagining an electron
being trapped inside the nucleus o size 1 0- 1 4 m. I confned
to a space this small, the electrons kinetic energy would be
estimated to be a actor o 1 08 times bigger. An electron with
an energy o the order o 1 00 MeV cannot be bound to a
nucleus and thus it would have enough energy to escape.
h
6.63 1 0
v _
= ___
4 1 .67 1 0 - 2 7 1 .0 1 0 - 1 1
4mx
- 34
= 3200 m s - 1
Thus uncertainty in position ater 1 .0 s = 3200 m = 3.2 km
The uncertainty principle can also be applied to illuminate some
general principles but, to quote Richard Feynman (Feynman
lectures on Physics, volume III, 1 963) , [the application] must
not be taken too seriously; the idea is right but the analysis is not very
accurate.
1.
Estimate o the energy o an electron in an atom.
When an electron is known to be confned within an atom,
then the uncertainty in its position x must be less than
the size o the atom, a. I we equate the two, this means the
uncertainty or its momentum can be estimated as:
h
h
p _
_
4a
4x
I we take this uncertainty in the momentum as a value or the
momentum o the electron (pp), the equations o classical
mechanics can estimate the kinetic energy o the electron:
p2
h2
EK = _ _
2m
32 2 ma2
The diameter o a hydrogen atom is approximately 1 0 - 1 0 m,
so the estimation o the kinetic energy is:
(6.6 1 0 - 3 4 ) 2
= 1 .5 1 0 - 1 9 J
EK ___
2
32 9.3 1 0 - 3 1 (1 0 - 1 0 ) 2
1 eV
126
Q u a n t u m a n d n u c l e a r p h ys i c s
Impossibility o an electron existing within a nucleus o
an atom.
3.
Estimate o lietime o an electron in an excited energy state.
The spectral linewidth associated with an atoms emission
spectrum is usually taken to be very small only discrete
wavelengths are observed. As a result o the uncertainty
principle, the linewidth is, however, not zero. Practically,
there will be a very limited range o wavelengths associated
with any given transition and thus the uncertainty associated
with the energy dierence between the two levels involved
is very small. An estimate o the lietime o an electron in
the excited state can be made using the uncertainty principle
as the uncertainty in energy, E, o a transition is inversely
proportional to the average lietime, t, in the excited state:
h
Et _
4
I E = 5 1 0 - 7 eV
6.6 1 0 - 3 4
t ___
4 5 1 0 - 7 1 .6 1 0 - 1 9
= 6.6 1 0 - 1 0 1 ns
tunneing, poenia barrier and acors afecing
unneing probabiiy
Heisenbergs uncertainty relationship can be used to explain
the quantum phenomenon o tunnelling. The situation being
considered is a particle that is trapped because its energy E
is less than the energy it needs to escape (U0 ) . In classical
physics, i a particle does not have enough energy to escape
rom the potential barrier then it will always remain trapped
inside the system. An example would be a 500 g tennis ball
with a total energy o 4 J bouncing up and down between
two walls that are 1 .2 m high. In order to get over one o
the walls, the tennis ball needs to have a potential energy o
mgh = 0.5 1 0 1 .2 = 6 J. Since it only has 4 J it must remain
trapped by the walls.
In an equivalent microscopic situation (e.g. an electron trapped
inside an atom with energy E which is less than the energy
U0 needed to escape) , the rules o quantum physics mean
that it is now possible or the particle to escape! The particles
waveunction is continuous and does not drop immediately
to zero when it meets the sides o the potential well but the
amplitude decreases exponentially. This means that i the barrier
has a nite width then the waveunction does continue on the
other side o the barrier (with reduced amplitude) . Thereore
there is a probability that particle will be able to escape despite
not having enough energy to do so. Escaping the potential well
does not use up any o the particles total energy.
classically forbidden region
U0
E
particle energy
incoming particle
wavefunction
particle wavefunction
past the barrier
The energy o the emitted alpha particle is 4.25 MeV which
is less than the total potential energy needed to escape the
strong orce within the nucleus. I we imagine an alpha particle
being ormed inside the uranium nucleus, it can only escape
by tunnelling through the potential barrier. In this example,
the very long hal-lie must mean that the probability o the
tunnelling process taking place (given by | | 2 ) must be very low.
wavefunction of
alpha particle
repulsive Coulomb
potential 1/r
U
energy V(r)
hl
X
r
nuclear
surface
attractive
nuclear
potential
Example 2 tunnelling electron microscope
In a scanning tunnelling microscope, a very ne metal tip
is scanned close to, but not touching (separated by a ew nm) ,
a sample metal surace. There is a potential dierence between
the probe and the surace but the electrons in the surace do not
have enough energy to escape the potential energy barrier as
represented by the work unction . Quantum tunnelling can,
however, take place and a tunnelling current will fow as the
waveunction o an electron at the surace will extend beyond
the metal surace. Some electrons will tunnel the gap and
electrical current will be measurable. The value o the current
depends on the separation o the tip and the surace and can be
used to visualize atomic structure.
exit
incident
Reduced probability, but not
reduced energy!
An explanation can be oered in terms o the uncertainty
principle. In order or the particle to escape it would need a
greater total energy (E + E = U0 ) . The particle can disobey
the law o conservation o energy by borrowing an amount o
energy E provided it pays it back in a time t such that the
uncertainty principle applies:
h
Et _
4
The longer the barrier, the more time it takes the particle to
tunnel. Increased tunnelling time will reduce the maximum
possible uncertainty in the energy.
Example 1 alpha decay
The protons and neutrons that orm alpha particles already exist
within nuclei and when emitted overall there is a release o
energy. For example uranium-238 has a hal-lie o about
4.5 billion years. It decays by emitting an alpha particle:
238
92
U
234
90
sample
tip
tunnelling
current
path of
the probe
scanning tunnelling
microscope (STM)
surface
of sample
Th + 42
Q u a n t u m a n d n u c l e a r p h ys i c s
127
t s
hl
the nucleuS Size
examPle
In the example below, alpha particles are allowed to bombard
gold atoms.
I the particles have an energy o 4.2 MeV, the closest
approach to the gold nucleus (Z = 79) is given by
As they approach the
gold nucleus, they eel a
orce o repulsion. I an
alpha particles
alpha particle is heading
directly or the nucleus, it
nucleus
will be refected straight
back along the same
closest approach, r
path. It will have got as
close as it can. Note that none o the alpha particles actually
collides with the nucleus they do not have enough energy.
Alpha particles are emitted rom their source with a known
energy. As they come in they gain electrostatic potential
energy and lose kinetic energy (they slow down) . At the
closest approach, the alpha particle is temporarily stationary
and all its energy is potential.
q1 q2
, and we know q 1 , the charge
Since electrostatic energy = ____
4 r
on an alpha particle and q2 , the charge on the gold nucleus we
can calculate r.
0
deviationS from rutherford Scattering in
high energy exPerimentS
Rutherord scattering is modelled in terms o the coulomb
repulsion between the alpha particle and the target nucleus.
At relatively low energies, detailed analysis o this model
accurately predicts the relative intensity o scattered alpha
particles at given angles o scattering. At high energies,
however, the scattered intensity departs rom predictions.
Fixed scattering angle,
range of alpha
particle energies.
relative intensity of scattered
alpha particles at 60
= 4.2 1 0 6 1 .6 1 0 - 1 9
2 1 .6 1 0 - 1 9 79
r = ___
4 8.85 1 0 - 1 2 4.2 1 0 6
= 5.4 1 0 - 1 4 m
nuclear Scattering exPeriment involving
electronS
Electrons, as leptons, do not eel the strong orce. High-energy
electrons have a very small de Broglie wavelength which can be
o the right order to diract around small objects such as nuclei.
The diraction pattern around a circular object o diameter D
has its rst minimum at an angle given by:
sin _
D
[Note that this small angle approximation is usually not
appropriate to use to determine the location o the minimum
intensity but this is being used to give an approximate answer
around a spherical object. A more exact expression that is
sometimes used or circular objects is sin = 1 .22 __
]
D
High energy (400 MeV) electrons are directed at a target
containing carbon-1 2 nuclei:
electron beam
xed
detector
thin sample
N+
+N
detector
60
scattering
angle
208
lead target 82 Pb
The results are shown below:
The scattered intensity
departs from the Rutherford
scattering formula at about
27.5 MeV
10 2
intensity of
diracted
electrons
(logarithmic
scale)
35
diraction angle ()
10
alpha
energy
in MeV
15
20 25 30 35 40
Eisberg, R. M. and Porter, C. E., Rev. Mod. Phys. 33, 1 90 (1 961 )
At these high energies the alpha particles are beginning to get
close enough to the target nucleus or the strong nuclear force to
begin to have an eect. In order to investigate the size o the
nucleus in more detail, high energy electrons can be used (see
box on the right) .
128
(2 1 .6 1 0 - 1 9 ) (79 1 .6 1 0 - 1 9 )
____
4 8.85 1 0 - 1 2 r
Q u a n t u m a n d n u c l e a r p h ys i c s
The rst minimum is = 35
The de Broglie wavelength or the electrons is eectively:
hc
6.6 1 0 - 3 4 3.0 1 0 8 = 3.1 1 0 - 1 5 m
=_
= ___
E
400 1 0 6 1 .6 1 0 - 1 9
= __
3.1 1 0 - 1 5 = 5.4 1 0 - 1 5 m
D_
sin 35
sin
So radius o nucleus 2.7 1 0 - 1 5 m
HL
n s
energy levelS
The decay o 2 2 6 Ra into
The energy levels in a nucleus are higher than the energy levels o
the electrons but the principle is the same. When an alpha particle or
a gamma photon is emitted rom the nucleus only discrete energies
are observed. These energies correspond to the dierence between
two nuclear energy levels in the same way that the photon energies
correspond to the dierence between two atomic energy levels.
3
1
H
3
2
He +
0
-1
The mass dierence or the decay is
1 9.5 keV c2 . This means that the beta
particles should have 1 9.5 keV o kinetic
energy. In act, a ew beta particles are
emitted with this energy, but all the others
have less than this. The average energy
is about hal this value and there is no
accompanying gamma photon. All beta
decays seem to ollow a similar pattern.
(4.78 MeV)
222
Rn*
excited state
86
ground
222
Rn
state
86
photon
(0.19 MeV)
confrmed its existence. The ull equation
or the decay o tritium is:
number of electrons
neutrinoS and antineutrinoS
226
88 Ra
Rn
(4.59 MeV)
Beta particles are observed to have a continuous spectrum o
energies. In this case there is another particle (the antineutrino in
the case o beta minus decay) that shares the energy. Once again the
amount o energy released in the decay is fxed by the dierence
between the nuclear energy levels involved. The beta particle and the
antineutrino can take varying proportions o the energy available. The
antineutrino, however, is very difcult to observe (see box below) .
Understanding beta decay properly requires
accepting the existence o a virtually
undetectable particle, the neutrino. It is
needed to account or the missing energy
and (angular) momentum when analysing
the decay mathematically. Calculations
involving mass dierence mean that we
know how much energy is available in
beta decay. For example, an isotope o
hydrogen, tritium, decays
as ollows:
3
222
3
1
3
2
H
0
-1
He +
_
+
_
where is an antineutrino
0
0.5
1
energy / MeV
The energy distribution o the electrons
emitted in the beta decay o bismuth-21 0.
The kinetic energy o these electrons is
between zero and 1 .1 7 MeV.
The neutrino (and antineutrino) must be
electrically neutral. Its mass would have
to be very small, or even zero. It carries
away the excess energy but it is very hard
to detect. One o the triumphs o the
particle physics o the last century was
to be able to design experiments that
As has been mentioned beore, another
orm o radioactive decay can also take
place, namely positron decay. In this
decay, a proton within the nucleus
decays into a neutron and the antimatter
version o an electron, a positron, which
is emitted.
1
1
p 10 n +
0
+1
+ +
In this case, the positron, + , is
accompanied by a neutrino.
The antineutrino is the antimatter orm
o the neutrino.
e.g.
14
6
19
10
Ne
C
14
7
N+
19
9
F+
0
-1
0
+1
+ +
_
+
mathematicS of exPonential decay
The basic relationship that defnes
exponential decay as a random process is
expressed as ollows:
dN
_
N
dt
The constant o proportionality between
the rate o decay and the number o nuclei
available to decay is called the decay
constant and given the symbol . Its units
are time - 1 i.e. s - 1 or yr- 1 etc.
dN
_
= -N
dt
The solution o this equation is:
N = N0 e - t
dN
The activity o a source, A, A = - _
dt
A = A0 e - t = N0 e - t
It is useul to take natural logarithms:
ln (N) = ln (N0 e - t)
= ln (N0 ) + ln (e - t)
In (N)
intercept =
ln (N 0 )
This is o the orm y = c + mx so a graph
o ln N vs t will give a straight-line graph.
gradient = -
t
The hal-lie o a radioactive isotope
is 1 0 days. Calculate the raction o a
sample that remains ater 25 days.
T__ = 1 0 days
1
2
ln 2
= _
T__
1
2
N = N0 e - t
I
= 6.93 1 0 - 2 day- 1
N = N0 e - t
t = T_1
2
N
N = _0
2
N0
- T
_
So
= N0 e
2
1 = e - T
_
2
N
Fraction remaining = _
N0
1
_
= e- ( 6 . 9 3
1 0-2 25)
2
= 0.1 87
1
_
2
= 1 8.7%
In = - T _1
2
- T _1 = - In
2
= ln (N0 ) - t ln (e)
ln (N) = ln (N0 ) - t (since ln (e) = 1 )
examPle
= ln 2
ln 2
T _1 = _
2
Q u a n t u m a n d n u c l e a r p h ys i c s
129
iB Questons quantum and nucear pyscs
hl
1.
The diagrams show the variation with distance x o the
waveunction o our dierent electrons. The scale on the
horizontal axis in all our diagrams is the same. For which
electron is the uncertainty in the momentum the largest?
a)
c)
x
0
d)
0
2.
e) From the graph, determine a value or the decay
constant .
x
0
x
x
4.
The diagram represents the available energy levels o an
atom. How many emission lines could result rom electron
transitions between these energy levels?
ground
state
A. 3
B. 6
C. 8
5.
D. 1 2
A medical physicist wishes to investigate the decay o a
radioactive isotope and determine its decay constant and hallie. A GeigerMller counter is used to detect radiation rom
a sample o the isotope, as shown.
[2]
h) Hence calculate the hal-lie o this radioactive
isotope.
[1 ]
This question is about the quantum concept.
a) Explain the term matter waves. State what quantity
determines the wavelength o such waves.
[2]
b) Electron diraction provides evidence to support
the existence o matter waves. What is electron
diraction?
[2]
Light is incident on a clean metal surace in a vacuum. The
maximum kinetic energy KEm a x o the electrons ejected rom
the surace is measured or dierent values o the requency f
o the incident light.
The measurements are shown plotted below.
voltage supply
and counter
GeigerMller
tube
[1 ]
Kmax (10 19 J)
radioactive
source
) Dene the hal-lie o a radioactive substance.
g) Derive a relationship between the decay constant
and the hal-lie .
A biography o Schrdinger contains the ollowing sentence:
Shortly ater de Broglie introduced the concept o matter waves
in 1 924, Schrdinger began to develop a new atomic theory.
energy
2.0
1.5
1.0
0.5
a) Dene the activity o a radioactive sample.
0.0
4.0
[1 ]
Theory predicts that the activity A o the isotope in the
sample should decrease exponentially with time t according
to the equation A = A 0 e - t, where A 0 is the activity at t = 0
and is the decay constant or the isotope.
[1 ]
5.0
5.5
6.0
6.5
7.0 7.5 8.0
f (10 14 H z)
[1 ]
b) Use the graph to determine
The Geiger counter detects a proportion o the particles
emitted by the source. The physicist records the count-rate R
o particles detected as a unction o time t and plots the data
as a graph o ln R versus t, as shown below.
c) Does the plot show that the experimental data are
consistent with an exponential law? Explain.
4.5
a) Draw a line o best t or the plotted data points.
b) Manipulate this equation into a orm which will give a
straight line i a semi-log graph is plotted with appropriate
variables on the axes. State what variables should be
plotted.
[2]
In (R / s -1 )
3.
[2]
The physicist now wishes to calculate the hal-lie.
0
b)
R o those particles that enter the Geiger tube. Explain
why this will not matter in determining the decay
constant o the sample.
[1 ]
(i) the Planck constant
[2]
(ii) the minimum energy required to eject an
electron rom the surace o the metal (the work
function) .
[3]
c) Explain briefy how Einsteins photoelectric theory
accounts or the act that no electrons are emitted
rom the surace o this metal i the requency o
the incident light is less than a certain value.
6.
2
[3]
Thorium-227 (Th-227) undergoes a-decay with a hal-lie o
1 8 days to orm radium-223 (Ra-223) . A sample o Th-227
has an initial activity o 3.2 1 0 5 Bq.
Determine the activity o the remaining thorium-227 ater
50 days.
[4]
1
7.
Explain:
a) The role o angular momentum in the Bohr model or
hydrogen
[3]
0
1
2
3
4
5
t / hr
b) Pair production and ahnihilation
[3]
c) Quantum tunnelling
[3]
d) The Geiger counter does not measure the total
activity A o the sample, but rather the count-rate
130
i B Q u e s t i o n s Q u a n t u m a n d n u c l e a r p h ys i c s
13 O p T i O n a R e l aT i v i T y
Rr r
ObseRveRs and fRames Of RefeRence
The proper treatment o large velocities involves an
understanding o Einsteins theory o relativity and this means
thinking about space and time in a completely dierent way.
The reasons or this change are developed in the ollowing
pages, but they are surprisingly simple. They logically ollow
rom two straightorward assumptions. In order to see why
this is the case we need to consider what we mean by an
obj ect in motion in the frst place.
A person sitting in a chair will probably think that they are at
rest. Indeed rom their point o view this must be true, but this
is not the only way o viewing the situation. The Earth is in orbit
around the Sun, so rom the Suns point o view the person
sitting in the chair must be in motion. This example shows that
an objects motion (or lack o it) depends on the observer.
Galilean TRansfORmaTiOns
It is possible to ormalize the relationship between two dierent
rames o reerence. The idea is to use the measurement in one
rame o reerence to work out the measurements that would
be recorded in another rame o reerence. The equations that
do this without taking the theory o relativity into consideration
are called Galilean transformations.
The simplest situation to consider is two rames o reerence
(S and S') with one rame (S') moving past the other one (S)
as shown below.
Is this person at rest
or moving at great velocity?
Each rame o reerence can record the position and time o
an event. Since the relative motion is along the x-axis, most
measurements will be the same:
y' = y; z' = z; t' = t
I an event is stationary according to one rame, it will be
moving according to the other rame the rames will record
dierent values or the x measurement. The transormation
between the two is given by
We can use these equations to ormalize the calculation o
velocities. The rames will agree on any velocity measured in
the y or z direction, but they will disagree on a velocity in the
x-direction. Mathematically,
u' = u - v
For example, i the moving rame is going at 4 m s - 1 , then
an object moving in the same direction at a velocity o
1 5 m s - 1 as recorded in the stationary rame will be measured as
travelling at 1 1 m s - 1 in the moving rame.
velocity v
t = later
y
frame S
Sun
x' = x - vt
t = zero (two frames on top of one another)
y
y frame S (stationary)
frame S
x
The calculation o relative velocity was considered on page 9.
This treatment, like all the mechanics in this book so ar,
assumes that the velocities are small enough to be able to
apply Newtons laws to dierent rames o reerence.
x
y
frame S
velocity v
x
Newtons 3 laws o motion describe how an objects motion is
eected. An assumption (Newton' s Postulates) underlying these
laws is that the time interval between two events is the same or
all observers. Time is the same or all rames and the separation
between events will also be the same in all rames. As a result,
the same physical laws will apply in all rames.
x
piOn decay eXpeRimenTs
failuRe Of Galilean TRansfORmaTiOn equaTiOns
In 1 964 an experiment at the European
Centre or Nuclear Reseach (CERN) measured
the speed o gamma-ray photons that had
been produced by particles moving close to
the speed o light and ound these photons
also to be moving at the speed o light. This
is consistent with the speed o light being
independent o the speed o its source, to a
high degree o accuracy.
I the speed o light has the same value or all observers (see box on let) then
the Galilean transormation equations cannot work or light.
The experiment analysed the decay o a
particle called the neutral pion into two
gamma-ray photons. Energy considerations
meant that the pions were known to be
moving aster than 99.9% o the speed o light
and the speed o the photons was measured to
be 2.9977 0.0040 1 0 8 m s - 1 .
velocity of bicycle, v
Light leaves the torch
at velocity c with respect
to the person on the bicycle.
Light arrives at the observer
at velocity c (not v + c) .
The theory o relativity attempts to work out what has gone wrong.
o p t i o n A r e l At i v i t y
131
mx quto
maXwell and The cOnsTancy Of The speed Of liGhT
In 1 864 James Clerk Maxwell presented a new theory at the
Royal Society in London. His ideas were encapsulated in a
mathematical orm that elegantly expressed not only what
was known at the time about the magnetic eld B and the
electric eld E, but it also proposed a uniying link between
the two electromagnetism. The rules o electromagnetic
interactions are summarized in our equations known as
Maxwells equations. These equations predict the nature o
electromagnetic waves.
The changing electric and magnetic elds move through space
the technical way o saying this is that the elds propagate
through space. The physics o how these elds propagate
allows the speed o all electromagnetic waves (including light)
to be predicted. It turns out that this can be done in terms o
the electric and magnetic constants o the medium through
which they travel.
____
c=
1
_
0
Most people know that light is an electromagnetic wave, but
it is quite hard to understand what this actually means. A
physical wave involves the oscillation o matter, whereas an
electromagnetic wave involves the oscillation o electric and
magnetic elds. The diagram below attempts to show this.
0
This equation does not need to be understood in detail. The
only important idea is that the speed o light is independent
o the velocity o the source o the light. In other words, a
prediction rom Maxwells equations is that the speed o light in
a vacuum has the same value or all observers.
z
This prediction o the constancy o the speed o light highlights
an inconsistency that cannot be reconciled with Newtonian
mechanics (where the resultant speed o light would be equal to
the addition o the relative speed o the source and the relative
speed o light as measured by the source) . Einsteins analysis
orced long-held assumptions about the independence o space
and time to be rejected.
oscillating electric eld
y
oscillating magnetic eld
cOmpaRinG elecTRic and maGneTic fields
Electrostatic orces and magnetic orces appear very dierent to
one another. Fundamentally, however, they are just dierent
aspects o one orce the electromagnetic interaction. The
nature o which eld is observed depends on the observer. For
example:
b) Two identically charged particles moving with parallel
velocities according to a laboratory rame o reerence.
An observer in a rame o reerence that is moving with
the charged particles will see the particles at rest. Thus
this observer sees the orce o repulsion between the two
charges as solely electrostatic in nature.
a) A charge moving at right angles to a magnetic eld.
An observer in a rame o reerence that is at rest with
respect to the magnetic feld will explain the orce acting
on the charge (and its acceleration) in terms o a magnetic
orce (FM = Bqv) that acts on the moving charge.
FE
+q
force between 2 stationary charges is electrostatic
+q
stationary
magnetic eld into paper
moving
charge
X
X
X
X
X
X
X
X
X
X
X
X
FE
initial force on moving
charge is magnetic
An observer in a rame o reerence that is at rest with
respect to the charge will explain the initial orce acting
on the charge (and its initial acceleration) in terms o
an induced electric orce that results rom the cutting o
magnetic fux.
stationary
charge
X
X
X
X
X
X
X
X
X
X
X
X
initial force on stationary
charge is electric
An observer in a rame o reerence where the laboratory
is at rest will see the total orce between the two charges
as a combination o electrostatic and magnetic. Moving
charges are currents and thus each moving charge creates
its own magnetic eld which is stationary in the laboratory
rame. Each charge is moving in the others stationary
magnetic eld and will experience a magnetic orce.
FE & FM
+q
+q
force between 2 moving objects is a combination
of electric and magnetic
moving magnetic eld
FE & FM
132
o p t i o n A r e l At i v i t y
s rtt
pOsTulaTes Of special RelaTiviTy
The special theory o relativity is based on two undamental
assumptions or postulates. I either o these postulates could
be shown to be wrong, then the theory o relativity would be
wrong. When discussing relativity we need to be even more
than usually precise with our use o technical terms.
One important technical phrase is an inertial frame of
reference. This means a rame o reerence in which the laws
o inertia (Newtons laws) apply. Newtons laws do not apply in
accelerating rames o reerence so an inertial rame is a rame
that is either stationary or moving with constant velocity.
An important idea to grasp is that there is no undamental
dierence between being stationary and moving at constant
velocity. Newtons laws link orces and accelerations. I there is
no resultant orce on an object then its acceleration will be zero.
This could mean that the object is at rest or it could mean that
the object is moving at constant velocity.
The two postulates o special relativity are:
the speed o light in a vacuum is the same constant or all
inertial observers
the laws o physics are the same or all inertial observers.
The rst postulate leads on rom Maxwells equations and
can be experimentally veried. The second postulate seems
completely reasonable particularly since Newtons laws do
not dierentiate between being at rest and moving at constant
velocity. I both are accepted as being true then we need to start
thinking about space and time in a completely dierent way. I
in doubt, we need to return to these two postulates.
simulTaneiTy
One example o how the postulates o relativity disrupt our
everyday understanding o the world around us is the concept
o simultaneity. I two events happen together we say that they
are simultaneous. We would normally expect that i two events
are simultaneous to one observer, they should be simultaneous
to all observers but this is not the case! A simple way to
demonstrate this is to consider an experimenter in a train.
The experimenter is positioned exactly in the middle o a
carriage that is moving at constant velocity. She sends out
two pulses o light towards the ends o the train. Mounted at
the ends are mirrors that refect the pulses back towards the
observer. As ar as the experimenter is concerned, the whole
carriage is at rest. Since she is in the middle, the experimenter
will know that:
pulses leave together
1st pulse hits back wall
2nd pulse hits front wall
the pulses were sent out simultaneously
the pulses hit the mirrors simultaneously
the pulses returned simultaneously.
pulses arrive together
pulses leave
together
pulses arrive
at mirrors
together
Interestingly, the observer on the platorm does see the beams
arriving back at the same time. The observer on the platorm
will know that:
the pulses were sent out simultaneously
the let-hand pulse hit the mirror beore the right-hand
pulse
pulses return
together
The situation will seem very dierent i watched by a stationary
observer (on the platorm) . This observer knows that light must
travel at constant speed both beams are travelling at the same
speed as ar as he is concerned, so they must hit the mirrors
at dierent times. The let-hand end o the carriage is moving
towards the beam and the right hand end is moving away. This
means that the refection will happen on the let-hand end rst.
the pulses returned simultaneously.
In general, simultaneous events that take place at the same
point in space will be simultaneous to all observers whereas
events that take place at dierent points in space can be
simultaneous to one observer but not simultaneous to another!
Do not dismiss these ideas because the experiment seems too
anciul to be tried out. The use o a pulse o light allowed us
to rely on the rst postulate. This conclusion is valid whatever
event is considered.
o p t i o n A r e l At i v i t y
133
lort trorto
lORenTz facTOR
lORenTz TRansfORmaTiOn eXample
The ormulae or special relativity all involve a actor
that depends on the relative velocity between dierent
observers, v.
We can apply the Lorentz transormation equations to the
situation shown on page 1 3 3 . Suppose the experiment on
the train measures the carriage to be 5 0. 0 m long and the
observer on the platorm measures the speed o the train
to be 2 . 7 1 0 8 m s - 1 ( 0.90 c) to the right. In this situation,
we know the times (t) and locations ( x) are measured
according to the experimenter on the train ( rame S) and
the experimenter on the platorm is rame S'.
We defne the Lorentz actor, as ollows:
1
= _
4
v2
1 -_
c2
Lorentz factor,
3
At low velocities,
the Lorentz actor is
approximately equal
to one relativistic
eects are negligible. It
approaches infnity near
the speed o light.
1.
2
Time taken or each pulse to reach mirror at end o
carriage is given by:
1
25.0 = 8.33 1 0 - 8 s
t = _
3.0 1 0 8
speed, v
speed of light, c
Total time taken or each pulse to complete the round
journey to the experimenter is:
5 0.0
tto ta l = _
= 1 .67 1 0 - 7 s
(3.0 10 8 )
lORenTz TRansfORmaTiOns
An observer defnes a rame o reerence and dierent events
can be characterized by dierent coordinates according to the
observers measurements o space and time. In a rame S, an event
will be associated with a given position (x, y and z coordinates)
and take place at a given time (t) . Observers in relative uniorm
motion disagree on the numerical values or these coordinates.
The Galilean transormations equations (page 1 31 ) allowed
us to calculate what an observer in a second rame will record
i we know the values in one rame but assume that the
measurement o time is the same in both rames. Einstein has
shown that this is not correct.
Time taken or LH pulse to reach mirror at end o carriage
is given by:
t' ( L H p u lse) = t - vx
c2
(
)
)
= 1 .9 1 0 - 8 s
Time taken or RH pulse to reach mirror at end o carriage
is given by:
time = t
y
frame S
(x, y, z, t)
t' ( RH pu lse) = t - vx
c2
(
(
Note that the time taken by each pulse is different they do not arrive
simultaneously according to the experimenter on the platform.
frame S
(x, y, z, t)
x
x
Because the rames were synchronized, the observers agree
on the measurements o y and z. To switch between the other
measurements made by dierent observers we need to use the
Lorentz transormations. These all involve the Lorentz actor, , as
defned above. The derivation o these equations is not required.
x' = (x - vt) ; x' = (x - vt) ;
vx ; t' = t - ____
vx
t' = t - _
c2
c2
The reverse transormations also apply. These are just a
consequence o the relative velocity o rame S (with respect
to rame S ' ) being in the opposite direction.
vx'
t = t' + _
c2
)
= 1 .91 1 0 - 7 + 1 .72 1 0 - 7 = 3.63 1 0 - 7 s
y
(
)
(- 2.7 1 0 8 ) 25.0
= 2.29 8.33 1 0 - 8 - __
(3.0 1 0 8 ) 2
time = t
velocity v
134
(
x
(
1
1
=_
= __
v2
(0.9c) 2
_
1 -_
1 c2
c2
1
1
_
__
=
=
= 2.29
1 - 0.81
0.1 9
= 1 .91 1 0 - 7 - 1 .72 1 0 - 7
x
x = (x' + vt') ;
According to the experimenter on the platorm (rame S') ,
(-2.7 1 0 8 ) (- 25.0)
t'( L H p u lse) = 2.29 8.33 1 0 - 8 - ___
(3.0 1 0 8 ) 2
velocity v
)
2.
where t = 8.33 1 0 - 8 s, v = -2.7 1 0 8 m s - 1 (relative
velocity o platorm is moving to the let) and x = -25.0 m
(pulse moving to let)
clock in frame S and clock in frame S are synchronized
to t = t = zero when frames coincide.
(two frames on top of one another)
y
y frame S (stationary)
frame S
(
According to the experimenter on the train (rame S) ,
)
)
o p t i o n A r e l At i v i t y
The return time or the LH pulse is the same as the time
taken or the RH to initially reach the mirror (in each
case, x = 25.0 m and t = 8.33 1 0 - 8 s)
So total time taken or LH pulse to return to centre o
carriage is
total time' ( L H p u lse) = 1 .9 1 0 - 8 + 3.63 1 0 - 7 = 3.82 1 0 - 7 s
This is the same as the total time taken or the RH pulse so
both experimenters observe the return o the pulses to be
simultaneous.
Check: The above calculates that or rame S', the total
time taken or the round trip is 3.82 1 0 - 7 s. The Lorentz
transormation, can also be applied to the pulses journey.
In this situation, x (in rame S) = 0 as the pulse returns
to its starting position.
total t' ( eith er p u lse) = t - vx
= t
c2
= 2.29 1 .67 1 0 - 7 = 3.82 1 0 - 7 s
(
)
vot to
cOmpaRisOn wiTh Galilean
equaTiOn
velOciTy addiTiOn
When two observers measure each others velocity, they will always agree on the
value. The calculation o relative velocity is not, however, normally straightorward.
For example, an observer might see two objects approaching one another, as shown
below.
velocity = 0.7c
The top line o the relativistic addition
o velocities equation can be compared
with the Galilean equation or the
calculation o relative velocities.
velocity = 0.7c
u' = u - v
At low values o v these two equations
give the same value. The Galilean
equation only starts to ail at high
velocities.
person A
stationary observer, C
(rst frame S)
At high velocities, the Galilean
equation can give answers o greater
than c, while the relativistic one always
gives a relative velocity that is less than
the speed o light.
person B
(second frame S)
I each object has a relative velocity o 0.7 c, the Galilean transormations would
predict that the relative velocity between the two objects would be 1 .4 c. This cannot
be the case as the Lorentz actor can only be worked out or objects travelling at less
than the speed o light.
The situation considered is one rame moving relative to another rame at velocity v.
y
frame S (stationary)
y
frame S (moving)
velocity v
x
x
Application o the Lorentz transormation gives the equation used to move between
rames:
u-v
u' = _
uv
1 -_
c2
u' the velocity under consideration in the x-direction as measured in the
second rame, S'
u the velocity under consideration in the x-direction as measured in the
frst rame, S
v the velocity o the second rame, S', as measured in the frst rame, S
In each o these cases, a positive velocity means motion along the positive
x-direction. I something is moving in the negative x-direction then a negative
velocity should be substituted into the equation.
Example
In the example above, two objects approached each other with 70% o the speed o
light. So u is person As velocity as measured in person Bs rame o reerence.
u' = relative velocity o approach to be calculated
u = 0.7 c
v = -0.7 c
1 .4 c
u' = _
(1 + 0.49)
note the sign in the brackets
1 .4 c
= _
1 .49
= 0.94 c
o p t i o n A r e l At i v i t y
135
irt tt
spaceTime inTeRval
pROpeR Time, pROpeR lenGTh & ResT mass
Relativity has shown that our Newtonian ideas o space
and time are incorrect. Two inertial observers will generally
disagree on their measurements o space and time but they
will agree on a measurement o the speed o light. Is there
anything else upon which they will agree?
In relativity, a good way o imagining what is going on is to
consider everything as dierent events in something called
spacetime. From one observers point o view, three
co-ordinates (x, y and z) can defne a position in space. One
urther coordinate is required to defne its position in
time (t) . An event is a given point specifed by these our
coordinates (x, y, z, t) .
a) Proper time interval t0
When expressing the time taken between events (or
example the length o time that a frework is giving out
light) , the proper time is the time as measured in a
rame where the events take place at the same point in
space. It turns out to be the shortest possible time that any
observer could correctly record or the event.
measuring how long a rework lasts
Moving frame measures a
longer time for the rework
since in this frame the
rework is moving.
As a result o the Lorentz transormation, another observer
would be expected to come up with totally dierent numbers
or all o these our measurements (x', y', z', t') . The amazing
thing is that these two observers will agree on something. This
is best stated mathematically:
(ct) 2 - x2 - y2 - z2 = (ct ') 2 - x' 2 - y ' 2 - z2
Clock that is stationary with the
rework measures the proper
time for which it lasted.
On normal __________
axes, Pythagorass theorem shows us that the
quantity (x2 + y2 + z2 ) is equal to the length o the line rom
the origin, so (x2 + y2 + z2 ) is equal to (the length o the line) 2 .
In other words, it is the separation in space.
(Separation in space) 2 = (x2 + y2 + z2 )
l 2 = x2 + y 2 + z 2
b) Proper length L 0
l
z
I A is moving past B then B will think that time is
running slowly or A. From As point o view, B is moving
past A. This means that A will think that time is running
slowly or B. Both views are correct!
x
y
The two observers agree about something very similar
to this, but it includes a coordinate o time. This can be
thought o as the separation in imaginary our-dimensional
spacetime.
As beore, dierent observers will come up with
dierent measurements or the length o the same object
depending on their relative motions. The proper length
o an object is the length recorded in a rame where the
object is at rest.
Moving frame measures
a shorter length for the
reworks diameter since
the rework is moving
in this frame.
(Separation in spacetime) 2 = (ct) 2 - x2 - y2 - z2
or
(Separation in spacetime) 2
= (time separation) 2 - (space separation) 2
In 1 dimension, this is simplifed to
(ct') 2 - (x') 2 = (ct) 2 - (x) 2
OTheR invaRianT quanTiTies
In addition to the spacetime interval between two events (see
box above) , all observers agree on the values o three other
quantities associated with the separation between two events
or with reerence to a given object. These are:
Proper time interval t0
Proper length L 0
Rest mass m 0
These our quantities are said to be invariant as they
are always constant and do not vary with a change o
observer. There are additional quantities, not associated with
mechanics, that are also invariant e.g. electric charge.
136
o p t i o n A r e l At i v i t y
Ruler that is stationary with the
rework measures the proper length
for its diameter.
c) Rest mass m 0
The measurement o mass depends on relative
velocity. Once again it is important to distinguish the
measurement taken in the rame o the object rom all
other possible rames. The rest mass o an object is its
mass as measured in a rame where the object is at rest.
A particles rest mass does not change.
T to
liGhT clOck
A light clock is an
imaginary device. A beam
o light bounces between
two mirrors the time
taken by the light between
bounces is one tick o the
light clock.
As shown in the derivation
the path taken by light in
a light clock that is moving
at constant velocity is
longer. We know that
the speed o light is fxed
so the time between the
ticks on a moving clock
must also be longer. This
eect that moving clocks
run slow is called time
dilation.
deRivaTiOn Of The effecT fROm fiRsT
pRinciples
tick
l
I we imagine a stationary observer with one light clock then
t is the time between ticks on their stationary clock. In this
stationary frame, a moving clock runs slowly and t' is the
time between ticks on the moving clock: t' is greater than t.
pulse leaves bottom mirror
l
l
l
tick
vt
In the time t',
the clock has moved on a distance = v t'
_________
pulse bounces o top mirror
Distance travelled by the light, l' = ((vt') 2 + l2 )
l'
t' = _
c
The time between bounces
t0 is the proper time or
this clock in the rame
where the clock is at rest.
________
tick
(vt') 2 + l2
= __
c
v2 t' 2 + l2
_
2
t' =
c2
2
2
v
l
_
_
2
t' 1 - 2 = 2
c
c
(
l2 = t2
_
c2
but
pulse returns to bottom mirror
deRivaTiOn Of effecT fROm lORenTz
TRansfORmaTiOn
I rame S is a rame where two events take place at the
same point in space, then the time interval between these
two events must be the proper time interval, t0 .
or
)
(
)
v2 = t2
t' 2 1 - _
c2
1
______ t or
t' = _
v2
1 -_
c2
t' = t
This equation is true or all measurements o time, whether
they have been made using a light clock or not.
Time dilation is then a direct consequence o the Lorentz
transormation:
x
t' = t - _
c2
(
)
Where t = t0 , ( the proper time interval) and x = zero
( same point in space)
time interval in rame S', t' = t0
o p t i o n A r e l At i v i t y
137
lgt otrto to ort
rtty
effecT Of lenGTh cOnTRacTiOn
eXample
Time is not the only measurement that is aected by relative motion. There is another
relativistic eect called length contraction. According to a (stationary) observer, the
separation between two points in space contracts i there is relative motion in that
direction. The contraction is in the same direction as the relative motion.
An unstable particle has a lietime o
4.0 1 0 - 8 s in its own rest rame. I it is
moving at 98% o the speed o light calculate:
a) Its lietime in the laboratory rame.
b) The length travelled in both rames.
_________
moving frame
a) =
1
__
1 - (0.98)
2
= 5.025
t = t0
= 5.025 4.0 1 0 - 8
= 2.01 1 0 - 7 s
Length contracts along direction
of motion when compared
with stationary frame.
b) In the laboratory rame, the particle moves
Length = speed time
= 0.98 3 1 0 8 2.01 1 0 7
= 59.1 m
stationary frame
In the particles rame, the laboratory
moves
59.1
l = _
Length contracts by the same proportion as time dilates the Lorentz actor is
once again used in the equation, but this time there is a division rather than a
multiplication.
= 1 1 .8 m
(alternatively: length = speed time
= 0.98 3 1 0 8 4.0 1 0 - 8
= 1 1 .8 m)
L0
L= _
deRivaTiOn Of lenGTh cOnTRacTiOn fROm lORenTz TRansfORmaTiOn
When we measure the length o a moving object, then we are
recording the position o each end o the object at one given
instant o time according to that rame o reerence. In other
words the time interval measured in rame S between these two
events will be zero, t = 0. In this case, the length measured x
is the length o the moving object L 0 .
Length contraction is then a direct consequence o the
Lorentz transormation, as, i we move into the rame, S' ,
The muOn eXpeRimenT
Muons are leptons (see page 78) they can be thought o as
a more massive version o an electron. They can be created in
the laboratory but they quickly decay. Their average lietime is
2.2 1 0 - 6 s as measured in the rame in which the muons are
at rest.
Muons are also created high up (1 0 km above the surace) in
the atmosphere. Cosmic rays rom the Sun can cause them
to be created with huge velocities perhaps 0.99 c. As they
travel towards the Earth some o them decay but there is still a
detectable number o muons arriving at the surace o the Earth.
shower of
particles
cosmic rays from Sun
atmos
ph ere
some muons decay
before reaching surface some muons
reach surface
Earth
138
o p t i o n A r e l At i v i t y
10 km
where the obj ect is at rest, we will be measuring the proper
length L 0 :
x' = (x - vt)
Where x' = L 0 (the proper length) and
t = zero (simultaneous measurements o position o end o object)
Length in rame S' , L 0 = (L)
L0
L= _
Without relativity, no muons would be expected to reach the
surace at all. A particle with a lietime o 2.2 1 0 - 6 s which
is travelling near the speed o light (3 1 0 8 m s - 1 ) would be
expected to travel less than a kilometre beore decaying
(2.2 1 0 - 6 3 1 0 8 = 660 m) .
The moving muons are eectively moving clocks. Their high
speed means that the Lorentz actor is high.
________
=
1
_
1 - 0.99
2
= 7.1
Thereore an average lietime o 2.2 1 0 - 6 s in the muons
rame o reerence will be time dilated to a longer time as ar
as a stationary observer on the Earth is concerned. From this
rame o reerence they will last, on average, 7.1 times longer.
Many muons will still decay but some will make it through to
the surace this is exactly what is observed.
In the muons rame they exist or 2.2 1 0 - 6 s on average.
They make it down to the surace because the atmosphere (and
the Earth) is moving with respect to the muons. This means
that the atmosphere will be length-contracted. The 1 0 km
distance as measured by an observer on the Earth will only
10
be ___
= 1 .4 km. A signifcant number o muons will exist long
7.1
enough or the Earth to travel this distance.
st gr (mnkowk gr) 1
spaceTime diaGRams
Spacetime separation was introduced on page 1 36. A spacetime
diagram is a visual way of representing the geometry.
Measurements can be taken from the diagram to calculate
actual values.
time
We cannot represent all four dimensions on the one diagram,
so we usually limit the number of dimensions of space that
we represent. The simplest representation has only one
dimension of space and one of time as shown below.
particle at rest
particle with constant speed
Whatever axes are being used, by convention, the path of a
beam of light is represented by a line at 45 to the axes.
The advance of proper time for any traveller can be
calculated from the overall separation in spacetime. In the
travellers frame of reference, they remained stationary so
the separation between two events can be calculated as
shown below.
eXample 1 Of spaceTime diaGRams
The advance of proper time for the journey between the
events ABCD can be calculated from the values on the
spacetime diagram.
time/yr
particle which starts fast
and then slows down
D
6
5
space
4
An object (moving or stationary) is always represented as a
line in spacetime.
C 3
light
Note that:
The values on the spacetime diagram are as would be
measured by an observer whose worldline is represented
by the vertical axis.
The vertical axis in the above spacetime diagram is time
t. An alternative is to plot (speed of light time) , ct. This
means that both axes can have the same units (m, lightyears or equivalent) .
2
B
1
A
1 0.5
1
2
3
4
space/ly
A journey through spacetime
Journey
Space separation
(x) /ly
Time separation
(t) /yr
(Spacetime separation) 2
(ct) 2 - (x) 2 /ly2
Advance of proper time according to
traveller / yr
_________
(ct) - (x)
_
c
2
t' =
AB
0.0
1 .0
1 2 - 02 = 1
BC
1 .5
2.0
4 2.25 = 1 .75
CD
2.5
3.0
9 - 6.25 = 2.75
2
____
1 .00
____
1 .75
____
2.75
= 1 .00
= 1 .32
= 1 .66
The total advance of proper time for the traveller is 1 .00 + 1 .32 + 1 .66 = 3.98 yr. This compares with the advance of 6.0 years
according to an observer whose worldline is a vertical line on this spacetime diagram. This difference is an example of time
dilation (see page 1 37) .
The alternative journey direct from A D shows a greater elapsed proper time.
Journey
AD
Space separation
(x) /ly
Time separation
(t) /yr
1 .0
6.0
(Spacetime separation) 2
(ct) 2 - (x) 2 /ly2
36 - 1 = 35
Advance of proper time according to traveller / yr
_________
(ct) - (x)
_
c
2
t' =
2
___
35
= 5.92
This is always true. A direct worldline always has a greater amount of elapsed proper time than an indirect worldline.
o p t i o n A r e l At i v i t y
139
st gr 2
calculaTiOn Of Time dilaTiOn and
lenGTh cOnTRacTiOn
Time dilation and length contraction
are quantitatively represented on spacetime
diagrams. Refer to diagram on page 1 39.
a) Time dilation: In the journey direct
from B C, the relative velocity between
the traveller and the stationary observer
1 .5 ly
is _____
= 0.75 c. The Lorentz gamma
2.0 yrs
factor is:
1
1
______ = __
________ = 1 .51
=_
1 - 0.75 2
v2
1 - _
c2
The journey takes 2 yrs according to the
observer at rest. This means the proper time as
measured by the traveller will be:
t = _
2.0 =1 .32 yr
t = t0 t0 = _
1 .51
as shown in the table on page 1 39.
b) Length contraction: The observer at rest
measures the journey length from BC
to be 1 .5 ly. The journey will be length
contracted to be
eXample 2 cuRved wORldline
B
16 13
15 12
14
11
13
10
12
9
11
10
8
time 9
7
8
6
7
5
6
5
4
4
3
3
2
2 1
2
2
increase in = advance - increase 2
1 1
proper time in time in space
O
L0
1 .5
_
L= _
= 1 .51 = 0.99 ly
1 2 3 4 5 6 7 8 9 10
space
The relative velocity of travel is 0.75 c, and
the time taken to go from from B C, in
the travellers frame of reference, is 1 .32 yr.
This makes the distance according to the
traveller to be 0.75 c 1 .32 yr = 0.99 ly as
shown above.
Proper time along a curved worldline from event O to event B is smaller
than the proper time along the straight line from O to B.
T t rox 1
As mentioned on page 1 36, the theory of relativity gives no
preference to different inertial observers the time dilation
effect (moving clocks run slowly) is always the same. This
leads to the twin paradox. In this imaginary situation, two
identical twins compare their views of time. One twin remains
on Earth while the other twin undergoes a very fast trip out to a
distant star and back again.
As far as the twin on the Earth is concerned the other twin is a
moving observer. This means that the twin that remains on the
Earth will think that time has been running slowly for the other
twin. When they meet up again, the returning twin should
have aged less.
before
after
This seems a very strange prediction, but it is correct according
to the time dilation formula. Remember that:
This is a relativistic effect time is running at different rates
because of the relative velocity between the two twins and
not because of the distance between them.
The difference in ageing is relative. Neither twin is getting
younger; as far as both of them are concerned, time has been
passing at the normal rate. Its just that the moving twin
thinks that she has been away for a shorter time than the
time as recorded by the twin on the Earth.
The paradox is that, according to the twin who made the
journey, the twin on the Earth was moving all the time and so
the twin left on the Earth should have aged less. Whose version
of time is correct?
The solution to the paradox comes from the realization that the
equations of special relativity are only symmetrical when the
two observers are in constant relative motion. For the twins to
meet back up again, one of them would have to turn around.
This would involve external forces and acceleration. If this is the
case then the situation is no longer symmetrical for the twins.
The twin on the Earth has not accelerated so her view of the
situation must be correct.
The resolution of the twin paradox using a spacetime diagram is
on page 1 41 .
140
o p t i o n A r e l At i v i t y
T rox 2
In order to check whose version o time is correct, they agree
to send light signals every year. The spacetime diagram or
this situation in the Earths rame o reerence is shown below
(let) .
ResOlvinG The Twin paRadOX usinG spaceTime
diaGRams
The diagram below is a spacetime diagram or a journey to a
distant planet ollowed by an immediate return.
Note that there is no paradox; they agree on the number o
signals sent and received; the travelling twin has aged less than
the twin that stayed on Earth.
According to the twin remaining on Earth:
the distance to the planet = 3.0 ly
relative velocity o traveller is 0.6 c
A more complicated spacetime diagram can be drawn or the
reerence rame o the outbound traveller (below right) . Note that:
3.0
each leg o the journey takes ___
= 5.0 yr
0.6
Total journey time = 1 0.0 yr
The frst our years has the travelling twins worldline
vertical i.e. stationary.
The gamma actor is
When the travelling twin turns round, she leaves her
original rame o reerence and changes to a rame where
the Earth is moving towards her at __35 c (= 0.6 c) .
1
1
______
_______ = 1 .25
= _
= __
1 - 0.6 2
v2
1-_
c2
Her relative velocity towards the Earth with respect to
her original rame o reerence can be calculated rom the
15
velocity transormation equations as __
c (= 0.88 c) back.
17
So according to the twin undertaking the journey:
5.0
= 4 .0 yr
each leg o the journey takes ____
1 .25
Total journey time = 8.0 yr
In this rame o reerence, the total time or the round trip
would be measured as 1 2.5 yr
3.0
= 2.4 ly
the distance to the planet = ____
1 .25
4.8
relative velocity o Earth = ___
= 0.6 c
8.0
reference frame
of Earth
10
10
reference frame
of outbound traveller
8
12
9.2
8
9
9
7
8
7.6
6
7
t (yr)
5
3
3
2
2.4
2
1.6
1
outbound
traveler
v = 3/5 c
5
6
5
4
annual
signals
from Earth
annual
signals
from
traveller
lines of
simultaneity
for traveller
3.2
8
7
Earth
5
v = -3/5 c
4
Earth 4
v= 0
inbound traveller
v = -15/17 c
6
6
5
6
10
9
7
6.8
light signals
from Earth
6.8
t (yr)
7
8
7.8
inbound traveler
v = 3/5 c
8.4
1
11
8.4
9.2
0.8
13
3.2
4
3
3
2.4
outbound
traveller
2 v= 0
2
1.6
0.3
1
1
x (ly)
x (ly)
-8
-7
-6
-5
-4
-3
-2
-1
o p t i o n A r e l At i v i t y
141
st gr 3
RepResenTinG mORe Than One ineRTial fRame
On The same spaceTime diaGRam
Mathematically for the above process to agree with the Lorentz
transformation calculations, the following must apply:
The Lorentz transformations describe how measurements
of space and time in one frame can be converted into the
measurements observed in another frame of reference. The
situation in each frame of reference can be visualized by using
separate spacetime diagrams for each frame of reference
(see page 1 41 for examples) .
The angle between the ct' axis (the worldline for the origin
of S') and the ct axis is the same as the angle between the x'
axis and the x axis. It is:
It is also possible to represent two inertial frames on the
same spacetime diagram. A frame S (coordinates x' and
ct') is moving at relative constant velocity +v according to
a frame S (coordinates x and ct) . The principles are as follows:
The same worldline applies to both sets of coordinate axes
(that is, to x and ct, as well as to x' and ct') .
The Lorentz transformation is made by changing the coordinate
system for frame S' rather than the position of the worldline.
The spacetime axes for frame S has x and ct at right angles to
one another as normal.
The spacetime axes for frame S' has its x' and ct' axes both
angled in towards the x = ct line (which represents a path of
a beam of light.
v
= tan- 1 _
c
()
The scales used by the axes in S' are different to the scales
used by the axes in S.
A given value is represented by a greater length on the ct'
axis when compared with the ct axis.
A given value is represented by a greater length on the x'
axis when compared with the x axis.
The ratio of the measurements on the axes depends on the
relative velocity between the frames. The equation (which
does not need to be recalled) is:
ct'
ratio of units _
=
ct
______
v
1 + __
c
_
v
1 - __
c
2
2
2
2
ct'
The coordinates of a spacetime event in S are read from the
x and ct axes directly.
ct
The coordinates of a spacetime event in S ' are measured by
drawing lines parallel to the ct' and x' axes until they hit the
x' and ct' axes.
Frame S
ct
x'
Frame S'
ct'
(x', ct') = (0, 1)
(x, ct) = (v/c, )
light
D
(x', ct') = (1, 0)
(x, ct) = (, v/c)
x
Summary
At greater speed:
the S' axes swing towards the x = ct line as the angle
increases.
C
the ct' and x' axes are more stretched when compared
with the ct and x axes.
x'
B
Events that are simultaneous in S are on the same horizontal
line.
A
x
1.
2.
Events A & B are simultaneous in frame S but are not
simultaneous in frame S' (A occurs before B)
2 = 0.25
tan = _
8
relative velocity of frames S and S = 0.25 c
Events C & D occur at same location in frame S' .
Events C & D occur at different locations in frame S.
3.
A pulse of light emitted by event A arrives at event D
according to both frames of reference. It cannot arrive at
events B or C.
142
o p t i o n A r e l At i v i t y
Events that are simultaneous in S' are on a line parallel to
the x' axis.
HL
m rg
mass and eneRGy
E = mc2
The most amous equation in all o physics is surely Einsteins
massenergy relationship E = mc2 , but where does it come
rom? By now it should not be a surprise that i time and
length need to be viewed in a dierent way, then so does
energy.
velocity
According to Newtons laws, a constant orce produces a
constant acceleration. I this was always true then any velocity
at all should be achievable even aster than light. All we
have to do is apply a constant orce and wait.
Mass and energy are equivalent. This means that energy can
be converted into mass and vice versa. Einsteins massenergy
equation can always be used, but one needs to be careul about
how the numbers are substituted. Newtonian equations (such
as KE = __12 mv2 or momentum = mv) will take dierent orms
when relativity theory is applied.
The energy needed to create a particle at rest is called the rest
energy E0 and can be calculated rom the rest mass:
E0 = m 0 c2
I this particle is given a velocity, it will have a greater total
energy.
constant force velocity as
predicted by Newton
E = m 0 c2
speed of light, c
constant acceleration
time
velocity
In practice, this does not happen. As soon as the speed o an
object starts to approach the speed o light, the acceleration
gets less and less even i the orce is constant.
constant force velocity as
predicted by Einstein
speed of light, c
acceleration decreases
as speed gets close to c
time
The orce is still doing work (= orce distance) , thereore
the object must still be gaining kinetic energy and a new
relativistic equation is needed or energy:
E = m 0 c2
Note that some textbooks compare this equation with the
defnition o rest energy (E0 = m 0 c2 ) in order to defne a
concept o relativistic mass that varies with speed (m = m 0 ) .
The current IB syllabus does not encourage this approach.
The preerred approach is to see rest mass as invariant and to
adopt a new relativistic ormula or kinetic energy:
Total energy = rest energy + kinetic energy = m0 c2
rest energy = m 0 c2
so, kinetic energy EK = ( - 1 ) m 0 c2
o p t i o n A r e l At i v i t y
143
HL
Rtvtc ot d rgy
equaTiOns
uniTs
The laws o conservation o momentum and conservation
o energy still apply in relativistic situations. However the
concepts oten have to be refned to take into account the new
ways o viewing space and time.
SI units can be applied in these equations. Sometimes,
however, it is useul to use other units instead.
For example, in Newtonian mechanics, momentum p is
defned as the product o mass and velocity.
p = mv
In relativity it has a similar orm, but the Lorentz actor needs
to be taken into consideration.
At the atomic scale, the joule is a huge unit. Oten the
electronvolt (eV) is used. One electronvolt is the energy
gained by one electron i it moves through a potential
dierence o 1 volt. Since
energy dierence
Potential dierence = __
charge
1 eV = 1 V 1 .6 1 0 - 1 9 C
= 1 .6 1 0 - 1 9 J
p = m0 v
In act the electronvolt is too small a unit, so the standard SI
multiples are used
The momentum o an object is related to its total energy. In
relativistic mechanics, the relationship can be stated as
1 keV = 1 000 eV
E2 = p 2 c2 + m 0 2 c4
1 MeV = 1 0 6 eV
In Newtonian mechanics, the relationship between energy
and momentum is
p2
E = _
2m
Do not be tempted to use the standard Newtonian equations
i the situation is relativistic, then you need to use the
relativistic equations.
etc.
Since mass and energy are equivalent, it makes sense to have
comparable units or mass. The equation that links the two
(E = mc2 ) defnes a new unit or mass the MeV c- 2 . The
speed o light is included in the unit so that no change o
number is needed when switching between mass and energy
I a particle o mass o 5 MeV c- 2 is converted completely into
energy, the energy released would be 5 MeV. It would also be
possible to use keV c- 2 or GeV c- 2 as a unit or mass.
In a similar way, the easiest unit or momentum is the
MeV c- 1 . This is the best unit to use i using the equation
which links relativistic energy and momentum.
eXample
The Large Electron / Positron (LEP) collider at the European
Centre or Nuclear Research (CERN) accelerates electrons to
total energies o about 90 GeV. These electrons then collide
with positrons moving in the opposite direction as shown below.
Positrons are identical in rest mass to electrons but carry a
positive charge. The positrons have the same energy as the
electrons.
Electron
Electron
b) For these two particles, estimate their relative velocity o
approach.
since so large
relative velocity c
c) What is the total momentum o the system (the two
particles) beore the collision?
zero
Total energy = 90 GeV
Total energy = 90 GeV
a) Use the equations o special relativity to calculate,
(i) the velocity o an electron (with respect to the
laboratory) ;
Total energy = 90 GeV = 90000 MeV
Rest mass = 0.5 MeVc- 2 = 18000 (huge)
vc
(ii) the momentum o an electron (with respect to the
laboratory) .
p2 c2 = E2 - m 0 2 c4
E2
p 90 GeVc- 1
144
o p t i o n A r e l At i v i t y
d) The collision causes new particles to be created.
(i) Estimate the maximum total rest mass possible or the
new particles.
Total energy available = 180 GeV
max total rest mass possible = 180 GeVc- 2
(ii) Give one reason why your answer is a maximum.
Above assumes that particles were created at rest
Rtvt x
HL
paRTicle acceleRaTiOn and elecTRic chaRGe
eXample
In a particle accelerator (e.g. a linear accelerator or cyclotron) ,
charged particles are accelerated up to very high energies. The
basic principle is to pass the charged particles through a series
of potential differences and each time, the particles total
energy increases as a result. The increase in kinetic energy
(EK) as a result of a charge q passing through a potential
difference V is given by:
An electron is accelerated through a pd of 1 .0 1 0 6 V.
Calculate its velocity.
qV = EK
Energy gained = 1 .0 1 0 6 1 .6 1 0 - 1 9 J
= 1 .6 1 0 - 1 3 J
E0 = m 0 c2 = 9.1 1 1 0 - 3 1 (3 1 0 8 ) 2
= 8.2 1 0 - 1 4 J
Total energy = 1 .6 1 0 - 1 3 + 8.2 1 0 - 1 4
= 2.42 1 0 - 1 3 J
2.42 1 0 - 1 3 = 2.95
= __
8.2 1 0 - 1 4
______
velocity =
1 - _1 c
2
= 0.94 c
phOTOns
Photons are particles that have a zero rest mass and travel at
the speed of light, c. Their total energy and their frequency f is
linked by Plancks constant h:
E = hf
eXample: decay Of a piOn
A neutral pion (0 ) is a meson of rest mass m0 = 1 35.0 MeV c- 2 .
A typical mode of decay is to convert into two photons:
2
0
The wavelength of these photons can be calculated:
a) Decay at rest
If the pion was at rest when it decayed, each photon would
have half the total energy of the pion:
E = 67.5 MeV = 67.5 1 0 6 1 .6 1 0 - 1 9 J
= 1 .08 1 0 - 1 1 J
The relativistic equation that links total energy, E and
momentum, p, must also apply to photons:
E2 = p 2 c2 + m 0 2 c4
The rest mass of a photon is zero so the momentum of a
photon is:
hf _
E=_
h
p=_
c
c =
Note that in this example, total energy 2 m 0 c2 , so = 2 so v
= 0.866 c
Each photon will have a total energy of
1 35 MeV = 2.1 6 1 0 - 1 1 J
and a momentum of 1 35 MeV c- 1 . The wavelengths of the
photons will be:
c = 6.63 1 0 - 3 4 __
3.0 1 0 8
= h_
E
2.1 6 1 0 - 1 1
= 9.21 1 0 - 1 5 m
Initial total momentum for the pion in the forward direction
can be calculated from
Plancks constant can be used to calculate the wavelength
of one of the photons:
c
E = h_
c = 6.63 1 0 - 3 4 __
3.0 1 0 8
= h_
E
1 .08 1 0 - 1 1
= 1 .84 1 0 - 1 4 m
The momentum of the pion was initially zero as it was at rest.
Conservation of momentum means that the photons will be
emitted in opposite directions. The total momentum of each
photon add together to give a total, once again, of zero.
E2 = p 2 c2 + m 0 2 c4
p c = E2 - m 0 2 c4 = (4 - 1 ) m 0 2 c4
2
p=
__
3
2
m 0 c = 1 .73 1 35.0 = 233.8 MeV c- 1
So conservation of momentum in forward direction is:
233.8 = 2 1 35 cos
233.8
cos = _
= 0.866
270
= 30
b) Decay while moving
Suppose the pion was moving forward when it decayed
with a total energy 270.0 MeV c- 2 ; the photons will be
emitted as shown below:
after
before
photon 1
pion
photon 2
o p t i o n A r e l At i v i t y
145
HL
Gr rtty t r
pRinciple Of equivalence
One o Einsteins thought experiments considers how an observers view o the world would change i they were accelerating. The
example below considers an observer inside a closed spaceship.
There are two possible situations to compare.
The rocket could be ar away rom any planet but accelerating orwards.
The rocket could be at rest on the surace o a planet.
dropped object
will fall
towards oor
dropped
accelerating object will fall
towards oor
forward
astronaut feels a force when rocket is
accelerating forward
rocket at rest
on planet
planet
astronaut feels a force when rocket is
at rest on the surface of a planet
Although these situations seem completely dierent, the observer inside the rocket would interpret these situations as being
identical.
This is Einsteins principle o equivalence a postulate that states that there is no dierence between an accelerating rame o
reerence and a gravitational feld.
From the principle o equivalence, it can be deduced that light rays are bent in a gravitational feld (see below) and that time slows
down near a massive body (see page 1 47) .
bendinG Of liGhT
Einsteins principle o equivalence suggests that a gravitational
feld should bend light rays! There is a small window high up in
the rocket that allows a beam o light to enter.
In both o the cases in diagrams 1 and 2, the observer is an
inertial observer and would see the light shining on the wall at
the point that is exactly opposite the small window. I, however,
the rocket was accelerating upwards (see diagram 3) then the
window
nal position of
window when
light
hits
light hits
wall
opposite original
window position
of window
1 rocket at rest
in space
146
But Einsteins principle o equivalence states that there is
no dierence between an accelerating observer and inertial
observer in a gravitational feld. I this is true then light should
ollow a curved path in a gravitational feld as shown in
diagram 4. This eect does happen!
light hits
wall
opposite
window nal position
of window
rocket
when
moves
light hits
upwards at
constant
velocity
2 rocket moving with
constant velocity
o p t i o n A r e l At i v i t y
beam o light would hit a point on the wall below the point
that is opposite the small window.
light hits
below window
as rocket has
speeded up
rocket
accelerating
3 rocket accelerating
upwards
light hits below
window in an
accelerating
rocket and in a
view
stationary rocket
inside
rocket in a gravitational
eld
4 rocket at rest in a
gravitational eld
HL
Grvtto rd ft
cOncepT
maThemaTics
The general theory o relativity makes other predictions that
can be experimentally tested. One such eect is gravitational
red shift clocks slow down in a gravitational eld. In other
words a clock on the ground foor o a building will run slowly
when compared with a clock in the attic the attic is urther
away rom the centre o the Earth.
This gravitational time dilation eect can be mathematically
worked out or a uniorm gravitational eld g. The change in
requency f is given by
gh
f _
___
= 2
f
c
where
f
is the requency emitted at the source
g
is the gravitational eld strength (assumed to be
constant)
h is the height dierence and
c
A clock on the ground
oor runs slow when
compared with a
clock in the attic
is the speed o light.
eXample
A UFO travels at such a speed to remain above one point on
the Earth at a height o 200 km above the Earths surace. A
radio signal o requency o 1 1 0 MHz is sent to the UFO.
(i) What is the requency received by the UFO?
(ii) I the signal was refected back to Earth, what would be
the observer requency o the return signal? Explain your
answer.
(i)
The same eect can be imagined in a dierent way. We have
seen that a gravitational eld aects light. I light is shone
away rom a mass (or example the Sun) , the photons o light
must be increasing their gravitational potential energy as they
move away. This means that they must be decreasing their
total energy. Since requency is a measure o the energy o a
photon, the observed requency away rom the source must
be less than the emitted requency.
f = 1 .1 1 0 8 Hz
g = 1 0 m s 2
h = 2.0 1 0 5 m
1 0 2.0 1 0 5 1 .1 1 0 8 Hz
f = __
(3 1 0 8 ) 2
= 2.4 1 0 3 Hz
f received = 1 .1 1 0 8 2.4 1 0 3
= 1 09999999.998 Hz
1 .1 1 0 8 Hz
(ii) The return signal will be gravitationally blue shited.
Thereore it will arrive back at exactly the same
requency as emitted.
At the top of the
building, the photon
has less energy, and
so a lower frequency,
than when it was at
the bottom.
The oscillations o the light can be imagined as the pulses o a
clock. An observer at the top o the building would perceive
the clock on the ground foor to be running slowly.
o p t i o n A r e l At i v i t y
147
HL
sortg
evidence TO suppORT GeneRal RelaTiviTy
Bending of star light
The predictions o general relativity, just like those o special relativity, seem so strange that we need strong experimental evidence.
One main prediction was the bending o light by a gravitational eld. One o the rst experiments to check this eect was done by a
physicist called Arthur Eddington in 1 91 9.
The idea behind the experiment was to measure the defection o light (rom a star) as a result o the Suns mass. During the day,
the stars are not visible because the Sun is so bright. During a solar eclipse, however, stars are visible during the ew minutes when
the Moon blocks all o the light rom the Sun. I the positions o the stars during the total eclipse were compared with the positions
o the same stars recorded at a dierent time, the stars that appeared near the edge o the Sun would appear to have moved.
actual position of star
Sun
apparent position
of star
Moon
not to scale!
Earth
apparent position usual position of star in sky
during eclipse
(compared with others)
The angle o the shit o these stars turned out to be exactly the angle as predicted by Einsteins general theory o relativity.
Gravitational lensing
The bending o the path o light or the warping o spacetime (depending on which description you preer) can also produce some
very extreme eects. Massive galaxies can defect the light rom quasars (or other very distance sources o light) so that the rays
bend around the galaxy as shown below.
image of quasar
massive galaxy
quasar
observer
image of quasar
not to scale!
In this strange situation, the galaxy is acting like a lens and we can observe multiple images o the distant quasar.
evidence TO suppORT GRaviTaTiOnal Red shifT
Atomic clock frequency shift
PoundRebkaSnider experiment
Because they are so sensitive, comparing the dierence in
time recorded by two identical atomic clocks can provide a
direct measurement o gravitational red shit. One o the clocks
is taken to high altitude by a rocket, whereas a second one
remains on the ground. The clock that is at the higher altitude
will run aster.
The decrease in the requency o a photon as it climbs out
o a gravitational eld can be measured in the laboratory.
The measurements need to be very sensitive, but they have
been successully achieved on many occasions. One o the
experiments to do this was done in 1 960 and is called the
PoundRebka experiment. The requencies o gamma-ray
photons were measured ater they ascended or descended
Jeerson Physical Laboratory Tower at Harvard University.
The original PoundRebka experiment was repeated with
greater accuracy by Pound and Snider.
148
o p t i o n A r e l At i v i t y
Global positioning system
For the global positioning system to be so accurate, general
relativity must be taken into account in calculating the details o
the satellite' s orbit.
HL
crtr o t
effecT Of GRaviTy On spaceTime
The Newtonian way o describing gravity is in terms o the
orces between two masses. In general relativity the way o
thinking about gravity is not to think o it as a orce, but as
changes in the shape (warping) o spacetime. The warping o
spacetime is caused by mass. Think about two travellers who
both set o rom dierent points on the Earths equator and
travel north.
On the surface of the Earth, two travellers
who set o parallel to one another
P
Q
may eventually meet:
P
Q
As they travel north they will get closer and closer together.
They could explain this coming together in terms o a orce
o attraction between them or they could explain it as a
consequence o the surace o the Earth being curved. The
travellers have to move in straight lines across the surace o the
Earth so their paths come together.
Einstein showed how spacetime could be thought o as being
curved by mass. The more matter you have, the more curved
spacetime becomes. Moving objects ollow the curvature o
spacetime or in other words, they take the shortest path in
spacetime. As has been explained, it is very hard to imagine
the our dimensions o spacetime. It is easier to picture what is
going on by representing spacetime as a fat two-dimensional
sheet.
spacetime represented by at sheet
Any mass present warps (or bends) spacetime. The more mass
you have the greater the warping that takes place. This warping
o spacetime can be used to describe the orbit o the Earth
around the Sun. The diagram below represents how Einstein
would explain the situation. The Sun warps spacetime around
itsel. The Earth orbits the Sun because it is travelling along
the shortest possible path in spacetime. This turns out to be a
curved path.
Earth
Sun
Mass tells spacetime how to curve.
Spacetime tells matter how to move.
applicaTiOns Of GeneRal RelaTiviTy TO The univeRse as a whOle
General relativity is now undamental to understanding how the objects in the Universe interact with spacetime and thus how they
aect each other. This allows ar-reaching predictions to be created about the uture development and ate o the Universe see
cosmology sections o the astrophysics option (option D) .
The development o the Universe can be modelled in detail. Many current aspects (e.g. its large-scale structure, the creation o the
elements and the presence o cosmic background radiation) are predicted.
Very large mass black holes may exist at the centres o many galaxies. General relativity predicts how these may interact with matter
and astronomers are searching or appropriate evidence.
General relativity predicts the existence o gravitational waves associated with high energy events such as the collision o two black
holes. Experimental evidence or the existence o these waves is being sought.
o p t i o n A r e l At i v i t y
149
HL
bk o
descRipTiOn
When a star has used up all o its nuclear uel, the orce o
gravity makes it collapse down on itsel (see the astrophysics
option or more details) . The more it contracts the greater the
density o matter and thus the greater the gravitational eld
near the collapsing star. In terms o general relativity, this
would be described in terms o the spacetime near a collapsing
star becoming more and more curved. The curvature o
spacetime becomes more and more severe depending on the
mass o the collapsing star.
At masses greater than this we do not know o any process that
can stop the contraction. Spacetime around the mass becomes
more and more warped until eventually it becomes so great that
it olds in over itsel. What is let is called a black hole. All the
mass is concentrated into a point the singularity.
I the collapsing star is less than about 1 .4 times the mass o the
Sun, then the electrons play an important part in eventually
stopping this contraction. The star that is let is called a white
dwar. I the collapsing star is greater than this, the electrons
cannot halt the contraction. A contracting mass o up to three
times the mass o the Sun can also be stopped this time the
neutrons play an important role and the star that is let is called
a neutron star. The curvature o spacetime near a neutron star is
more extreme than the curvature near a white dwar.
spacetime with
extreme curvature
schwaRzchild Radius
sin gu la rity
The curvature o spacetime near a black hole is so
extreme that nothing, not even light, can escape. Matter
can be attracted into the hole, but nothing can get out
since nothing can travel aster than light. The gravitational
orces are so extreme that light would be severely
defected near a black hole.
e ve n t h o rizo n
p h o to n sp h e re
At a particular distance rom the centre, called the
Schwarzchild radius, we get to a point where the escape
velocity is equal to the speed o light. Newtonian mechanics
predicts that the escape velocity v rom a mass M o radius r is
given by the ormula
_____
v=
photon sphere
black hole
I you were to approach a black hole, the gravitational orces
on you would increase. The rst thing o interest would be
the photon sphere. This consists o a very thin shell o light
photons captured in orbit around the black hole. As we all
urther in, the gravitational orces increase and so the escape
velocity at that distance also increases.
eXample
Calculate the size o a black hole that has the same mass as
our Sun (1 .99 1 0 3 0 kg) .
RS ch
2 6.67 1 0 - 1 1 1 .99 1 0 3 0
= ___
(3 1 0 8 ) 2
= 2949.6 m
= 2.9 km
150
o p t i o n A r e l At i v i t y
2GM
_
r
I the escape velocity is the speed o light, c, then the
Schwarzchild radius would be given by
2GM
RS = _
c2
It turns out that this equation is also correct i we use
the proper equations o general relativity. I we cross the
Schwarzchild radius and get closer to the singularity, we would
no longer be able to communicate with the Universe outside.
For this reason crossing the Schwarzchild radius is sometimes
called crossing the event horizon. An observer watching an
object approaching a black hole would see time slowing down
or the object.
The observed time dilation is worked out rom
t0
______
t = _
RS
1 -_
r
where r is the distance rom the black hole.
ib questons opton a reltvty
1.
2.
In the laboratory rame o reerence, a slow moving alpha
particle travels parallel to stationary metal wire that carries
an electric current. In this rame o reerence, the velocity
o the alpha particle and the drit velocity o the electrons in
the wire are identical. Explain the origin o the orce on the
alpha particle in the rame o reerence o
a) The alpha particle
[2]
b) The laboratory
[2]
and the station platorm. Observer S is standing on the
station platorm midway between the two strikes, while
observer T is sitting in the middle o the train. Light rom
each strike travels to both observers.
0.5 c
Two identical rockets are moving along the same straight
line as viewed rom Earth. Rocket 1 is moving away rom
the Earth at speed 0.80 c relative to the Earth and rocket 2
is moving away rom rocket 1 at speed 0.60 c relative to
rocket 1.
Earth
rocket 1
rocket 2
0.80 c relative
to Earth
0.60 c relative
to rocket 1
a) Calculate the velocity o rocket 2 relative to the Earth,
using the
(i)
Galilean transormation equation.
[1 ]
d) What will be the distance between the scorch marks
on the train, according to T and according to S?
[3]
e) What will be the distance between the scorch marks
on the platform, according to T and according to S?
[2]
[2]
b) Comment on your answers in (a) .
[2]
5.
c) The rest mass o rocket 1 is 1 .0 1 0 3 kg. Determine the
relativistic kinetic energy o rocket 1 , as measured by
an observer on Earth.
[3]
3.
[4]
c) Which strike will T conclude occurred frst?
HL
[1 ]
(ii) relativistic transormation equation.
b) I observer S on the station concludes rom his
observations that the two lightning strikes occurred
simultaneously, explain why observer T on the train
will conclude that they did not occur simultaneously.
In a laboratory experiment two identical particles (P and Q), each
o rest mass m0 , collide. In the laboratory frame of reference,
they are both moving at a velocity o 2/3 c. The situation beore
the collision is shown in the diagram below.
Beore:
The spacetime diagram below shows two events, A and B,
as observed in a reerence rame S. Each event emits a light
signal.
P
Use the diagram to calculate, according to rame S,
a) In the laboratory rame o reerence,
a) The time between event A and event B
[2]
b) The time taken or the light signal leaving event A to
arrive at the position o event B.
[2]
2/3 c
Q
(i)
P (rest)
[4]
Q
b) In Ps rame o reerence,
(i)
5
what is Qs velocity, v?
[3]
(ii) what is the total momentum o P and Q?
[3]
(iii) what is the total energy o P and Q?
[3]
c) As a result o the collision, many particles and photons are
ormed, but the total energy o the particles depends on
the rame o reerence. Do the observers in each rame o
reerence agree or disagree on the number o particles and
photons ormed in the collision? Explain your answer. [2]
B
3
A
6.
1
The concept o gravitational red-shit indicates that clocks run
slower as they approach a black hole.
a) Describe what is meant by
(i)
4.
[3]
velocity = v
ct / ly
6
0
[1 ]
(ii) what is the total energy o P and Q?
d) The velocity o a moving rame o reerence in which
event A and event B occurred simultaneously.
2
what is the total momentum o P and Q?
The same collision can be viewed according to Ps frame
of reference as shown in the diagrams below.
c) The location o a stationary observer who receives the
light signal rom events A simultaneously with
receiving the light signal rom event B.
[2]
4
2/3 c
1
2
3
4
5
6
7
x / ly
Relativity and simultaneity
a) State two postulates o the special theory o relativity.
[2]
Einstein proposed a thought experiment along the
ollowing lines. Imagine a train o proper length 1 00 m
passing through a station at hal the speed o light. There
are two lightning strikes, one at the ront and one at the
rear o the train, leaving scorch marks on both the train
gravitational red-shit.
[2]
(ii) spacetime.
[1 ]
(iii) a black hole with reerence to the concept o
spacetime.
[2]
b) A particular black hole has a Schwarzschild radius R. A
person at a distance o 2R rom the event horizon o the
black hole measures the time between two events to be 1 0 s.
Deduce that or a person a very long way rom the black hole
the time between the events will be measured as 1 2 s.
[1 ]
i B Q u e s t i o n s o p t i o n A r e l At i v i t y
151
14 o p t i o n B e n g i n e e r i n g p h ys i C s
t d
ConCepts
Translational motion
Rotational motion
The complex motion o a rigid body can be analysed as a
combination o two types o motion: translation and rotation.
Both these types o motion are studied separately in this study
guide (pages 9 and 65) .
Every particle in the object
has the same instantaneous
velocity
Every particle in the object
moves in a circle around the
same axis o rotation
Displacement, s, measured
in m
Angular displacement, ,
measured in radians [rad]
Velocity, v, is the rate o
change o displacement
measured in m s - 1
ds
v= _
dt
Angular velocity, , is the
rate o change o angle
measured in rad s - 1
Acceleration, a, is the rate o
change o velocity measured
in m s - 2
dv
a= _
dt
Angular acceleration, , is
the rate o change o angular
velocity measured in rad s - 2
d
=_
dt
mg
A bottle thrown through the air the centre o mass o the
bottle ollows a path as predicted by projectile motion. In
addition the bottle rotates about one (or more) axes.
Translational motion is described using displacements, velocities
and linear accelerations; all these quantities apply to the centre
of mass o the object. Rotational motion is described using
angles (angular displacement), angular velocities and angular
accelerations; all these quantities apply to circular motion about a
given axis o rotation.
The concept o angular velocity, , has already been introduced
with the mechanics o circular motion (see page 66) and is
linked to the requency o rotation by the ollowing ormula:
Comparison o linear and rotational motion
example: BiCyCle wheel
When a bicycle is moving orward at constant velocity v, the
dierent points on the wheel each have dierent velocities.
The motion o the wheel can be analysed as the addition o
the translational and the rotational motion.
a)
Translational motion
The bicycle is moving orward at velocity v so the
wheels centre o mass has orward translational motion
o velocity v. All points on the wheels rim have a
translational component orward at velocity v.
= 2 f
angular velocity
requency
translational component of
velocity
equations of uniform angular aCCeleration
The defnitions o average linear velocity and average linear
acceleration can be rearranged to derive the constant acceleration
equations (page 1 1 ). An equivalent rearrangement derives the
equations o constant angular acceleration.
Translational motion
b) Rotational motion
The wheel is rotating around the central axis o rotation
at a constant angular velocity . All points on the wheels
rim have a tangential component o velocity v (= r)
Rotational motion
Displacement
s
Initial velocity
u
Initial angular velocity
i
Final velocity
v
Final angular velocity
f
Time taken
t
Time taken
t
Acceleration
a
Angular acceleration
Angular displacement
[constant]
tangential
component of
velocity
[constant]
v = u + at
f = i + t
1 at2
s = ut + _
2
1 t2
= it + _
2
v2 = u 2 + 2as
f2 = i2 + 2
(v + u) t
s= _
2
( f + i ) t
= _
2
c)
Combined motion
The motion o the dierent points on the wheels rim is
the vector addition o the above two components:
Point at top of wheel is
moving with instantaneous
velocity of 2, forward
Point in contact with
ground is at rest.
Instantaneous
velocity is zero
152
d
=_
dt
o p t i o n B E n g i n E E r i n g p h ys i c s
Point at side of wheel is moving
with instantaneous velocity of
2, at 45 to the horizontal
t
relationship Between linear and rotational
quantities
When an object is just rotating about a fxed axis, and there is no
additional translational motion o the object, all the individual
particles that make up that object have dierent instantaneous
values o linear displacement, linear velocity and linear
acceleration. They do, however, all share the same instantaneous
values o angular displacement, angular velocity and angular
acceleration. The link between these values involves the distance
rom the axis o rotation to the particle.
instantaneous velocity
V1
particle 1
r2
m1
r1
V2 , instantaneous
m 2 velocity
rigid body
Rotation about
axis. All particles
have same
instantaneous
angular velocity
axis
of rotation
(into the page) particle 2
V1 V2
c)
Accelerations
The total linear acceleration o any particle is made up o
two components:
a) The centripetal acceleration, ar, (towards the axis
o rotation see page 65) , also known as the radial
acceleration.
Tangential velocity
Angular velocity
2
v
a r = r = r 2
_
Centripetal acceleration
Distance rom axis o
rotation to particle
(along the radius)
b) An additional tangential acceleration, at , which results
rom an angular acceleration taking place. I = 0, then
at = 0.
Instantaneous acceleration
(along the tangent)
Angular acceleration
a t = r
a) Displacements
Distance travelled
on circular path
Angular displacement
s = r
Distance rom axis o
rotation to particle
The total acceleration o the particle can be______
ound by vector
addition o these two components: a = r 4 + 2
Distance rom axis o
rotation to particle
b) Instantaneous velocities
Linear instantaneous
velocity (along the
tangent)
Angular velocity
v = r
Distance rom axis o
rotation to particle
o p t i o n B E n g i n E E r i n g p h ys i c s
153
t b
the moment of a forCe: the torque
A particle is in equilibrium i its acceleration is zero. This occurs
when the vector sum o all the external orces acting on the
particle is zero (see page 1 6) . In this situation, all the orces
pass through a single point and sum to zero. The orces on
real objects do not always pass through the same point and can
create a turning eect about a given axis. The turning eect is
called the moment of the force or the torque. The symbol or
torque is the Greek uppercase letter gamma, .
force F
axis of
rotation
r
The moment or torque o a orce, F about an axis is defned as
the product o the orce and the perpendicular distance rom
the axis o rotation to the line o action o the orce.
moment or torque
r
O
perpendicular
distance from O
to line of action of F
orce
line of action of F
= Fr
Note:
The torque and energy are both measured in N m, but only
energy can also be expressed as joules.
perpendicular distance
= Fr sin
The direction o any torque is clockwise or anticlockwise
about the axis o rotation that is being considered. For the
purposes o calculations, this can be treated as a vector
quantity with the direction o the torque vector considered
to be along the axis o rotation. In the example above, the
torque vector is directed into the paper. I the orce F was
applied in the opposite direction, the torque vector would be
directed out o the paper.
Couples
rotational and translational equiliBrium
A couple is a system o orces that has no resultant orce but
which does produce a turning eect. A common example is
a pair o equal but anti-parallel orces acting with dierent
points o application. In this situation, the resultant torque
is the same about all axes drawn perpendicular to the plane
defned by the orces.
I a resultant orce acts on an object then it must accelerate
(page 1 7) . When there is no resultant orce acting on an
object then we know it to be in translational equilibrium
(page 1 6) as this means its acceleration must be zero.
F
x
O
arbitrary
axis
d
Similarly, i there is a resultant torque acting on an object then
it must have an angular acceleration, . Thus an object will
be in rotational equilibrium only i the vector sum o all the
external torques acting on the object is zero.
I an object is not moving and not rotating then it is said to be
in static equilibrium. This must mean that the object is in
both rotational and translational equilibrium.
For rotational equilibrium:
=0=0
F
In 2D problems (in the x-y plane) , it is sufcient to show
that there is no torque about any one axis perpendicular
to the plane being considered (parallel to the z-axis) . In 3D
problems, three axis directions (x, y and z) would need to be
considered.
For translational equilibrium:
a = 0 F= 0
Torque of forces = F(x + d) - Fx
= Fd clockwise
This result is independent of position of axis, O
In 2D problems, it is sufcient to show that there is no
resultant orce in two dierent directions. In 3D problems
three axis directions (x, y and z) would need to be considered.
5N
3N
axis into
page
3m
2m
fN
2m
In the example above, or rotational equilibrium:
f = 2.25 N
154
o p t i o n B E n g i n E E r i n g p h ys i c s
equbu s
Centre of gravity
(a) plank balances if pivot is in middle
centre of gravity
The eect o gravity on all the dierent parts o the
object can be treated as a single orce acting at the
objects centre of gravity.
W
I an object is o uniorm shape and density, the centre
o gravity will be in the middle o the object. I the
object is not uniorm, then fnding its position is not
trivial it is possible or an objects centre o gravity to
be outside the object. Experimentally, i you suspend
an object rom a point and it is ree to move, then the
centre o gravity will always end up below the point o
suspension.
There is no moment about
the centre of gravity.
(b) plank rotates clockwise if pivot is to the left
W
(c) plank rotates anticlockwise if pivot is to the right
W
example 1
All orces at an axis have zero moment about that axis.
10 m
6m
4m
R1
R2
You do not have to choose the pivot as the axis about which
you calculate torques, but it is oten the simplest thing to do
(or the reason above) .
You need to remember the sense (clockwise or anticlockwise) .
Wc , weight of car
Wb , weight of bridge
When a ca r go es a cross a brid ge, the forces ( on the brid ge ) a re
a s shown .
Ta kin g m om en ts abou t righ t-ha n d su pport:
clo ckwise m om en t = a n ticlo ckwise m om en t
( R 1 2 0 m ) = ( W b 10 m ) + ( W c 4 m )
Ta kin g m om en ts abou t left-ha n d su pport:
When solving two-dimensional problems it is sufcient to show
that an object is in rotational equilibrium about any ONE axis.
Newtons laws still apply. Oten an object is in rotational
AND in translational equilibrium. This can provide a simple
way o fnding an unknown orce.
The weight o an object can be considered to be concentrated
at its centre o gravity.
I the problem only involves three non-parallel orces, the
lines o action o all the orces must meet at a single point in
order to be in rotational equilibrium.
( R 2 2 0 m ) = ( W b 10 m ) + ( W c 16 m )
Also, sin ce bridge is n ot a ccelera tin g:
R 1 + R 2 = Wb + Wc
P
R
When solving problems to do with rotational equilibrium
remember:
W
3 orces must meet at a point i in equilibrium
example 2
A ladder o length 5.0 m leans against a smooth wall (no
riction) at an angle o 30 to the vertical.
b) What is the minimum coefcient o static raction
between the ladder and the ground or the ladder to
stay in place?
a) Explain why the ladder can only stay in place i there is
riction between the ground and the ladder.
Rw
P
wall 30
5m
h
Rg
W
60
x
Q
ground
RH
(a) The reaction from the wall,
Rw and the ladders weight Rg
meet at point P. For
equilibrium the force from
the ground, Rg must also
pass through this point
(for zero torque about P) .
Rg is as shown and has
a horizontal component
(i.e. friction must be acting)
Rv
(b) Equilibrium conditions:W = Rv
( )
RH = Rw
( )
moments Rwh = Wx
about Q
1
2
3
Ff s R
R H s R v
using 1 & 2
3
Rw
W
s x = 2.5 cos 60
h
5.0 sin 60
s 0.29
s
o p t i o n B E n g i n E E r i n g p h ys i c s
155
n c
newtons seCond law definition of moment of inertia
F
xed axis of
rotation
angular
acceleration
O
Every particle in the object has the same angular
acceleration, .
tangential
acceleration
at
The moment o inertial, I, o an object about a particular axis is
defned by the summation below:
rigid body
the distance o the particle
rom the axis or rotation
moment o inertia
particle
I = mr2
Newtons second law as applied to one particle in a rigid body
mass o an individual
particle in the object
Newtons second law applies to every particle that makes up a large
object and must also apply i the object is undergoing rotational
motion. In the diagram above, the object is made up o lots o
small particles each with a mass m. F is the tangential component o
the resultant orce that acts on one particle. The other component,
the radial component, cannot produce angular acceleration so it is
not included. For this particle we can apply Newtons second law:
Note that moment o inertia, I, is
A scalar quantity
Measured in kg m2 (not kg m- 2 )
Dependent on:
The mass o the object
F = m at = mr
The way this mass is distributed
so torque = (mr) r = mr
2
The axis o rotation being considered.
Similar equations can be created or all the particles that make
up the object and summed together:
Using this defnition, equation 1 becomes:
= mr2
or ext = mr2
angular acceleration in rad s - 2
resultant external
torque in N m
(1 )
= I
Note that:
Newtons third law applies and, when summing up all the
torques, the internal torques (which result rom the internal
orces between particles) must sum to zero. Only the
external torques are let.
moment o inertia in kg m2
This is Newtons second law or rotational motion and can be
compared to F = ma
moments of inertia for different oBjeCts
Equations or moments o inertia in dierent situations do not need to be memorized.
Object
Axis of
rotation
moment of
inertia
through centre,
perpendicular to
plane
mr
Obj ect
thin ring (simple wheel)
r
m
Axis of
rotation
moment of
inertia
through centre
2 mr2
_
5
Through the
centre of mass,
perpendicular to
the plane of the
lamina
l2 + h 2
m _
12
Sphere
2
thin ring
r
through a
diameter
1 mr2
_
2
through centre,
perpendicular to
plane
1 mr2
_
2
m
disc and cylinder (solid ywheel)
r
m
thin rod, length d
m
through centre,
perpendicular
to rod
Rectangular lamina
l
1 md2
_
12
h
d
example
i. what is the angular velocity o the wheel?
A torque o 30 N m acts on a wheel with moment o inertia
600 kg m2 . The wheel starts o at rest.
ii. how ast is a point on the rim moving?
a)
What angular acceleration is produced?
b) The wheel has a radius o 40 cm. Ater 1 .5 minutes:
= _
30 = 5.0 1 0 - 2 rad s - 2
a) = I = _
I
600
b) i. = t = 5.0 1 0 - 2 90 = 4.5 rad s - 1
ii. v = r = 0.4 4.5 = 1 .8 m s - 1
156
o p t i o n B E n g i n E E r i n g p h ys i c s
(
)
r dc
energy of rotational motion
Conservation of angular momentum
Energy considerations oten provide simple solutions to
complicated problems. When a torque acts on an object, work is
done. In the absence o any resistive torque, the work done on
the object will be stored as rotational kinetic energy.
In exactly the same way that Newtons laws can be applied to
linear motion to derive:
F
the concept o the impulse o a orce
the relationship between impulse and change in
momentum
the law o conservation o linear momentum,
F
r
then Newtons laws can be applied to angular situations
to derive:
P
axis of rotation
Calculation o work done by a torque
The concept o the angular impulse:
Angular impulse is the product o torque and the time or
which the torque acts:
angular impulse = t
In the situation above, a orce F is applied and the object
rotates. As a result, an angular displacement o occurs. The
work done, W, is calculated as shown below:
W = F (distance along arc) = F r =
Using = I we know that W = I
We can apply the constant angular acceleration equation to
substitute or :
= i + 2
2
f
2
(
)
f2
i2
1 I 2 - _
1 I 2
W = I _ - ___
=_
2
2
2 f
2 i
This means that we have an equation or rotational KE:
1 I 2
EK = _
2
Work done by the torque acting on object = change in
rotational KE o object
ro t
The total KE is equal to the sum o translational KE and the
rotational KE:
Total KE = translational KE + rotational KE
1 Mv2 + _
1 I2
Total KE = _
2
2
angular momentum
For a single particle
The linear momentum, p, o a particle o mass m which has a
tangential speed v is m v.
I the torque varies with time then the total angular
impulse given to an object can be estimated rom the area
under the graph showing the variation o torque with
time. This is analogous to estimating the total impulse
given to an object as a result o a varying orce (see
page 23) .
The relationship between angular impulse and change in
angular momentum:
angular impulse applied to an object = change o
angular momentum experienced by the object
The law o conservation o angular momentum.
The total angular momentum o a system remains constant
provided no resultant external torque acts.
Examples:
a) A skater who is spinning on a vertical axis down their
body can reduce their moment o inertia by drawing in
their arms. This allows their mass to be redistributed so
that the mass o the arms is no longer at a signifcant
distance rom the axis o rotation thus reducing mr2 .
Extended arms mean
larger radius and smaller
velocity of rotation.
Bringing in her arms
decreases her moment
of inertia and therefore
increases her rotational
velocity.
The angular momentum, L, is defned as the moment o the
linear momentum about the axis o rotation
Angular momentum, L = (mv) r = (mr) r = (mr2 )
For a larger obj ect
The angular momentum L o an object about an axis o
rotation is defned as
Angular momentum, L = (mr2 )
L = I
Note that total angular momentum, L, is:
a vector (in the same way that a torque is considered to be
a vector or calculations)
measured in kg m2 s - 1 or N m s
dependent on all rotations taking place. For example, the
total angular momentum o a planet orbiting a star would
involve:
the spinning o the planet about an axis through the
planets centre o mass and
the orbital angular momentum about an axis through
the star.
b) The EarthMoon system produces tides in the oceans. As
a result o the relative movement o water, riction exists
between the oceans and Earth. This provides a torque that
acts to reduce the Earths spin on its own axis and thus
reduces the Earths angular momentum. The conservation
o angular momentum means that there must be a
corresponding increase in the orbital angular momentum
o the EarthMoon system. As a result, the EarthMoon
separation is slowly increasing.
o p t i o n B E n g i n E E r i n g p h ys i c s
157
s b
summary Comparison of equations of linear and rotational motion
Every equation for linear motion has a corresponding angular equivalent:
Linear motion
Rotational motion
Physics principles
A resultant external force on a point object
causes acceleration. The value of the
acceleration is determined by the mass and
the resultant force.
A resultant external torque on an extended object
causes rotational acceleration. The value of the
angular acceleration is determined by the moment
of inertia and the resultant torque.
Newtons second law
F= ma
= I
Work done
W= F s
W=
Kinetic energy
1 m v2
EK = _
2
1 I 2
EK = _
2
P= Fv
P=
Power
ro t
Momentum
p=mv
L = I
Conservation of momentum
The total linear momentum of a system
remains constant provided no resultant
external force acts.
The total angular momentum of a system remains
constant provided no resultant external torque
acts.
Symbols used
Resultant force
F
Resultant torque
Mass
m
Moment of inertia
I
Acceleration
a
Angular acceleration
Displacement
s
Angular displacement
Velocity
v
Angular velocity
Linear momentum
p
Angular momentum
L
proBlem solving and graphiCal work
example
When analysing any rotational situation, the simplest
approach is to imagine the equivalent linear situation and use
the appropriate equivalent relationships.
A solid cylinder, initially at rest, rolls down a 2.0 m long slope
of angle 30 as shown in the diagram below:
a) Graph of angular displacement vs time
2.0 m
This graph is equivalent to a graph of linear displacement vs
time. In the linear situation, the area under the graph does
not represent any useful quantity and the gradient of the line
at any instant is equal to the instantaneous velocity (see page
1 0). Thus the gradient of an angular displacement vs
time graph gives the instantaneous angular velocity.
b) Graph of angular velocity vs time
This graph is equivalent to a graph of linear velocity vs
time. In the linear situation, the area under the graph
represents the distance gone and the gradient of the line at
any instant is equal to the instantaneous acceleration (see
page 1 0) . Thus the area under an angular velocity vs
time graph gives the total angular displacement and
the gradient of an angular velocity vs time graph
gives the instantaneous angular acceleration.
c)
Graph of torque vs time
This graph is equivalent to a graph of force vs time. In
the linear situation, the area under the graph represents
the total impulse given to the object which is equal to the
change of momentum of the object (see page 23) . Thus
the area under the torque vs time graph represents
the total angular impulse given to the object which
is equal to the change of angular momentum.
30
The mass of the cylinder is m and the radius of the cylinder is R.
Calculate the velocity of the cylinder at the bottom of the slope.
Answer:
Vertical height fallen by cylinder = 2.0 sin30 = 1 .0 m
PE lost = mgh
1 mv2 + _
1 I2
KE gained = _
2
2
1 mR2
but I = _
(cylinder) see page 1 56
2
v
and = _
R
1 mv2 + _
mR2 _
v2
1 _
KE gained = _
2
2 2
R2
1 mv2 + _
1 mv2
= _
2
4
3 mv2
= _
4
Conservation of energy
3 mv2
mgh = _
4 ____
gh
v = 4_
3
___________
=
4 9.8 1 .0
__
3
= 3.61 m s - 1
158
o p t i o n B E n g i n E E r i n g p h ys i c s
thrmyamc ym a ccp
definitions
Historically, the study of the behaviour of ideal gases led to some very fundamental concepts that are applicable to many other
situations. These laws, otherwise known as the laws of thermodynamics, provide the modern physicist with a set of very
powerful intellectual tools.
The terms used need to be explained.
Thermodynamic
system
Most of the time when studying the behaviour of an ideal gas in particular situations, we focus on the
macroscopic behaviour of the gas as a whole. In terms of work and energy, the gas can gain or lose
thermal energy and it can do work or work can be done on it. In this context, the gas can be seen as a
thermodynamic system.
The
surroundings
If we are focusing our study on the behaviour of an ideal gas, then everything else can be called
its surroundings. For example the expansion of a gas means that work is done by the gas on the
surroundings (see below) .
Heat Q
In this context heat refers to the transfer
of a quantity of thermal energy between
the system and its surroundings.
This transfer must be as a result of a
temperature difference.
HOT
HOT
thermal
energy
ow
thermal energy ow
COLD
HOT
Work W
In this context, work refers to the macroscopic transfer of energy. For example
1. work done = force distance
2. work done = potential difference current time
F
heater
F
compression
When a gas is compressed, work is done on the gas
power supply
When a gas is compressed, the surroundings do
work on it. When a gas expands it does work on
the surroundings.
Internal energy
U (U = change
in internal
energy)
This is just another
example of work being
done on the gas.
The internal energy can be thought of as the energy held within a system. It is the sum of the PE due to
the intermolecular forces and the kinetic energy due to the random motion of the molecules. See
page 26.
This is different to the total energy of
the system, which would also include
the overall motion of the system and
any PE due to external forces.
In thermodynamics, it is the changes
in internal energy that are being
considered. If the internal energy of
a gas is increased, then its temperature
must increase. A change of phase
(e.g. liquid gas) also involves a
change of internal energy.
Internal energy
of an ideal
monatomic gas
thermal energy ow
system
with
internal
energy U
v
velocity (system also has kinetic energy)
h
height (system also has
gravitational potential energy)
The total energy of a system is not the same as
its internal energy
The internal energy of an ideal gas depends only on temperature. When the temperature of an ideal
gas changes from T to (T + T) its internal energy changes from U to (U + U) . The same U always
produces the same T. Since the temperature is related to the average kinetic energy per molecule (see
__
R
page 30) , EK = __32 kB T = __32 __
T, the internal energy U, is the sum of the total random kinetic energies of the
N
molecules:
A
__
3 nRT
U = nNA EK = _
2
[n = number of moles; NA = Avogadros constant]
o p t i o n B E n g i n E E r i n g p h ys i c s
159
w by l
work done during expansion
at Constant pressure
Work done W = force distance
= Fx
F
Whenever a gas expands, it is doing
work on its surroundings. If the pressure
of the gas is changing all the time, then
calculating the amount of work done
is complex. This is because we cannot
assume a constant force in the equation
of work done (work done = force
distance) . If the pressure changes
then the force must also change. If the
pressure is constant then the force is
constant and we can calculate the
work done.
force
Since pressure = _
area
F = pA
constant
pressure p
therefore
x
so work done = pV
F
p V diagrams and work done
pressure p
It is often useful to represent the changes that happen to a gas during a
thermodynamic process on a pV diagram. An important reason for choosing to do this
is that the area under the graph represents the work done. The reasons for this are
shown below.
A
p
B
area of strip
= pV
= work done
in expansion
area under graph
= work done in expanding
from state A to state B
V
volume V
A
B
pressure p
pressure p
This turns out to be generally true for any thermodynamic process.
work done by gas
expanding from
state A to state B
to state C
A
work done by
atmosphere as
gas contracts
from state C to
state D to state A
D
C
C
volume V
160
W = pAx
but Ax = V
volume V
o p t i o n B E n g i n E E r i n g p h ys i c s
So if a gas increases its volume (V
is positive) then the gas does work (W is
positive)
t f c
irst law o thermodynamiCs
There are three undamental laws o thermodynamics. The
frst law is simply a statement o the principle o energy
conservation as applied to the system. I an amount o thermal
energy Q is given to a system, then one o two things must
happen (or a combination o both) . The system can increase its
internal energy U or it can do work W.
Q
I this is positive, then thermal energy is going into the
system.
I it is negative, then thermal energy is going out o the
system.
U
I this is positive, then the internal energy o the system
is increasing. (The temperature o the gas is increasing.)
I it is negative, the internal energy o the system is
decreasing.(The temperature o the gas is decreasing.)
W
I this is positive, then the system is doing work on
the surroundings.(The gas is expanding.)
I it is negative, the surroundings are doing work on
the system. (The gas is contracting.)
As energy is conserved
Q = U + W
It is important to remember what the signs o these symbols
mean. They are all taken rom the systems point o view.
ideal gas proCesses
A gas can undergo any number o dierent types o change or process. Four important processes are considered below. In each case
the changes can be represented on a pressurevolume diagram and the frst law o thermodynamics must apply. To be precise, these
diagrams represent a type o process called a reversible process.
pressure p
In an isochoric
process, also called an
isovolumetric process,
the gas has a constant
volume. The diagram
below shows an
isochoric decrease in
pressure.
Isobaric
3.
Isothermal
4.
In an isobaric process
the gas has a constant
pressure. The diagram
below shows an isobaric
expansion.
In an isothermal process
the gas has a constant
temperature. The
diagram below shows an
isothermal expansion.
A
B
A
B
A
volume V
Isobaric change
B
volume V
Isochoric (volumetric)
change
V = constant, or
p
_
= constant
T
Q negative
volume V
Isothermal change
p = constant, or
V = constant
_
T
Q positive
T = constant, or
pV = constant
Q positive
U positive (T )
U zero
W positive
W positive
Adiabatic
In an adiabatic process
there is no thermal
energy transer between
the gas and the
surroundings. This means
that i the gas does work it
must result in a decrease
in internal energy. A rapid
compression or expansion
is approximately
adiabatic. This is because
done quickly there is not
sufcient time or thermal
energy to be exchanged
with the surroundings.
The diagram below shows
an adiabatic expansion.
pressure p
2.
pressure p
Isochoric
(isovolumetric)
pressure p
1.
A
B
U negative (T)
volume V
W zero
Adiabatic change
Q zero
example
A monatomic gas doubles its volume as a result o an
adiabatic expansion. What is the change in pressure?
5
_
5
_
p 1 V1 3 = p 2 V2 3
( )
p2
V1
_
_
p 1 = V2
5
_
3
U negative (T)
W positive
For a monatomic gas, the
equation or an adiabatic
process is
5
_
pV3 = constant
5
_
= 0.5 3
= 0.31
fnal pressure = 31 % o initial pressure
o p t i o n B E n g i n E E r i n g p h ys i c s
161
sc c
seCond law of thermodynamiCs
entropy and energy degradation
Historically the second law o thermodynamics has been
stated in many dierent ways. All o these versions can be
shown to be equivalent to one another.
Entropy is a property that expresses the disorder in the
system.
The details are not important but the entropy S o a system
is linked to the number o possible arrangements W o the
system. [S = kB ln(W) ]
In principle there is nothing to stop the complete conversion
o thermal energy into useul work. In practice, a gas can not
continue to expand orever the apparatus sets a physical
limit. Thus the continuous conversion o thermal energy
into work requires a cyclical process a heat engine.
Because molecules are in random motion, one would expect
roughly equal numbers o gas molecules in each side o a
container.
W
Thot
Q hot
Q cold
An arrangement
like this
is much
more likely
than one like
this.
Tcold
The number o ways o arranging the molecules to get the
set-up on the right is greater than the number o ways o
arranging the molecules to get the set-up on the let. This
means that the entropy o the system on the right is greater
than the entropy o the system on the let.
Carnot showed In other words there
that Q hot > W. must be thermal energy
wasted to the cold reservoir.
This realization leads to possibly the simplest ormulation
o the second law o thermodynamics (the KelvinPlanck
ormulation) .
In any random process the amount o disorder will tend
to increase. In other words, the total entropy will always
increase. The entropy change S is linked to the thermal
Q
)
energy change Q and the temperature T. (S = ___
T
No heat engine, operating in a cycle, can take in heat
rom its surroundings and totally convert it into work.
Other possible ormulations include the ollowing:
thermal energy ow
No heat pump can transer thermal energy rom a
low-temperature reservoir to a high-temperature
reservoir without work being done on it (Clausius) .
Q
Thot
Heat fows rom hot objects to cold objects.
The concept o entropy leads to one nal version o the
second law.
decrease of entropy =
The entropy o the Universe can never decrease.
examples
3.
source of work
is the electric
energy supply
A rerigerator
It should be possible to design a
theoretical system or propelling a
boat based around a heat engine. The
atmosphere could be used as the hot
reservoir and cold water rom the sea
could be used as the cold reservoir.
The movement o the boat through
the water would be the work done.
This is possible BUT it cannot continue
to work or ever. The sea would be
warmed and the atmosphere would
be cooled and eventually there would
be no temperature dierence.
162
Q
Tcold
In many situations the idea o energy degradation is a useul
concept. The more energy is shared out, the more degraded
it becomes it is harder to put it to use. For example, the
internal energy that is locked up in oil can be released when
the oil is burned. In the end, all the energy released will be in
the orm o thermal energy shared among many molecules.
It is not easible to get it back.
A rerigerator is an example o a heat pump.
thermal energy taken from
ice box and ejected to
surroundings
2.
increase of entropy =
When thermal energy fows rom a hot object to a colder
object, overall the total entropy has increased.
The rst and second laws o thermodynamics both must apply
to all situations. Local decreases o entropy are possible so
long as elsewhere there is a corresponding increase.
1.
Q
Thot
Tcold
Water reezes at 0 C because this is the temperature at
which the entropy increase o the surroundings ( when
receiving the latent heat) equals the entropy decrease o
the water molecules becoming more ordered. It would not
reeze at a higher temperature because this would mean
that the overall entropy o the system would decrease.
increasing temperature of surroundings
-2 C
ICE
since
entropy
entropy
decrease
increase
of ice <
of
formation surroundings
o p t i o n B E n g i n E E r i n g p h ys i c s
0 C
ICE/WATER
MIX
since
entropy
entropy
decrease
increase
of ice =
of
formation surroundings
2 C
WATER
since
entropy
entropy
decrease
increase
of ice >
of
formation surroundings
h
work
done W
HOT
reservoir
Thot
thermal
energy
Q hot
ENGINE
COLD
reservoir
thermal
energy
Q cold
In order to do this, some thermal energy must have been
taken rom a hot reservoir (during the isovolumetric increase
in pressure and the isobaric expansion) . A dierent amount
o thermal energy must have been ejected to a cold reservoir
(during the isovolumetric decrease in pressure and the isobaric
compression) .
pressure p
heat engines
A central concept in the study o thermodynamics is the heat
engine. A heat engine is any device that uses a source o
thermal energy in order to do work. It converts heat into work.
The internal combustion engine in a car and the turbines that
are used to generate electrical energy in a power station are
both examples o heat engines. A block diagram representing a
generalized heat engine is shown below.
A
isobaric expansion
B
isovolumetric
increase in
pressure
total work
done by
the gas
isovolumetric
decrease in
pressure
Tcold
C isobaric compression D
volume V
The thermal efciency o a heat engine is defned as
Heat engine
In this context, the word reservoir is used to imply a constant
temperature source (or sink) o thermal energy. Thermal energy can
be taken rom the hot reservoir without causing the temperature o
the hot reservoir to change. Similarly thermal energy can be given to
the cold reservoir without increasing its temperature.
An ideal gas can be used as a heat engine. The pV diagram right
represents a simple example. The our-stage cycle returns the gas
to its starting conditions, but the gas has done work. The area
enclosed by the cycle represents the amount o work done.
work done
= ____
(thermal energy taken rom hot reservoir)
This is equivalent to
rate o doing work
= ____
(thermal power taken rom hot reservoir)
useul work done
= __
energy input
The cycle o changes that results in a heat engine with the
maximum possible efciency is called the Carnot cycle.
Carnot CyCles and Carnot theorem
The Carnot cycle represents the cycle o processes or a
theoretical heat engine with the maximum possible efciency.
Such an idealized engine is called a Carnot engine.
input
work W
HOT
reservoir
COLD
reservoir
pressure p
heat pumps
A heat pump is a heat engine being run in reverse. A
heat pump causes thermal energy to be moved rom a cold
reservoir to a hot reservoir. In order or this to be achieved,
mechanical work must be done.
A
Q hot
thermal energy taken in
B
HEAT
PUMP
Thot
thermal
energy
Q hot
Tcold
D
thermal energy
given out
Q cold
thermal
energy
Q cold
area = work done
by gas during
Carnot cycle
C
volume V
Carnot cycle
Once again an ideal gas can be used as a heat pump. The
thermodynamic processes can be exactly the same ones as were
used in the heat engine, but the processes are all opposite. This
time an anticlockwise circuit will represent the cycle o processes.
It consists o an ideal gas undergoing the ollowing processes.
pressure p
Heat pump
isobaric compression
A
D
isovolumetric
decrease in
pressure
Isothermal expansion (A B)
Adiabatic expansion (B C)
Isothermal compression (C D)
total work
done on
the gas
isovolumetric
increase in
pressure
Adiabatic compression (D A)
The temperatures o the hot and cold reservoirs fx the
maximum possible efciency that can be achieved.
The efciency o a Carnot engine can be shown to be
Tco ld
C a rn o t = 1 - _
(where T is in kelvin)
Th o t
An engine operates at 300 C and ejects heat to the surroundings
at 20 C. The maximum possible theoretical efciency is
B isobaric expansion C
volume V
293 = 0.49 = 49%
C a rn o t = 1 - _
573
o p t i o n B E n g i n E E r i n g p h ys i c s
163
HL
f
definitions of density and pressure
average density
m
=_
V
normal orce
pressure
F
p = _
A
The symbol representing density is the Greek letter rho, . The
average density o a substance is dened by the ollowing equation:
area
mass
Pressure is a scalar quantity the orce has a direction but
the pressure does not. Pressure acts equally in all directions.
volume
Density is a scalar quantity.
The SI unit o pressure is N m- 2 or pascals (Pa). 1 Pa = 1 N m- 2
The SI units o density are kg m- 3 .
Atmospheric pressure 1 0 5 Pa
Densities can also be quoted in g cm- 3 (see conversion actor
below)
Absolute pressure is the actual pressure at a point in a
fuid. Pressure gauges oten record the difference between
absolute pressure and atmospheric pressure. Thus i a
dierence pressure gauge gives a reading o 2 1 0 5 Pa or a
gas, the absolute pressure o the gas is 3 1 0 5 Pa.
The density o water is 1 g cm- 3 = 1 ,000 kg m- 3
Pressure at any point in a fuid (a gas or a liquid) is dened
in terms o the orce, F, that acts normally (at 90) to a
small area, A, that contains the point.
variation of fluid pressure
BuoyanCy and arChimedes prinCiple
The pressure in a fuid increases with depth. I two points are
separated by a vertical distance, d, in a fuid o constant density,
f, then the pressure dierence, p, between these two points is:
Archimedes principle states that when a body is immersed
in a fuid, it experiences a buoyancy upthrust equal in
magnitude to the weight o the fuid displaced. B = fVf g
density o fuid
gravitational eld strength
p = fgd
pressure dierence due to depth
22N
depth
17N
The total pressure at a given depth in a liquid is the addition
o the pressure acting at the surace (atmospheric pressure)
and the additional pressure due to the depth:
Atmospheric pressure
P = P0 + fgd
gravitational eld strength
W
W
(a)
Note that:
Pressure can be expressed in terms o the equivalent
depth (or head) in a known liquid. Atmospheric pressure
is approximately the same as exerted by a 760 mm high
column o mercury (Hg) or a 1 0 m column o water.
volume of
uid displaced
(w = 5N)
AB
B2
atmospheric pressure
the water column exerts
a pressure at B equal to
the excess pressure of
the gas supply: P = hg
W
volume of
uid displaced
(w = 10N)
A consequence o this
principle is that a foating
object displaces its own
weight o fuid.
As pressure is dependent on depth, the pressures at two
points that are at the same horizontal level in the same
liquid must be the same provided they are connected by
that liquid and the liquid is static.
h
density
of uid
B1
density o fuid
depth
Total pressure
excess gas
pressure P
12N
weight of uid displaced
= total weight of duck
pasCals prinCiple
Pascals principle states that the pressure applied to an
enclosed liquid is transmitted to every part o the liquid,
whatever the shape it takes. This principle is central to the
design o many hydraulic systems and is dierent to how
solids respond to orces.
The pressure is independent o the cross-sectional area
this means that liquids will always nd their own level.
When a solid object (e.g. an incompressible stick) is pushed at
one end and its other end is held in place, then the same orce
will be exerted on the restraining object.
hydrostatiC equiliBrium
Incompressible solids transmit orces whereas incompressible
liquids transmit pressures.
A fuid is in hydrostatic equilibrium when it is at rest. This
happens when all the orces on a given volume o fuid are
balanced. Typically external orces (e.g. gravity) are balanced
by a pressure gradient across the volume o fuid (pressure
increases with depth see above) .
downward force due to
pressure from uid above
volume of uid
weight of uid
W
contained in volume
164
upward force due to
pressure from uid below
o p t i o n B E n g i n E E r i n g p h ys i c s
A2
A1
load platform
load = F
applied force F
(eort)
piston of
area A 1
piston of area A 2
hydraulic liquid
HL
B fc
the ideal luid
In most real situations, fuid fow is extremely complicated. The
ollowing properties dene an ideal fuid that can be used to
create a simple model. This simple model can be later rened to
be more realistic.
Is non-viscous as a result o fuid fow, no energy gets
converted into thermal energy. See page 1 67 or the denition
o the viscosity o a real fuid.
An ideal fuid:
Involves a steady fow (as opposed to a turbulent, or chaotic,
fow) o fuid. Under these conditions the fow is laminar (see
box below). See page 1 67 or an analysis o turbulent fow.
Is incompressible thus its density will be constant.
Does not have angular momentum it does not rotate.
laminar low, streamlines and the
Continuity equation
When the fow o a liquid is steady or laminar, dierent parts
o the fuid can have dierent instantaneous velocities. The
fow is said to be laminar i every particle that passes through
a given point has the same velocity whenever the observation
is made. The opposite o laminar fow, turbulent fow, takes
place when the particles that pass through a given point have a
wide variation o velocities depending on the instant when the
observation is made (see page 1 67) .
A streamline is the path taken by a particle in the fuid and
laminar fow means that all particles that pass through a
given point in the fuid must ollow the same streamline. The
direction o the tangent to a streamline gives the direction o
the instantaneous velocity that the particles o the fuid have at
that point. No fuid ever crosses a streamline. Thus a collection
o streamlines can together dene a tube o fow. This is
tubular region o fuid where fuid only enters and leaves the
tube through its ends and never through its sides.
speed 2
speed 1
area A 1
density 1
boundary
(streamlines)
area A 2
density 2
In a time t, the mass, m 1 , entering the cross-section A 1 is
m1 = 1 A 1 v1 t
Similarly the mass, m 2 , leaving the cross-section A 2 is
m2 = 2 A 2 v2 t
Conservation o mass applies to this tube o fow, so
1 A 1 v1 = 2 A 2 v2
This is an ideal fuid and thus incompressible meaning 1 = 2 , so
A1 v1 = A 2 v2 or Av = constant
This is the continuity equation.
the Bernoulli eeCt
the Bernoulli equation
When a fuid fows into a narrow section o a pipe:
The Bernoulli equation results rom a consideration o the
work done and the conservation o energy when an ideal fuid
changes:
The fuid must end up moving at a higher speed (continuity
equation).
This means the fuid must have been accelerated
orwards.
its speed (as a result o a change in cross-sectional area)
its vertical height as a result o work done by the fuid pressure.
The equation identies a quantity that is always constant
along any given streamline:
density
o fuid
higher pressure
lower speed
lower pressure
higher speed
higher pressure
lower speed
This means there must be a pressure dierence orwards with
a lower pressure in the narrow section and a higher pressure
in the wider section.
Thus an increase in fuid speed must be associated with a
decrease in fuid pressure. This is the Bernoulli eect the
greater the speed, the lower the pressure and vice versa.
speed o fuid
particles
vertical height
fuid pressure
1 v2 + gz + p = constant
2 density
gravitational
_
o fuid
eld strength
Note that:
1 2
The rst term (__
v ), can be thought o as the dynamic pressure.
2
The last two terms (gz + p) , can be thought o as the static
pressure.
Each term in the equation has several possible units:
N m- 2 , Pa, J m- 3 .
The last o the above units leads to a new interpretation or
the Bernoulli equation:
KE
gravitational PE
per unit +
per unit
+ pressure = constant
volume
volume
o p t i o n B E n g i n E E r i n g p h ys i c s
165
B xm
HL
appliCations of the Bernoulli equation
a) Flow out o a container
d) Pitot tube to determine the speed o a plane
A
liquid
density
A pitot tube is attached acing orward on a plane. It has
two separate tubes:
streamline
direction
of airow
h
small static
pressure openings
B
arbitrary zero
impact
opening
To calculate the speed o fuid fowing out o a container, we
can apply Bernoullis equation to the streamline shown above.
total
pressure
tube
At A, p = atmospheric and v = zero
At B, p = atmospheric and v = ?
1 v2 + gz + p = constant
_
2
1 v2 + 0 + p
0 + hg + p = _
2
___
v = 2gh
b) Venturi tubes
A Venturi meter allows the rate o fow o a fuid to be
calculated rom a measurement o pressure dierence
between two dierent cross-sectional areas o a pipe.
to metal end
area A
constriction
of area a
A
The ront hole (impact opening) is placed in the
airstream and measures the total pressure (sometimes
called the stagnation pressure) , PT.
The side hole(s) measures the static pressure, Ps .
The dierence between PT and Ps , is the dynamic
pressure. The Bernoulli equation can be used to calculate
airspeed:
1 v2
PT - Ps = _
2
________
v=
2(P - P )
_
T
s
e) Aerooil (aka airoil)
dynamic lift F
B
pressure P1
ow of (e.g.) water,
density 1
h
1
manometer liquid
(e.g. mercury) ,
density 2
aerofoil
2
The pressure dierence between A and B can be
calculated by taking readings o h and 2 rom the
attached manometer:
air ow
pressure P2
PA - PB = h2 g
This value and measurements o A, a and 1 allows the
fuid speed at A to be calculated by using Bernoullis
equation and the equation o continuity
v=
c)
[ ( )
2h2 g
________
A 2 - 1
1 __
a
static
pressure
tube
]
Note that:
Streamlines closer together above the aerooil imply a
decrease in cross-sectional area o equivalent tubes o fow
above the aerooil.
The rate o fow o fuid through the pipe is equal to A v
Decrease in cross-sectional area o tube o fow implies
increased velocity o fow above the aerooil (equation o
continuity) . v1 > v2
Fragrance spray
Since v1 > v2 , P1 < P2
b. Constriction in tube causes low pressure
region as air travels faster in this section
below-pressure zone
squeezebulb
c. Liquid is drawn up tube
by pressure dierence
and forms little droplets
a. Squeezing
as it enters the air jet
bulb
d.
Fine
spray of fragrance
forces air
is emitted from nozzle
through
tube
166
o p t i o n B E n g i n E E r i n g p h ys i c s
Bernoulli equation can be used to calculate the pressure
dierent (height dierence not relevant) which can support
the weight o the aeroplane.
When angle o attack is too great, the fow over the upper
surace can become turbulent. This reduces the pressure
dierence and leads to the plane stalling.
HL
vc
definition of visCosity
A) Tangential stress
An ideal fuid does not resist the relative motion between
dierent layers o fuid. As a result there is no conversion o work
into thermal energy during laminar fow and no external orces
are needed to maintain a steady rate o fow. Ideal fuids are nonviscous whereas real fuids are viscous. In a viscous fuid, a steady
external orce is needed to maintain a steady rate o fow (no
acceleration) . Viscosity is an internal riction between dierent
layers o a fuid which are moving with dierent velocities.
The denition o the viscosity o a fuid, , (Greek letter Nu) is
in terms o two new quantities, the tangential stress, , and
v
the velocity gradient, ___
(see RH side).
y
The coecient o viscosity is dened as:
relative
velocity v
area of contact A
retarding force
-F
accelerating force
F
The tangential stress is dened as:
F
=_
A
Units o tangential stress are N m- 2 or Pa
B) Velocity gradient
y
velocity
tangential stress
FA
= __ = _
velocity gradient
vy
(v + v)
v
y
The units o are N s m- 2 or kg m- 1 s - 1 or Pa s
v
Typical values at room temperature:
Water: 1 .0 1 0 - 3 Pa s
Thick syrup: 1 .0 1 0 2 Pa s
Viscosity is very sensitive to changes o temperature.
For a class o fuid, called Newtonian fuids, experimental
measurements show that tangential stress is proportional to velocity
gradient (e.g. many pure liquids). For these fuids the coecient o
viscosity is constant provided external conditions remain constant.
stokes law
Stokes law predicts the viscous drag orce FD that acts on a
perect sphere when it moves through a fuid:
sphere has
uniform
velocity
v
-F
equal opposing
viscous drag
Drag orce acting on sphere in N
v
velocity gradient = _
y
Units o velocity gradient are s- 1
The fuid is innite in volume. Real spheres alling through
columns o fuid can be aected by the proximity o the
walls o the container.
The size o the particles o the fuid is very much smaller
than the size o the sphere.
uid at this point moves
with body (boundary layer)
F
r
driving
force
innite expanse
of uid
The velocity gradient is dened as:
The orces on a sphere alling through a fuid at terminal
velocity are as shown below:
uid upthrust
sphere
velocity v
sphere
density
uid
density
W
velocity o sphere in m s - 1
At terminal velocity vt ,
Note Stokes law assumes that:
W = U + FD
The speed o the sphere is small so that:
FD = U - W
the fow o fuid past the sphere is streamlined
there is no slipping between the fuid and the sphere
4 r3 ( - ) g
6rvt = _
3
2
( - ) g
2r
vt = __
9
turBulent flow the reynolds numBer
Streamline fow only occurs at low fuid fow rates. At high
fow rates the fow becomes turbulent:
laminar
viscous drag
r
pull of
Earth
viscosity o fuid in Pa s
FD = 6 rv
radius o sphere in m
U FD
Reynolds
number
turbulent
It is extremely dicult to predict the exact conditions when
fuid fow becomes turbulent. When considering fuid fow
down a pipe, a useul number to consider is the Reynolds
number, R, which is dened as:
speed o
bulk fow
vr
R= _
radius o pipe
density o fuid
viscosity o fuid
Note that:
The Reynolds number does not have any units it is just a ratio.
Experimentally, fuid fow is oten laminar when R < 1 000 and
turbulent when R > 2000 but precise predictions are dicult.
o p t i o n B E n g i n E E r i n g p h ys i c s
167
fc c c (1)
damping
displacement, x
Damping involves a rictional orce that is always in the
opposite direction to the direction o motion o an oscillating
particle. As the particle oscillates, it does work against this
resistive (or dissipative) orce and so the particle loses
energy. As the total energy o the particle is proportional to the
(amplitude) 2 o the SHM, the amplitude decreases exponentially
with time.
exponential envelope
2
4
time, t
Heavy damping or overdamping involves large resistive
orces (e.g. the SHM taking place in a viscous liquid) and can
completely prevent the oscillations rom taking place. The time
taken or the particle to return to zero displacement can again
be long.
Critical damping involves an intermediate value or resistive
orce such that the time taken or the particle to return to zero
displacement is a minimum. Eectively there is no overshoot.
Examples o critically damped systems include electric meters
with moving pointers and door closing mechanisms.
displacement
HL
overdamped
critical
damping
time
0.2
The above example shows the eect o light damping (the
system is said to be underdamped) where the resistive orce
is small so a small raction o the total energy is removed each
cycle. The time period o the oscillations is not aected and the
oscillations continue or a signifcant number o cycles. The time
taken or the oscillations to die out can be long.
I a system is temporarily displaced rom its equilibrium
position, the system will oscillate as a result. This oscillation will
be at the natural frequency of vibration o the system. For
example, i you tap the rim o a wine glass with a knie, it will
oscillate and you can hear a note or a short while. Complex
systems tend to have many possible modes o vibration each
with its own natural requency.
It is also possible to orce a system to oscillate at any requency
that we choose by subjecting it to a changing orce that varies
with the chosen requency. This periodic driving orce must be
provided rom outside the system. When this driving frequency
is frst applied, a combination o natural and orced oscillations
take place which produces complex transient oscillations. Once
the amplitude o the transient oscillations die down, a steady
condition is achieved in which:
0.6
0.8
1.0
1.2
1.4
1.6
overshoot
underdamped
The amplitude o the orced oscillations depends on:
the comparative values o the natural requency and the
driving requency
the amount o damping present in the system.
amplitude of oscillation
natural frequenCy and resonanCe
0.4
light damping
increased damping
heavy damping
The system oscillates at the driving requency.
The amplitude o the orced oscillations is fxed. Each cycle
energy is dissipated as a result o damping and the driving
orce does work on the system. The overall result is that the
energy o the system remains constant.
q faCtor and damping
The degree o damping is measured by a quantity called the
quality actor or Q actor. It is a ratio (no units) and the
defnition is:
energy stored
Q = 2 __
energy lost per cycle
Since the energy stored is proportional to the square o
amplitude o the oscillation, measurements o decreasing
amplitude with time can be used to calculate the Q actor. The
Q actor is approximately equal to the number o oscillations
that are completed beore damping stops the oscillation.
168
o p t i o n B E n g i n E E r i n g p h ys i c s
driving frequency, fdriving
natural frequency, fnatural
Resonance occurs when a system is subject to an oscillating
orce at exactly the same requency as the natural requency o
oscillation o the system.
Typical orders o magnitude or dierent Q-actors:
Car suspension:
Simple pendulum:
Guitar string:
Excited atom:
1
1 03
1 03
1 07
When a system is in resonance and its amplitude is constant,
the energy provided by the driving requency during one cycle
is all used to overcome the resistive orces that cause damping.
In this situation, the Q actor can be calculated as:
energy stored
Q = 2 resonant requency __
power loss
HL
rc (2)
phase of forCed osCillations
Ater transient oscillations have died down, the requency o the orced oscillations equals the driving requency. The phase
relationship between these two oscillations is complex and depends on how close the driven system is to resonance:
phase lag
/rad
driven vibration
period behind
1
2
2
heavy damping
light damping
in phase
driven vibration
period behind
1
4
0
natural
frequency
f/Hz
forcing
frequency
examples of resonanCe
Comment
Vibrations in machinery
When in operation, the moving parts o machinery provide regular driving orces on the
other sections o the machinery. I the driving requency is equal to the natural requency, the
amplitude o a particular vibration may get dangerously high. e.g. at a particular engine speed
a trucks rear view mirror can be seen to vibrate.
Quartz oscillators
A quartz crystal eels a orce i placed in an electric feld. When the feld is removed, the
crystal will oscillate. Appropriate electronics are added to generate an oscillating voltage rom
the mechanical movements o the crystal and this is used to drive the crystal at its own natural
requency. These devices provide accurate clocks or microprocessor systems.
Microwave generator
Microwave ovens produce electromagnetic waves at a known requency. The changing
electric feld is a driving orce that causes all charges to oscillate. The driving requency o the
microwaves provides energy, which means that water molecules in particular are provided
with kinetic energy i.e. the temperature is increased.
Radio receivers
Electrical circuits can be designed (using capacitors, resistors and inductors) that have their
own natural requency o electrical oscillations. The ree charges (electrons) in an aerial will
eel a driving orce as a result o the requency o the radio waves that it receives. Adjusting
the components o the connected circuit allows its natural requency to be adjusted to equal
the driving requency provided by a particular radio station. When the driving requency
equals the circuits natural requency, the electrical oscillations will increase in amplitude and
the chosen radio stations signal will dominate the other stations.
Musical instruments
Many musical instruments produce their sounds by arranging or a column o air or a string to
be driven at its natural requency which causes the amplitude o the oscillations to increase.
Greenhouse eect
The natural requency o oscillation o the molecules o greenhouse gases is in the inra-red
region. Radiation emitted rom the Earth can be readily absorbed by the greenhouse gases in
the atmosphere. See page 92 or more details.
o p t i o n B E n g i n E E r i n g p h ys i c s
169
iB questons opton B engneerng physcs
1.
A sphere o mass m and radius r rolls, without slipping, rom
rest down an inclined plane. When it reaches the base o the
plane, it has allen a vertical distance h. Show that the speed
o the sphere, v, when it arrives at the base o the incline is
given by:
4.
1 An adiabatic compression to 1 /20th o its original volume.
_____
v=
1 0gh
_
7
In a diesel engine, air is initially at a pressure o 1 1 0 5 Pa
and a temperature o 27 C. The air undergoes the cycle o
changes listed below. At the end o the cycle, the air is back at
its starting conditions.
2 A brie isobaric expansion to 1 /1 0th o its original volume.
[4]
3 An adiabatic expansion back to its original volume.
2.
A fywheel o moment o inertia 0.75 kg m2 is accelerated
uniormly rom rest to an angular speed o 8.2 rad s 1 in 6.5 s.
a) Calculate the resultant torque acting on the fywheel
during this time.
4 A cooling down at constant volume.
a) Sketch, with labels, the cycle o changes that the gas
undergoes. Accurate values are not required.
[2]
b) I the pressure ater the adiabatic compression has risen
to 6.6 1 0 6 Pa, calculate the temperature o the gas.
[2]
b) Calculate the rotational kinetic energy o the fywheel
when it rotates at 8.2 rad s 1
[2]
c) In which o the our processes:
c) The radius o the fywheel is 1 5 cm. A breaking orce
applied on the circumerence and brings it to rest rom
an angular speed o 8.2 rad s 1 in exactly 2 revolutions.
Calculate the value o the breaking orce.
[2]
(i)
A xed mass o a gas undergoes various changes o
temperature, pressure and volume such that it is taken round
the pV cycle shown in the diagram below.
pressure/10 5 Pa
3.
[3]
is work done on the gas?
[1 ]
(ii) is work done by the gas?
[1 ]
(iii) does ignition o the air-uel mixture take place?
[1 ]
d) Explain how the 2nd law o thermodynamics applies
to this cycle o changes.
[2]
HL
2.0
5.
X
With the aid o diagrams, explain
a) What is meant by laminar fow
b) The Bernoulli eect
c) Pascals principle
1.0
Z
Y
1.0 2.0 3.0 4.0 5.0
d) An ideal fuid
[8]
6.
Oil, o viscosity 0.35 Pa s and density 0.95 g cm- 3 , fows
through a pipe o radius 20 cm at a velocity o 2.2 m s - 1 .
Deduce whether the fow is laminar or turbulent.
[4]
7.
A pendulum clock maintains a constant amplitude by means
o an electric power supply. The ollowing inormation is
available or the pendulum:
volume/10 3 m 3
The ollowing sequence o processes takes place during
the cycle.
X Y the gas expands at constant temperature and the gas
absorbs energy rom a reservoir and does 450 J o
work.
Y Z the gas is compressed and 800 J o thermal energy is
transerred rom the gas to a reservoir.
Maximum kinetic energy:
5 1 0- 2 J
Frequency o oscillation:
2 Hz
Q actor: 30
Z X the gas returns to its initial stage by absorbing energy
rom a reservoir.
Calculate:
a) The driving requency o the power supply
[3]
a) Is there a change in internal energy o the gas during
the processes X Y? Explain.
b) The power needed to drive the clock.
[3]
[2]
b) Is the energy absorbed by the gas during the process
X Y less than, equal to or more than 450 J? Explain. [2]
c) Use the graph to determine the work done on the gas
during the process Y Z.
[3]
d) What is the change in internal energy o the gas
during the process Y Z?
[2]
e) How much thermal energy is absorbed by the gas
during the process Z X? Explain your answer.
[2]
) What quantity is represented by the area enclosed by
the graph? Estimate its value.
[2]
g) The overall eciency o a heat engine is dened as
net work done by the gas during a cycle
Eciency = ____
total energy absorbed during a cycle
I this pV cycle represents the cycle or a particular heat
engine determine the eciency o the heat engine.
[2]
170
i B Q u E s t i o n s o p t i o n B E n g i n E E r i n g p h ys i c s
15 o p t I o n C I m a g I n g
I fri
Ray dIagRams
In order to fnd the location and nature o this image a ray
diagram is needed.
I an object is placed in ront o a plane mirror, an image will
be ormed.
image
object
upright
same size
as object
laterally
inverted
x
x
The process is as ollows:
The image ormed by reection in a plane mirror is always:
Light sets o in all directions rom every part o the object.
(This is a result o diuse reections rom a source o light.)
the same distance behind the mirror as the object is
in front
Each ray o light that arrives at the mirror is reected
according to the law o reection.
upright (as opposed to being inverted)
the same size as the object (as opposed to being magnifed
or diminished)
These rays can be received by an observer.
The location o the image seen by the observer arises because
the rays are assumed to have travelled in straight lines.
laterally inverted (i.e. let and right are interchanged)
virtual (see below) .
Real and vIRtual Images
object
real image
(a) real image
(b) virtual image
converging
rays
O
point
object
Upside down
Diminished
Real.
optical system
The opposite o a virtual image is a real
image. In this case, the rays o light do
actually pass through a single point.
Real images cannot be ormed by plane
mirrors, but they can be ormed by
concave mirrors or by lenses. For
example, i you look into the concave
surace o a spoon, you will see an image
o yoursel. This particular image is
concave
mirror
I
virtual point
image
I
real
point
image
O
point
object
optical system
The image ormed by reection in a plane
mirror is described as a virtual image. This
term is used to describe images created when
rays o light seem to come rom a single point
but in act they do not pass through that point.
In the example above, the rays o light seem
to be coming rom behind the mirror. They do
not, o course, actually pass behind the mirror
at all.
diverging
rays
stICk In wateR
The image ormed as a result o the reraction o light
leaving water is so commonly seen that most people orget
that the objects are made to seem strange. A straight stick
will appear bent i it is placed in water. The brain assumes
that the rays that arrive at ones eyes must have been
travelling in a straight line.
air
water
A straight stick appears bent
when placed in water
The image o the end o the pen is:
Nearer to the surace than the
pen actually is.
Virtual.
o pti o n C i m ag i n g
171
Cri
ConveRgIng lenses
air
air
A converging lens brings parallel rays into one ocus point.
normal
converging lens
normal
parallel rays
refraction at
1st surface
glass
refraction at
2nd surface
The rays o light are all brought together in one point because o
the particular shape o the lens. Any one lens can be thought o
as a collection o dierent-shaped glass blocks. It can be shown
that any thin lens that has suraces ormed rom sections o
spheres will converge light into one ocus point.
focal point
The reason that this happens is the reraction that takes place at
both suraces o the lens.
poweR of a lens
A converging lens will always be thicker at the centre when
compared with the edges.
wave model of Image foRmatIon
The power o a lens measures the extent to
which light is bent by the lens. A higher power
lens bends the light more and thus has a smaller
ocal length. The defnition o the power o a
lens, P, is the reciprocal o the ocal length, f:
region in which waves are
made to travel more slowly
O
I
f is the ocal length measured in m
object (source
of wave energy)
real image (point to which
wave energy is concentrated)
P is the power o the lens measured in m- 1 or
dioptres (dpt)
Formation o a real image by reraction (ignoring diraction)
1
P= _
f
A lens o power = +5 dioptre is converging
and has a ocal length o 20 cm. When two thin
lenses are placed close together their powers
approximately add.
defInItIons
When analysing lenses and the images that they orm, some
technical terms need to be defned.
The curvature o each surace o a lens makes it part o a
sphere. The centre o curvature or the lens surace is the
centre o this sphere.
The ocal point (principal ocus) o a lens is the point on
the principal axis to which rays that were parallel to the
principal axis are brought to ocus ater passing through the
lens. A lens will thus have a ocal point on each side.
The ocal length is the distance between the centre o the
lens and the ocal point.
The linear magnifcation, m, is the ratio between the size
(height) o the image and the size (height) o the object. It
has no units.
h
image size
linear magnifcation, m = _ = _i
object size
ho
The principal axis is the line going directly through the
middle o the lens. Technically it joins the centres o
curvature o the two suraces.
lens
centre of
curvature
focal point
c
f
f
focal length
172
o pti o n C i m ag i n g
c principal axis (PA)
Ig fri i cvx
ImpoRtant Rays
In order to determine the nature and
position o the image created o a given
object, we need to construct a scaled
ray diagram o the set-up. In order to
do this, we concentrate on the paths
taken by three particular rays. As soon
as the paths taken by two o these rays
have been constructed, the paths o all
the other rays can be inerred. These
important rays are described below.
lens
distant object
O
f
Any ray that was travelling
parallel to the principal axis will be
reracted towards the ocal point
on the other side o the lens.
I
f
f
O
f
I real image
inverted
same size
object between 2f and f
f
PA
PA
real inverted
magnied
I
object at f
f
O
2.
PA
f
O
f
f
PA
object at 2f
Converging lens
1.
real image
inverted
diminished
f
Any ray that travelled through the
ocal point will be reracted parallel
to the principal axis.
PA
virtual image
upright
image at innity
object closer than f
f
f
3.
I
PA
f
O
PA
f
virtual image
upright
magnied
Any ray that goes through
the centre o the lens will be
undeviated.
Converging lens images
f
f
PA
possIble sItuatIons
A ray diagram can be constructed as ollows:
An upright arrow on the principal axis represents the object.
The paths o two important rays rom the top o the object are constructed.
This locates the position o the top o the image.
The bottom o the image must be on the principal axis directly above (or below)
the top o the image.
A ull description o the image created would include the ollowing inormation:
i it is real or virtual
i it is upright or inverted
i it is magnifed or diminished
its exact position.
It should be noted that the important rays are just used to locate the image. The real
image also consists o all the other rays rom the object. In particular, the image will
still be ormed even i some o the rays are blocked o.
An observer receiving parallel rays sees an image located in the ar distance (at
infnity) .
o pti o n C i m ag i n g
173
thi i
lens equatIon
There is a mathematical method o locating the image ormed by a lens. An analysis o the angles
involved shows that the ollowing equation can be applied to thin spherical lenses:
1 = _
1 + _
1
_
f v u
lIneaR
magnIfICatIon
In all cases, linear
magnifcation,
h
v
m = _i = - _
u
h
o
height o image
m = __
height o object
h
= _i
ho
object
f
f
v
= - _
u
image
f
object distance u
Suppose
For real images, m is
negative and image is
inverted
f
For virtual images m is
positive and image is
upright
image distance v
u = 25 cm
f = 1 0 cm
1 = _
5 - _
3
1 - _
1 = _
1 - _
1 =_
2 = _
This would mean that _
v
u
10
25
50
50
50
f
50
In other word, v = _
= 1 6.7 cm i.e. image is real
3
- 1 6.7
In this case m = _
= - 1 .67 and inverted.
10
Real Is posItIve
Care needs to be taken with virtual images. The equation does work but or this to be the case, the ollowing
convention has to be ollowed:
Distances are taken to be positive i actually traversed by the light ray (i.e. distances to real object and image) .
Distances are taken to be negative i apparently traversed by the light ray (distances to virtual objects and images) .
Thus a virtual image is represented by a negative value or v in other words, it will be on the same side o
the lens as the object.
image
object
f
object
distance u
negative image
distance v
Suppose
u = 1 0 cm
f = 25 cm
1 = _
5 = - _
3
1 - _
1 = _
1 - _
1 = _
2 -_
This would mean that _
v
u
25
10
50
50
50
f
50
In other word, v = - _
= - 1 6.7 cm i.e. image is virtual
3
1 6.7 = +1 .67 and upright
In this case m = + _
10
174
o pti o n C i m ag i n g
f
diri
dIveRgIng lenses
A diverging lens spreads parallel rays apart. These rays appear
to all come rom one ocus point on the other side o the lens.
concave lens
When constructing ray diagrams or diverging lenses, the
important rays whose paths are known (and rom which all
other ray paths can be inerred) are:
1 . Any ray that was travelling parallel to the principal axis will
be reracted away rom a ocal point on the incident side o
the lens.
focal point
f
focal length
f
PA
2. Any ray that is heading towards the ocal point on the
other side o the lens, will be reracted so as to be parallel to
the principal axis.
The reason that this happens is the reraction that takes place
at both suraces. A diverging lens will always be thinner at the
centre when compared with the edges.
f
f
PA
defInItIons and ImpoRtant Rays
Diverging lenses have the same analogous defnitions as
converging lenses or all o the ollowing terms:
3. Any ray that goes through the centre o the lens will be
undeviated.
Centre o curvature, principal axis, ocal point, ocal length, linear
magnifcation.
f
Note that:
f
The ocal point is the point on the principal axis from
which rays that were parallel to the principal axis appear
to come ater passing through the lens.
PA
As the ocal point is behind the diverging lens, the focal
length of a diverging lens is negative.
Images CReated by a dIveRgIng lens
Whatever the position o the object, a diverging lens will always create an upright, diminished and virtual image located between
the ocal point and the lens on the same side o the lens as the object.
If you look at an object through a concave lens,
it will look smaller and closer.
If you move the object further out, the image will not
move as much.
object
f
f
u
image
f
v
object inside focal length
object
image
object outside focal length
The thin lens equation will still work providing one remembers the negative ocal length o a diverging lens.
For example, i an object is placed at a distance 2l away rom a diverging lens o ocal length l, the image can be calculated as
ollows:
Given: u = 2l, = - l, v = ?
1 + _
1 = _
1
_
u
v
1 = _
-3
1 - _
1 = _
1 - _
1 = _
_
v
u
-l
2l
2l
2l
_
v = -
3
1
This is a virtual diminished and upright image with m = + _
3
o pti o n C i m ag i n g
175
Cvri ivri irrr
geometRy of mIRRoRs and lenses
Image foRmatIon In mIRRoRs
The geometry o the paths o rays ater reection by a
spherical concave or convex mirror is exactly analogous to the
paths o rays through converging or diverging lenses. The only
dierence is that mirrors reect all rays backwards whereas
rays pass through lenses and continue orwards.
(1) Concave
object at innity
F
2f
(a) Convex lens
real
PA inverted
diminished
I
object between innity and 2f
PA
O
2f
f
(b) Concave mirror
real
PA inverted
diminished
F
I
object at 2f
O
2f
PA
F
real
PA inverted
same size
F
real
PA inverted
magnifed
I
object between 2f and f
f
O
(c) Concave lens
2f
I
object at f
PA
O
2f
f
PA
F
virtual
upright
image at
infnity
object between f and mirror
(d) Convex mirror
2f
F
O
I
PA
virtual
upright
magnifed
PA
f
(2) Convex
object at innity
This analogous behaviour means that all the defnitions and
equations or lenses can be used (with suitable attention to
detail with the sign conventions) with mirrors.
I
F
An additional important ray or mirrors is the ray that travels
through (or towards) the centre o curvature o the mirror
(located at twice the ocal length) . This ray will be reected
back along the same path.
object near lens
O
176
o pti o n C i m ag i n g
2f
virtual
upright
PA
diminished
I
F
virtual
PA upright
2f
diminished
th i iyi
neaR and faR poInt
angulaR sIze
The human eye can ocus objects at dierent
distances rom the eye. Two terms are useul to
describe the possible range o distances the near
point and the ar point distance.
I we bring an object closer to us (and our eyes are still able to ocus on it)
then we see it in more detail. This is because, as the object approaches, it
occupies a bigger visual angle. The technical term or this is that the object
subtends a larger angle.
The distance to the near point is the distance
between the eye and the nearest object that
can be brought into clear ocus (without strain
or help rom, or example, lenses) . It is also
known as the least distance o distinct vision.
By convention it is taken to be 25 cm or
normal vision.
The distance to the ar point is the distance
between the eye and the urthest object that
can be brought into ocus. This is taken to be
infnity or normal vision.
objects are the
same size
angle subtended
by close object
distant
object
angulaR magnIfICatIon
close
object
angle subtended
by distant object
1.
The angular magnifcation, M, o an optical instrument is
defned as the ratio between the angle that an object subtends
normally and the angle that its image subtends as a result o
the optical instrument. The normal situation depends on the
context. It should be noted that the angular magnifcation is not
the same as the linear magnifcation.
Image ormed at infnity
In this arrangement, the object is placed at the ocal point.
The resulting image will be ormed at infnity and can be
seen by the relaxed eye.
i = hf
top
rays from object at
specied distance
o
h
i
bottom
f
i
top
eye focused
on innity
f
rays from nal image
formed by optical
instrument
i
In this case the angular magnifcation would be
h
__
f
D
Min f n ity = _i = _
= _
h
__
f
o
D
bottom
This is the smallest value that the angular magnifcation
can be.
Angular magnifcation, M = _i
o
2.
The largest visual angle that an object can occupy is when it
is placed at the near point. This is oten taken as the normal
situation.
o = h
D
Image ormed at near point
In this arrangement, the object is placed nearer to the lens.
The resulting virtual image is located at the near point. This
arrangement has the largest possible angular magnifcation.
M=
i h i /D h i D
=
= =
o h/D
h
a
1 1
1
+ =
u v
f
1 1
1
- =
a D
f
D D
= + 1
a
f
h
o
hi
D
A simple lens can increase the angle subtended. It is usual to
consider two possible situations.
h
f
i
a
f
i
D
D +1
So the magnitude o Mn e a r p o in t = _
f
o pti o n C i m ag i n g
177
abrrin
spheRICal
A lens is said to have an aberration if, for
some reason, a point object does not produce
a perfect point image. In reality, lenses that
are spherical do not produce perfect images.
Spherical aberration is the term used to
describe the fact that rays striking the outer
regions of a spherical lens will be brought to a
slightly different focus point from those striking
the inner regions of the same lens. This is not
to be confused with barrel distortion.
ray striking outer regions
ray striking inner regions
In general, a point object will focus into a small
circle of light, rather than a perfect point. There
are several possible ways of reducing this effect:
the shape of the lens could be altered in
such a way as to correct for the effect.
The lens would, of course, no longer be
spherical. A particular shape only works for
objects at a particular distance away.
ray striking outer regions
range of focal points
outer sections
of lens not used
the effect can be reduced for a given lens by
decreasing the aperture. The technical term
for this is stopping down the aperture. The
disadvantage is that the total amount of light
is reduced and the effects of diffraction (see
page 46) would be made worse.
aperture
The effect for mirrors can be eliminated for all
point objects on the axis by using a parabolic
(as opposed to a spherical) mirror. For mirrors,
the effect can again be reduced by using a
smaller aperture.
Spherical aberration
ChRomatIC
Chromatic aberration is the term used
to describe the fact that rays of different
colours will be brought to a slightly
different focus point by the same lens.
The refractive index of the material used
to make the lens is different for different
frequencies of light.
white light
A point object will produce a blurred
image of different colours.
The effect can be eliminated for two
given colours (and reduced for all) by
using two different materials to make up
a compound lens. This compound lens is
called an achromatic doublet. The two
types of glass produce equal but opposite
dispersion.
white light
converging lens
of crown glass
(low dispersion)
Achromatic doublet
o pti o n C i m ag i n g
violet
R
violet
R
red
Mirrors do not suffer from chromatic
aberration.
178
V
red
V
violet
focus
Canada balsam
cement
diverging lens
of int glass
( high dispersion)
red
focus
th c icrc ric c
Compound mICRosCope
A compound microscope consists o two lenses the objective lens and the eyepiece lens. The frst lens (the objective lens)
orms a real magnifed image o the object being viewed. This real image can then be considered as the object or the second lens
(the eyepiece lens) which acts as a magniying lens. The rays rom this real image travel into the eyepiece lens and they orm a
virtual magnifed image. In normal adjustment, this virtual image is arranged to be located at the near point so that maximum
angular magnifcation is obtained.
objective
lens fo
top half
of object
B
h
O
real image formed
by objective lens
h 1 i
fo
virtual image
of top half of
object
construction
line
eyepiece lens fe
construction
line
i
fe
eye focused on near point to
see virtual image (in practice
it would be much nearer to the
eyepiece lens than implied here)
h2
M
D
h2
__
h
h
h
D
M = _i = _
= _2 = _2 _1 = linear magnifcation produced by eyepiece linear magnifcation produced by objective
h
__
h
h1
h
o
D
astRonomICal telesCope
An astronomical telescope also consists o two lenses. In this case, the objective lens orms a real but diminished image o the
distant object being viewed. Once again, this real image can then be considered as the object or the eyepiece lens acting as a
magniying lens. The rays rom this real image travel into the eyepiece lens and they orm a virtual magnifed image. In normal
adjustment, this virtual image is arranged to be located at infnity.
objective
lens
eyepiece lens
fo
parallel rays all from
top of distant object
o
o
fe
real image formed in mutual
focal plane of lenses
fe
fo
h 1 i
construction
line
i
eye focused on innity
virtual image
at innity
h1
__
f
f
M = _i = _
= _o
h
f
o
__
e
e
1
fo
The length o the telescope fo + fe .
o pti o n C i m ag i n g
179
aric rfci c
CompaRIson o ReleCtIng and ReRaCtIng
telesCopes
A reracting telescope uses an objective (converging) lens to
orm a real diminished image o a distant object. This image is
then viewed by the eyepiece lens (converging) which, acting
as a simple magniying glass, produces a virtual but magnifed
fnal image.
newtonIan mountIng
A small at mirror is placed on the principal axis o the mirror
to reect the image ormed to the side:
concave
mirror
small at mirror
Fo
In an analogous way, a reecting telescope uses a concave
mirror set up so as to orm a real, diminished image o a
distant object. This image, however, would be difcult to
view as it would be produced in ront o the concave mirror.
Thus mirrors are used to produce a viewable image that
can, like the reracting telescope, be viewed by the eyepiece
lens (converging) . Once again the eyepiece acts as a simple
magniying glass and produces the virtual, but magnifed, fnal
image. Two common mountings or reecting telescopes are
the Newtonian mounting and the Cassegrain mounting.
F'o
eyepiece lens
All telescopes are made to have large apertures in order to:
CassegRaIn mountIng
a) reduce diraction eects, and
A small convex mirror is mounted on the principal axis o the
mirror. The mirror has a central hole to allow the image to be
viewed.
b) collect enough light to make bright images o low power
sources.
Large telescopes are reecting because:
Mirrors do not suer rom chromatic aberration
The convex mirror will add to the angular magnifcation
achieved.
concave
m irror
It is difcult to get a uniorm reractive index throughout a
large volume o glass
Mounting a large lens is harder to achieve than mounting
a large mirror.
Fext
Only one surace needs to be the right shape.
Reecting telescopes can easily suer damage to the mirror
surace.
180
o pti o n C i m ag i n g
eyepiece
lens
small convex m irror
Fo
Ri c
sIngle dIsh RadIo telesCopes
RadIo InteRfeRometRy telesCopes
A single dish radio telescope operates in a very similar way
to a reecting telescope. Rather than reecting visible light
to orm an image, the much longer wavelengths o radio
waves are reected by the curved receiving dish. The antenna
that is the receiver o the radio waves can be tuned to pick
up specifc wavelengths under observation and are used to
study naturally occurring radio emission rom stars, galaxies,
quasars and other astronomical objects between wavelengths
o about 1 0 m and 1 mm.
The angular resolution o a radio telescope can be improved
using a principle called intererometry. This process analyses
signals received at two (or more) radio telescopes that are
some distance apart but pointing in the same direction. This
eectively creates a virtual radio telescope that is much larger
than any o the individual telescopes.
Radio telescope
incoming radio
waves
Radio waves reect
o the dish and
focus at the tip.
Receivers detect and
amplify radio signals.
Diraction eects can signifcantly limit the accuracy with
which a radio telescope can locate individual sources o radio
signals. Increasing the diameter o a radio telescope improves
the telescopes ability to resolve dierent sources and ensure
that more power can be received (see resolution on page 1 01 ) .
The technique is complex as it involves collecting signals
rom two or more radio telescopes (an array telescope)
in one central location. The arrival o each signal at an
individual antenna needs to be careully calibrated against a
single shared reerence signal so that dierent signals can be
combined as though they arrived at one single antenna. When
the signals rom the dierent telescopes are added together,
they interere. The result is to create a combined telescope
that is equivalent in resolution (though not in sensitivity) to a
single radio telescope whose diameter is approximately equal
to the maximum separations o the antennae.
The principle can be extended, in a process called Very
Long Baseline Interferometry, to allow recordings o
radio signals (originally made hundreds o km apart) to be
synchronized to within a millionth o a second thus allowing
scientists rom dierent countries to collaborate to create a
virtual radio telescope o huge size and high resolving power.
CompaRatIve peRfoRmanCe of eaRth-bound and satellIte-boRne telesCopes
The ollowing points about Earth-based (EB) and satellite-borne (SB) telescopes can be made:
SB observations are ree rom intererence and/or absorptions due to the Earths atmosphere that hinder EB observations, giving
better resolution or SB telescopes.
Modern computer techniques can eectively correct or many atmospheric eects making new ground-based telescopes similar
in resolution to some SB telescopes.
Many signifcant wavelengths o EM radiation (UV, IR and long wavelength radio) are absorbed by the Earths atmosphere so SB
telescopes are the only possibility in their wavelengths.
SB observations do not suer rom light pollution / radio intererence as a result o nearby human activity.
SB acilities are not subject to continual wear and tear as a result o the Earths atmosphere (storms etc.) .
The possibility o damage rom space debris exists or SB telescopes.
There is a great deal o added cost in getting the telescope into orbit and controlling it remotely, meaning that SB telescopes are
signifcantly more expensive to build and this places a limit on their size and weight.
There is an added difcultly o eecting repairs / alterations to a SB telescope once operational.
SB telescopes need to withstand wider temperature variations than EB telescopes.
EB optical telescopes can only operate at night whereas SB telescopes can operate at all times.
o pti o n C i m ag i n g
181
fir ic
optIC fIbRe
Optic fbres use the principle o total internal reection (see
page 45) to guide light along a certain path. The idea is to
make a ray o light travel along a transparent fbre by bouncing
between the walls o the fbre. So long as the incident angle o
the ray on the wall o the fbre is always greater than the critical
angle, the ray will always remain within the fbre even i the
fbre is bent (see right) .
In the medical world. Bundles o optic fbres can be used to
carry images back rom inside the body. This instrument is
called an endoscope.
This type o optic fbre is known as a step-index optic
fbre. Cladding o a material with a lower reractive index
surrounds the fbre. This cladding protects and strengthens
the fbre.
As shown on page 45, the relation between critical angle, c, and
reractive index n is given by
1
n=_
sin c
Two important uses o optic fbres are:
In the communication industry. Digital data can be encoded
into pulses o light that can then travel along the fbres. This
is used or telephone communication, cable TV etc.
types of optIC fIbRes
The simplest fbre optic is a step-index fbre.
Technically this is known as a multimode stepindex fbre. Multimode reers to the act that light can
take dierent paths down the fbre which can result in
some distortion o signals (see waveguide dispersion,
page 1 83) . The (multimode) graded-index fbre is
an improvement. This uses a graded reractive index
profle in the fbre meaning that rays travel at dierent
speeds depending on their distance rom the centre.
This has the eect o reducing the spreading out o the
pulse. Most fbres used in data communications have a
graded index. The optimum solution is to have a very
narrow core a singlemode step-index fbre.
index index output
prole pulse pulse
n2
n1
multimode step-index
n2
n1
multimode graded-index
n2
n1
singlemode step-index
182
o pti o n C i m ag i n g
diri, i i i ic fbr
mateRIal dIspeRsIon
waveguIde dIspeRsIon
The reractive index o any substance depends on the
requency o electromagnetic radiation considered. This is the
reason that white light is dispersed into dierent colours when
it passes through a triangular prism.
I the optical fbre has a signifcant diameter, another process
called waveguide dispersion that can cause the stretching o
a pulse is multipath or modal dispersion. The path length
along the centre o a fbre is shorter than a path that involves
multiple reections. This means that rays rom a particular
pulse will not all arrive at the same time because o the dierent
distances they have travelled.
As light travels along an optical fbre, dierent requencies will
travel at slightly dierent speeds. This means that i the source
o light involves a range o requencies, then a pulse that starts
out as a square wave will tend to spread out as it travels along
the fbre. This process is known as material dispersion.
B
C
A
cladding
after transmission
attenuatIon
As light travels along an optic fbre, some o the energy can be
scattered or absorbed by the glass. The intensity o the light energy
that arrives at the end o the fbre is less than the intensity that
entered the fbre. The signal is said to be attenuated.
The amount o attenuation is measured on a logarithmic scale
in decibels (dB) . The attenuation is given by
I
attenuation (dB) = 1 0log _
Io
I is the intensity o the output power measured in W
Io is the intensity o the original input power measured in W
A negative attenuation means that the signal has been reduced
in power. A positive attenuation would imply that the signal
has been amplifed.
See page 1 88 or another example o the use o the decibel scale.
It is common to quote the attenuation per unit length as
measured in dB km- 1 . For example, 5 km o fbre optic cable
causes an input power o 1 00 mW to decrease to 1 mW.
The attenuation per unit length is calculated as ollows:
attenuation = 1 0 log (1 0 /1 0 ) = 1 0 log (1 0 )
= -20 dB
-3
-1
-2
core
The problems caused by modal dispersion have led to the
development o monomode (or singlemode) step-index
fbres. These optical fbres have very narrow cores (o the
same order o magnitude as the wavelength o the light being
used (approximately 5 m) so that there is only one eective
transmission path directly along the fbre.
The attenuation o a 1 0 km length o this fbre optic cable
would thereore be - 40 dB. The overall attenuation resulting
rom a series o actors is the algebraic sum o the individual
attenuations.
The attenuation in an optic fbre is a result o several
processes: those caused by impurities in the glass, the general
scattering that takes place in the glass and the extent to which
the glass absorbs the light. These last two actors are aected by
the wavelength o light used. A typical the overall attenuation is
shown below:
6
5
4
3
2
1
attenuation per
unit length / dB km -1
before transmission
paths
0.6
0.8
1.0
1.2
1.4
1.6
1.8
wavelength / m
attenuation per unit length = -20 dB/5 km
= -4 dB km- 1
CapaCIty
noIse, amplIIeRs and ReshapeRs
Attenuation causes an upper limit to the amount o digital
inormation that can be sent along a particular type o optical
fbre. This is oten stated in terms o its capacity.
Noise is inevitable in any electronic circuit. Any dispersions or
scatterings that take place within an optical fbre will also add
to the noise.
capacity o an optical fbre = bit rate distance
A fbre with a capacity o 80 Mbit s km can transmit
80 Mbit s - 1 along a 1 km length o fbre but only 20 Mbit s - 1
along a 4 km length.
-1
An amplifer increases the signal strength and thus will tend
to correct the eect o attenuation these are also sometimes
called regenerators. An amplifer will also increase any noise
that has been added to the electrical signal.
A reshaper can reduce the eects o noise on a digital signal
by returning the signal to a series o 1 s and 0s with sharp
transitions between the allowed levels.
o pti o n C i m ag i n g
183
Channels of communication
The table below shows some common communication links.
Wire pairs
(twisted pair)
copper wire
insulation
Coaxial cables
copper
wire
insulation
copper mesh
outside insulation
Options for communication
Uses
Advantages and disadvantages
Two wires can connect
the sender and receiver o
inormation. For example
a simple link between a
microphone, an amplifer and
a loudspeaker.
Very simple communication
systems e.g. intercom
Very simple and cheap.
This arrangement o two wires
reduces electrical intererence.
A central wire is surrounded
by the second wire in the orm
o an outer cylindrical copper
tube or mesh. An insulator
separates the two wires.
Coaxial cables are used
to transer signals rom
TV aerials to TV receivers.
Historically they are
standard or underground
telephone links.
Susceptible to noise and intererence.
Unable to transer inormation at the
highest rates.
Simple and straightorward.
Less susceptible to noise compared
to simple wire pair but noise still a
problem.
Wire links can carry
requencies up to about 1 GHz
but the higher requencies will
be attenuated more or a given
length o wire. A typical 1 00
MHz signal sent down low-loss
cable would need repeaters
at intervals o approximately
0.5 km.
The upper limit or a single
coaxial cable is approximately
1 40 Mbit s - 1 .
Optical fbres
Laser light can be used to send
signals down optical fbres
with approximately the same
requency limit as cables 1 GHz.
The attenuation in an optical
fbre is less than in a coaxial
cable. The distance between
repeaters can easily be tens (or
even hundreds) o kilometres.
Long-distance
telecommunication and
high volume transer o
digital data including video
data.
Compared to coaxial cables with
equivalent capacity, optical fbres:
have a higher transmission capacity
are much smaller in size and weight
cost less
allow or a wider possible spacing o
regenerators
oer immunity to electromagnetic
intererence
suer rom negligible cross talk
(signals in one channel aecting
another channel)
are very suitable or digital data
transmission
provide good security
are quiet they do not hum even
when carrying large volumes o
data.
There are some disadvantages:
the repair o fbres is not a simple
task
regenerators are more complex and
thus potentially less reliable.
184
o pti o n C i m ag i n g
x-r
HL
IntensIty, qualIty and attenuatIon
basIC x-Ray dIsplay teChnIques
The eects o X-rays on matter depend on two things, the
intensity and the quality o the X-rays.
The basic principle o X-ray imaging is that some body parts
(or example bones) will attenuate the X-ray beam much
more than other body parts (or example skin and tissue) .
Photographic flm darkens when a beam o X-rays is shone on
them so bones show up as white areas on an X-ray picture.
The intensity, I, is the amount o energy per unit area that
is carried by the X-rays.
The quality o the X-ray beam is the name given to the
spread o wavelengths that are present in the beam.
Low-energy photons will be absorbed by all tissues and
potentially cause harm without contributing to orming the
image. It is desirable to remove these rom the beam.
I the energy o the beam is absorbed, then it is said to be
attenuated. I there is nothing in the way o an X-ray beam, it
will still be attenuated as the beam spreads out. Two processes o
attenuation by matter, simple scattering and the photoelectric
eect are the dominant ones or low-energy X-rays.
scattering
photoelectric
eect
X-ray
photon
low X-ray
photon
electron
electron
low X-ray
photon
X-ray beam
X-ray
photograph
The sharpness o an X-ray image is a measure o how easy it is
to see edges o dierent organs or dierent types o tissue.
X-ray beams will be scattered in the patient being scanned
and the result will be to blur the fnal image and to reduce
the contrast and sharpness. To help reduce this eect, a metal
flter grid is added below the patient:
X-ray beam
light
photon
Simple scattering aects X-ray photons that have energies
between zero and 30 keV.
In the photoelectric eect, the incoming X-ray has enough
energy to cause one o the inner electrons to be ejected rom
the atom. It will result in one o the outer electrons alling
down into this energy level. As it does so, it releases some
light energy. This process aects X-ray photons that have
energies between zero and 1 00 keV.
X transmission
Both attenuation processes result in a near exponential
transmission o radiation as shown in the diagram below. For
a given energy o X-rays and given material there will be a
certain thickness that reduces the intensity o the X-ray by
50% . This is known as the hal-value thickness.
100%
50%
I = e -x
I0
half-value
thickness
thickness of absorber, x
The attenuation coefcient is a constant that mathematically
allows us to calculate the intensity o the X-rays given any
thickness o material. The equation is as ollows:
I = I0 e - x
The relationship between the attenuation coefcient and the
hal-value thickness is
x_1 = ln 2
2
x_1
The hal-value thickness o the material (in m)
2
ln 2 The natural log o 2. This is the number 0.6931
The attenuation coefcient (in m- 1 )
depends on the wavelength o the X-rays short wavelengths
are highly penetrating and these X-rays are hard. Sot x-rays
are easily attenuated and have long wavelengths.
patient
metal grid
X-ray lm
Alternatively computer sotware can be used to detect and
enhance edges.
Since X-rays cause ionizations, they are dangerous. This
means that the intensity used needs to be kept to an absolute
minimum. This can be done by introducing something to
intensiy (to enhance) the image. There are two simple
techniques o enhancement:
When X-rays strike an intensiying screen the energy is reradiated as visible light. The photographic flm can absorb
this extra light. The overall eect is to darken the image in
the areas where X-rays are still present (see page 1 87) .
In an image-intensifer tube, the X-rays strike a uorescent
screen and produce light. This light causes electrons to be
emitted rom a photocathode. These electrons are then
accelerated towards an anode where they strike another
uorescent screen and give o light to produce an image.
mass attenuatIon CoeffICIent
An alternative way o writing the equation or the attenuation
coefcient is shown below:
- _ x
I = I0 e ( )
Where is the density o the substance. In this ormat, _
is
known as the mass attenuation coefcient _
,
and
x
is
known as the area density or mass thickness.
2
-1
Units o mass attenuation coefcient, _
= m kg
-2
Units o area density, x = kg m
I = I0 e
()
- _
x
o pti o n C i m ag i n g
185
HL
x-ray imaging techniques
1 ) Intensiying screens
3) Tomography
The arrangement o the intensiying screens described on
page 1 85 are shown below.
X-rays emerging from patient
cassette front (plastic)
front intensifying screen: phosphor
double-sided lm
rear intensifying screen: phosphor
felt padding
cassette
Tomography is a technique that makes the X-ray
photograph ocus on a certain region or slice through
the patient. All other regions are blurred out o ocus. This
is achieved by moving the source o X-rays and the flm
together.
motion
about
12 mm
X-ray tube
pivot point
plane of cut
A
B
With a simple X-ray photograph it is hard to identiy
problems within sot tissue, or example in the gut. There
are two general techniques aimed at improving this
situation.
2) Barium meals
In a barium meal, a dense substance is swallowed and
its progress along the gut can be monitored. The contrast
between the gut and surrounding tissue is increased.
Typically the patient is asked to swallow a harmless solution
o barium sulate. The result is an increase in the sharpness
o the image.
186
o pti o n C i m ag i n g
X-ray table
lm
A B B
A B
motion
An extension o basic tomography is the computed
tomography scan or CT scan. In this set-up a tube sends
out a pulse o X-rays and a set o sensitive detectors collects
inormation about the levels o X-radiation reaching each
detector. The X-ray source and the detectors are then rotated
around a patient and the process is repeated. A computer
can analyse the inormation recorded and is able to
reconstruct a 3-dimensional map o the inside o the body
in terms o X-ray attenuation.
HL
uric igig
ultRasound
a- and b-sCans
The limit o human hearing is about 20 kHz. Any sound that
is o higher requency than this is known as ultrasound.
Typically ultrasound used in medical imaging is just within
the MHz range. The velocity o sound through sot tissue is
approximately 1 500 m s - 1 meaning that typical wavelengths
used are o the order o a ew millimetres.
There are two ways o presenting the inormation gathered
rom an ultrasound probe, the A-scan or the B-scan. The
A-scan (amplitude-modulated scan) presents the inormation
as a graph o signal strength versus time. The B-scan
(brightness-modulated scan) uses the signal strength to aect
the brightness o a dot o light on a screen.
pulse
vertebra
echo
A-scan display
reections from
boundaries
abdominal wall
organ
to scan display
strength
of signal
Unlike X-rays, ultrasound is not ionizing so it can be used very
saely or imaging inside the body with pregnant women or
example. The basic principle is to use a probe that is capable
o emitting and receiving pulses o ultrasound. The ultrasound
is reected at any boundary between dierent types o tissue.
The time taken or these reections allows us to work out
where the boundaries must be located.
vertebra
organ
time
B-scan display
probe
path of ultrasound
aCoustIC ImpedanCe
The acoustic impedance o a substance is the product o the
density, , and the speed o sound, c.
Z = c
A-scans are useul where the arrangement o the internal
organs is well known and a precise measurement o distance
is required. I several B-scans are taken o the same section
o the body at one time, all the lines can be assembled into
an image which represent a section through the body. This
process can be achieved using a large number o transducers.
(i)
probe
placenta
(ii)
(iii)
unit o Z = kg m- 2 s - 1
Very strong reections take place when the boundary is
between two substances that have very dierent acoustic
impedances. This can cause some difculties.
In order or the ultrasound to enter the body in the frst
place, there needs to be no air gap between the probe and
the patients skin. An air gap would cause almost all o the
ultrasound to be reected straight back. The transmission o
ultrasound is achieved by putting a gel or oil (o similar density
to the density o tissue) between the probe and the skin.
Very dense objects (such as bones) can cause a strong
reection and multiple images can be created. These need
to be recognized and eliminated.
original
reection
2nd reection reected
by bone back to probe
path of
ultrasound
organ
beam reected
from bone
2nd reection back
ward b n
foetal
skull
limbs
(i)
( ii)
(iii)
Building a picture rom a series o B-scan lines
ChoICe of fRequenCy
The choice o requency o ultrasound to use can be seen as
the choice between resolution and attenuation.
Here, the resolution means the size o the smallest object
that can be imaged. Since ultrasound is a wave motion,
diraction eects will be taking place. In order to image a
small object, we must use a small wavelength. I this was
the only actor to be considered, the requency chosen
would be as large as possible.
pIezoeleCtRIC CRystals
Unortunately attenuation increases as the requency o
ultrasound increases. I very high requency ultrasound
is used, it will all be absorbed and none will be reected
back. I this was the only actor to be considered, the
requency chosen would be as small as possible.
These quartz crystals change shape when an electric current ows
and can be used with an alternating pd to generate ultrasound.
They also generate pds when receiving sound pressure waves so
one crystal is used or generation and detection.
On balance the requency chosen has to be somewhere
between the two extremes. It turns out that the best choice o
requency is oten such that the part o the body being imaged
is about 200 wavelengths o ultrasound away rom the probe.
o pti o n C i m ag i n g
187
HL
Igig ci
RelatIve IntensIty levels of ultRasound
The relative intensity levels o ultrasound between two points
are compared using the decibel scale (dB) . As its name suggests,
the decibel unit is simply one tenth o a base unit that is called
the bel (B) . The decibel scale is logarithmic.
Mathematically,
Relative intensity level in bels,
intensity level o ultrasound at measurement point
L I = log _____
intensity level o ultrasound at reerence point
I
I0
or Relative intensity level in bels = log _1
Since 1 bel = 1 0 dB,
I
I0
Relative intensity level in decibels, LI = 1 0 log _1
nmR
Nuclear Magnetic Resonance (NMR) is a very complicated
process but one that is extremely useul. It can provide detailed
images o sections through the body without any invasive
or dangerous techniques. It is o particular use in detecting
tumours in the brain. It involves the use o a non-uniorm
magnetic feld in conjuction with a large uniorm feld.
In outline, the process is as ollows:
I a pulse o radio waves is applied at the Larmor requency,
the nuclei can absorb this energy in a process called
resonance. The protons make a spin transition.
Ater the pulse, the nuclei return to their lower energy state
by emitting radio waves.
The time over which radio waves are emitted is called the
relaxation time.
The nuclei o atoms have a property called spin.
The radio waves emitted and their relaxation times can be
processed to produce the NMR scan image.
The spin o these nuclei means that they can act like tiny
magnets.
The signal analysis is targeted at the hydrogen nuclei
(protons) present.
These nuclei will tend to line up in a strong magnetic feld.
The number o H nuclei varies with the chemical
composition so dierent tissues extract dierent amounts o
energy rom the applied signal.
RF generator
eld
gradient
coils
Thus RF signal orces protons to make a spin transition and
S
The gradient feld allows determination o the point rom
which the photons are emitted.
RF coil
receiver
body
The proton spin relaxation time depends on the type o
tissue at the point where the radiation is emitted.
N
permanent magnet
relaxation time
oscilloscope
They do not, however, perectly line up they oscillate in a
particular way that is called precession. This happens at a very
high requency the same as the requency o radio waves.
The particular requency o precession depends on the
magnetic feld and the particular nucleus involved. It is
called the Larmor frequency.
CompaRIson between ultRasound and nmR
The ollowing points can be noted:
Ultrasound waves do not enter the body easily and multiple
reections can reduce the clarity o the image.
NMR imaging is very expensive when compared with
ultrasound equipment and is very bulky patient needs to be
brought to the NMR machine and process is time consuming.
Both wave energies carry energy but the energy associated
with the ultrasound is greater that the energy associated
with the radio requencies used in NMR.
Ultrasound measurements are easy to perorm (equipment can
be brought to patient at the point o care) and can be repeated
as required but quality o image can rely on skill o operator.
At the radio requencies used in NMR there is no danger o
resonance but some ultrasound energy can cause heating.
NMR produces a 3-dimensional scan, ultrasound typically
produces a 2-dimensional scan.
Detail produced by NMR is greater than by ultrasound.
NMR particularly useul or very delicate areas o body e.g. brain.
NMR patients have to remain very still, ultrasound images
can be more dynamic.
188
o pti o n C i m ag i n g
Ultrasound can cause cavitation the production o small
gas bubbles which will absorb energy and can damage
surrounding tissue. The requencies and intensities used or
diagnostics avoid this possibility as much as possible.
The strong magnetic felds used in NMR present problems
or patients with surgical implants and / or pacemakers.
Ib questions option C imaging
1.
For each o the ollowing situations, locate and describe
the fnal image ormed. Solutions should ound using scale
diagrams and mathematically.
The purpose o this particular scan is to fnd the depth d
o the organ labelled O below the skin and also to fnd its
length, I.
a) An object is placed 7 cm in ront o a concave mirror o
ocal length 1 4 cm.
[4]
b) (i) Suggest why a layer o gel is applied between the
ultrasound transmitter/receiver and the skin.
b) A diverging lens o ocal length 1 2.0 cm is placed at the
ocal point o a converging lens o ocal length 8.0 cm. An
object is placed 1 6.0 cm in ront o the converging lens. [4]
On the graph below the pulse strength o the reected pulses
is plotted against the time lapsed between the pulse being
transmitted and the time that the pulse is received, t.
2.
pulse strength /
relative units
c) An object is placed 1 8.0 cm in ront o a convex lens o
ocal length 6.0 cm. A second convex lens o ocal length
3.0 cm is an additional 1 8 cm behind the frst lens.
[4]
A student is given two converging lenses, A and B, and a
tube in order to make a telescope.
a) Describe a simple method by which she can determine
the ocal length o each lens.
[2]
b) She fnds the ocal lengths to be as ollows:
1 0 cm
Focal length o lens B
50 cm
(iii) The mean speed in tissue and muscle o the
ultrasound used in this scan is 1 .5 1 0 3 ms - 1 .
Using data rom the above graph, estimate the
depth d o the organ beneath the skin and the
length l o the organ O.
labels or each lens;
(iii) the position o the eye when the telescope is in use. [4]
c) On your diagram, mark the location o the intermediate
image ormed in the tube.
[1 ]
d) Is the image seen through the telescope upright or
upside-down?
[1 ]
e) Approximately how long must the telescope tube be?
[1 ]
Explain what is meant by
a) Material dispersion
e) A Cassegrain mounting
b) Waveguide dispersion
) Total Internal reection
c) Spherical aberrations
g) Step-index fbres
d) Chromatic aberrations
4.
gain = 20 dB
[1 ]
d) State one advantage and one disadvantage o using
ultrasound as opposed to using X-rays in medical
diagnosis.
[2]
a) State and explain which imaging technique is normally
used
output
[2]
20
15
10
gain = 30 dB
5
0
a) the overall gain o the system
b) the output power.
[2]
The graph below shows the variation o the intensity I o
a parallel beam o X-rays ater it has been transmitted
through a thickness x o lead.
Calculate
0
2
4
6
8
[2]
b) ( i)
HL
5.
to detect a broken bone
(ii) to examine the growth o a etus.
[2 each]
optical bre
6.
[4]
c) The above scan is known as an A-scan. State one
way in which a B-scan diers rom an A-scan.
(i)
A 1 5 km length o optical fbre has an attenuation o
4 dB km- 1 . A 5 mW signal is sent along the wire using two
amplifers as represented by the diagram below.
input power
= 5 mW
C
(ii) Indicate on the diagram the origin o the reected
pulses A, B and C and D.
[2]
(ii) the ocal points or each lens;
3.
D
B
0 25 50 75 100 125 150 175 200 225 250 275 300
t / s
Draw a diagram to show how the lenses should be
arranged in the tube in order to make a telescope. Your
diagram should include:
(i)
A
intensity
Focal length o lens A
[2]
10 12
x / mm
D efne half-value thickness, x _1 .
[2 ]
2
This question is about ultrasound scanning.
a) State a typical value or the requency o ultrasound
used in medical scanning.
The diagram below shows an ultrasound transmitter and
receiver placed in contact with the skin.
d
ultrasound transmitter
and receiver
layer of skin and fat
[1 ]
( ii) Use the graph to estimate x _1 or this beam in
2
lead.
[2 ]
( iii) D etermine the thickness o lead required to
reduce the intensity transmitted to 2 0% o its
initial value.
[2 ]
( iv) A second metal has a hal- value thickness x _1
2
or this radiation o 8 mm. C alculate what
thickness o this metal is required to reduce the
intensity o the transmitted beam by 80% .
[3 ]
O
l
i B Q u esti o n s o pti o n C i m ag i n g
189
16 o p t i o n D a S t r o p h yS i c S
ojs vs (1)
Solar SyStem
We live on the Earth. This is one o eight planets that orbit the Sun collectively this system is known as the Solar System.
Each planet is kept in its elliptical orbit by the gravitational attraction between the Sun and the planet. Other smaller masses
such as dwarf planets like Pluto or planetoids also exist.
diameter / km
distance to Sun / 10 8 m
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
4,880
1 2,1 04
1 2,756
6,787
1 42,800
1 20,000
51 ,800
49,500
58
1 07.5
1 49.6
228
778
1 ,427
2,870
4,497
Relative positions of the planets
Jupiter
Venus
Mars
Sun
Earth
Mercury
Uranus
Saturn
Neptune
Some o these planets (including the Earth) have other small
objects orbiting around them called moons. Our Moon is
3.8 1 0 8 m away and its diameter is about 1 /4 o the Earths.
Mercury
Earth
An asteroid is a small rocky body that drits around the Solar
System. There are many orbiting the Sun between Mars and
Jupiter the asteroid belt. An asteroid on a collision course
with another planet is known as a meteoroid.
Mars
Small meteors can be vaporized due to the riction with the
atmosphere (shooting stars) whereas larger ones can land
on Earth. The bits that arrive are called meteorites.
Venus
Sun
Jupiter
Comets are mixtures o rock and ice (a dirty snowball) in very
elliptical orbits around the Sun. Their tails always point away
rom the Sun.
Saturn
View from one place on earth
Uranus
Neptune
Relative sizes of the planets
nebulae
In many constellations there are diuse but relatively large
structures which are called nebulae. These are interstellar
clouds o dust, hydrogen, helium and other ionized gases. An
example is M42 otherwise known as the Orion Nebula.
I we look up at the night sky we see the stars many o these
stars are, in act, other galaxies but they are very ar away.
The stars in our own galaxy appear as a band across the sky
the Milky Way.
Patterns o stars have been identifed and 88 dierent regions
o the sky have been labelled as the dierent constellations.
Stars in a constellation are not necessarily close to one
another.
Over the period o a night, the constellations seem to rotate
around one star. This apparent rotation is a result o the
rotation o the Earth about its own axis.
On top o this nightly rotation, there is a slow change in
the stars and constellations that are visible rom one night
to the next. This variation over the period o one year is due
to the rotation o the Earth about the Sun.
Planetary systems have been discovered around many stars.
View from place to place on earth
I you move rom place to place around the Earth, the section
o the night sky that is visible over a year changes with
latitude. The total pattern o the constellations is always the
same, but you will see dierent sections o the pattern.
190
O p t i O n D A s t r O p h ys i c s
objcs vs (2)
During one Day
The most important observation is that the pattern o the
stars remains the same rom one night to the next. Patterns
o stars have been identifed and 88 dierent regions o the
sky have been labelled as the dierent constellations. A
particular pattern is not always in the same place, however.
The constellations appear to move over the period o one night.
They appear to rotate around one direction. In the Northern
Hemisphere everything seems to rotate about the pole star.
It is common to reer measurements to the fxed stars the
patterns o the constellations. The fxed background o stars
always appears to rotate around the pole star. During the night,
some stars rise above the horizon and some stars set beneath it.
maximum height at midday. At this time in the Northern
Hemisphere the S un is in a southerly direction.
Sun
looking south
east
west
During the year
Every night, the constellations have the same relative positions
to each other, but the location o the pole star (and thus the
portion o the night sky that is visible above the horizon)
changes slightly rom night to night. Over the period o a year
this slow change returns back to exactly the same position.
pole star
west
looking north
east
The same movement is continued during the day. The
S un rises in the east and sets in the west, reaching its
The Sun continues to rise in the east and set in the west, but
as the year goes rom winter into summer, the arc gets bigger
and the Sun climbs higher in the sky.
unitS
When comparing distances on the astronomical scale, it can be quite unhelpul to remain in SI units. Possible other units include
the astronomical unit (AU) , the parsec (pc) or the light year (ly) . See page 1 93 or the defnition o the frst two o these.
The light year is the distance travelled by light in one year (9.5 1 0 1 5 m) . The next nearest star to our Sun is about 4 light years
away. Our galaxy is about 1 00,000 light years across. The nearest galaxy is about a million light years away and the observable
Universe is 1 3.7 billion light years in any given direction.
the milky way galaxy
When observing the night sky a aint
band o light can be seen crossing the
constellations. This path (or way)
across the night sky became known as
the Milky Way. What you are actually
seeing is some o the millions o stars
that make up our own galaxy but they
are too ar away to be seen as individual
stars. The reason that they appear to be
in a band is that our galaxy has a
spiral shape.
The centre o our galaxy lies in the
direction o the constellation Sagittarius.
The galaxy is rotating all the stars
are orbiting the centre o the galaxy as
a result o their mutual gravitational
attraction. The period o orbit is about
250 million years.
plan view
side view
100 000 light years
25 000 light years central bulge
galactic nucleus
Sun
Sun
direction
of
rotation
disc
globular clusters
The Milky Way galaxy
the uniVerSe
Stars are grouped together in stellar clusters. These
can be open containing 1 0 3 stars e.g. located in the disc
o our galaxy or globular containing 1 0 5 stars. Our Sun
is just one o the billions o stars in our galaxy (the
Milky Way galaxy) . The galaxy rotates with a period o
about 2.5 1 0 8 years.
Beyond our galaxy, there are billions o other galaxies.
Some o them are grouped together into clusters or
super clusters o galaxies, but the vast majority o
space (like the gaps between the planets or between
stars) appears to be empty essentially a vacuum.
Everything together is known as the Universe.
1 .5 1 0 2 6 m
(= 1 5 billion light years)
the visible
Universe
5 1 02 2 m
(= 5 million light years)
local group
o galaxies
1 02 1 m
(= 1 00,000 light years)
our galaxy
1 01 3 m
(= 0.001 light years)
our Solar
System
O p t i O n D A s t r O p h ys i c s
191
th ss
energy flow for StarS
equilibrium
The stars are emitting a great deal o energy. The source or all this
energy is the usion o hydrogen into helium. See page 1 96. Sometimes
this is reerred to as hydrogen burning but it this is not a precise
term. The reaction is a nuclear reaction, not a chemical one (such as
combustion) . Overall the reaction is
The Sun has been radiating energy or the past 4
billion years. It might be imagined that the powerul
reactions in the core should have orced away the
outer layers o the Sun a long time ago. Like other
stars, the Sun is stable because there is a hydrostatic
equilibrium between this outward pressure and the
inward gravitational orce (see page 1 64) .
4 11 p 42 He + 2 01 e + + 2
The mass o the products is less than the mass o the reactants. Using
E = m c2 we can work out that the Sun is losing mass at a rate o
4 1 0 9 kg s - 1 . This takes place in the core o a star. Eventually all this
energy is radiated rom the surace approximately 1 0 2 6 J every second.
The structure inside a star does not need to be known in detail.
outward radiation
pressure
convective zone
inward pull
of gravity
surface
radiative
zone
core
(nuclear reactions)
binary StarS
Our Sun is a single star. Many stars actually turn out to be two
(or more) stars in orbit around each other. (To be precise they
orbit around their common centre o mass.) These are called
binary stars.
A stable star is in equilibrium
B
observer
night 12
centre of mass
binary stars two stars in orbit around
their common centre of mass
2.
A spectroscopic binary star is identifed rom the analysis
o the spectrum o light rom the star. Over time the
wavelengths show a periodic shit or splitting in requency.
An example o this is shown (below) .
wavelength
Both stars are
moving at 90
to observer
Star A is moving
towards observer
B
whereas star B is
moving away
light from B
will be red shifted
A
observer
3.
An eclipsing binary star is identifed rom the analysis o the
brightness o the light rom the star. Over time the brightness
shows a periodic variation. An example o this is shown below.
brightness
A visual binary is one that can be distinguished as two
separate stars using a telescope.
A
light from A
will be blue shifted observer
B
night 24
There are dierent categories o binary star visual,
spectroscopic and eclipsing.
1.
A
night 0
night 0
time (nights)
night 12
AB
ABAB
Each wavelength
splits into two
A B separate
wavelengths.
star B
When one star blocks the light
coming from the other star, the
overall brightness is reduced
night 24
The explanation or the shit in requencies involves the Doppler
eect. As a result o its orbit, the stars are sometimes moving
towards the Earth and sometimes they are moving away. When
a star is moving towards the Earth, its spectrum will be blue
shited. When it is moving away, it will be red shited.
192
The explanation or the dip in brightness is that as a result
o its orbit, one star gets in ront o the other. I the stars
are o equal brightness, then this would cause the total
brightness to drop to 50% .
O p t i O n D A s t r O p h ys i c s
star A
observer
S
principleS of meaSurement
mathematicS unitS
As you move rom one position to another objects change
their relative positions. As ar as you are concerned, near
objects appear to move when compared with ar objects.
Objects that are very ar away do not appear to move at
all. You can demonstrate this eect by closing one eye and
moving your head rom side to side. An object that is near to
you (or example the tip o your fnger) will appear to move
when compared with objects that are ar away (or example a
distant building) .
The situation that gives rise to a change in apparent position
or close stars is shown below.
This apparent movement is known as parallax and the eect
can used to measure the distance to some o the stars in our
galaxy. All stars appear to move over the period o a night, but
some stars appear to move in relation to other stars over the
period o a year.
distant stars
close star
stellar
distance
d
parallax angle
If ca refu lly obser ved , over th e perio d
of a y ea r som e sta rs ca n appea r to m ove
be tween two extrem es.
Earth
(July)
Sun
Earth (January)
1 AU
orbit of Earth
The reason or this apparent movement is that the Earth has
moved over the period o a year. This change in observing
position has meant that a close star will have an apparent
movement when compared with a more distant set o stars.
The closer a star is to the Earth, the greater will be the
parallax shit.
The parallax angle, , can be measured by observing the changes
in a stars position over the period o a year. From trigonometry,
i we know the distance rom the Earth to the Sun, we can work
out the distance rom the Earth to the star, since
Since all stars are very distant, this eect is a very small one
and the parallax angle will be very small. It is usual to quote
parallax angles not in degrees, but in seconds. An angle o 1
second o arc (' ' ) is equal to one sixtieth o 1 minute o arc (' )
and 1 minute o arc is equal to one sixtieth o a degree.
Since is a very small angle, tan sin (in radians)
In terms o angles, 3600' ' = 1
360 = 1 ull circle.
example
The star alpha Eridani (Achemar) is 1 .32 1 0 1 8 m away.
Calculate its parallax angle.
d = 1 .32 1 0 1 8 m
1 .32 1 0 1 8 pc
= ___________
3.08 1 0 1 6
= 42.9 pc
1
parallax angle = _____
42.9
= 0.02' '
(distance rom Earth to Sun)
tan = __________________________
(distance rom Sun to Star)
1
This means that __________________________
(distance rom Earth to star)
In other words, parallax angle and distance away are inversely
proportional. I we use the right units we can end up with a
very simple relationship. The units are defned as ollows.
The distance rom the Sun to the Earth is defned to be one
astronomical unit (AU). It is 1 .5 1 0 1 1 m. Calculations
show that a star with a parallax angle o exactly one second
o arc must be 3.08 1 0 1 6 m away (3.26 light years) . This
distance is defned to be one parsec (pc). The name parsec
represents parallax angle o one second.
I distance = 1 pc, = 1 second
I distance = 2 pc, = 0.5 second etc.
1
Or, distance in pc = _________________________
(parallax angle in seconds)
1
d = __
p
The parallax method can be used to measure stellar distances
that are less than about 100 parsecs. The parallax angle
or stars that are at greater distances becomes too small to
measure accurately. It is common, however, to continue to
use the unit. The standard SI prefxes can also be used even
though it is not strictly an SI unit.
1 000 parsecs = 1 kpc
1 0 6 parsecs = 1 Mpc etc.
O p t i O n D A s t r O p h ys i c s
193
ls
luminoSity anD apparent brightneSS
The total power radiated by a star is called its luminosity (L) .
The SI units are watts. This is very different to the power
received by an observer on the Earth. The power received per
unit area is called the apparent brightness of the star. The SI
units are W m- 2 .
If two stars were at the same distance away from the Earth
then the one with the greater luminosity would be brighter.
Stars are, however, at different distances from the Earth. The
brightness is inversely proportional to the (distance) 2 .
area 4A
It is thus possible for two very different stars to have the same
apparent brightness. It all depends on how far away the stars are.
close star
( small luminosity)
distant star
(high luminosity)
Two stars can have the same apparent brightness even if they
have different luminosities
area A
alternatiVe unitS
x
x
As d ista n ce in crea ses, th e brigh tn ess d ecrea ses sin ce
the ligh t is spread over a bigger a rea .
d ista n ce
x
The magnitude scale can also be used to compare the luminosity
of different stars, provided the distance to the star is taken into
account. Astronomers quote values of absolute magnitude in
order to compare luminosities on a familiar scale.
brigh tn ess
b
2x
3x
4x
5x
a n d so on
b
4
b
9
b
16
b
25
The SI units for luminosity and brightness have already
been introduced. In practice astronomers often compare the
brightness of stars using the apparent magnitude scale.
A magnitude 1 star is brighter than a magnitude 3 star. This
measure of brightness is sometimes shown on star maps.
inverse sq u a re
L
apparent brightness b = _____
4r2
example on luminoSity
black-boDy raDiation
The star Betelgeuse has a parallax angle of 7.7 1 0 - 3 arc
seconds and an apparent brightness of 2.0 1 0- 7 W m- 2 .
Calculate its luminosity.
Stars can be analysed as perfect emitters, or black bodies. The
luminosity of a star is related to its brightness, surface area
and temperature according to the StefanBoltzmann law.
Wiens law can be used to relate the wavelength at which the
intensity is a maximum to its temperature. See page 90 for
more details.
Distance to Betelgeuse d
1
= __
p
1
__________
=
pc
7.7 1 0 - 3
= 1 29.9 pc
= 1 29.9 3.08 1 0 1 6 m
= 4.0 1 0 1 8 m
L = b 4d2 = 4.0 1 0 3 1 W
194
O p t i O n D A s t r O p h ys i c s
Example:
e.g. our suns temperature is 5,800k
So the wavelength at which the intensity of its radiation is at a
2.9 1 0 - 3 = 500 nm
maximum is m a x = __________
5800
S s
abSorption lineS
The radiation from stars is not a perfect continuous spectrum there are particular wavelengths that are missing.
bands of wavelengths
emitted by the Sun
missing wavelength
wavelength
red
violet
The missing wavelengths correspond to the absorption spectra
of a number of elements. Although it seems sensible to assume
that the elements concerned are in the Earths atmosphere, this
assumption is incorrect. The wavelengths would still be absent if
light from the star was analysed in space.
A star that is moving relative to the Earth will show a Doppler
shift in its absorption spectrum. Light from stars that are
receding will be red shifted whereas light from approaching
stars will be blue shifted.
The absorption is taking place in the outer layers of the star.
This means that we have a way of telling what elements exist in
the star at least in its outer layers.
claSSification of StarS
Stefanboltzmann law
Different stars give out different spectra of light. This allows
us to classify stars by their spectral class. Stars that emit
the same type of spectrum are allocated to the same spectral
class. Historically these were just given a different letter, but
we now know that these different letters also correspond to
different surface temperatures.
The StefanBoltzmann law links the total power radiated by
a black body (per unit area) to the temperature of the black
body. The important relationship is that
The seven main spectral classes (in order of decreasing
surface temperature) are O, B, A, F, G, K and M. The main
spectral classes can be subdivided.
Class
Effective surface temperature/K
Colour
Total power radiated T4
In symbols we have,
Total power radiated = A T 4
Where
is a constant called the StefanBoltzmann constant.
= 5.67 1 0 - 8 W m- 2 K- 4
A is the surface area of the emitter (in m2 )
T is the absolute temperature of the emitter (in kelvin)
e.g. The radius of the Sun = 6.96 1 0 8 m.
O
30,00050,000
blue
B
1 0,00030,000
blue-white
A
7,5001 0,000
white
Surface area
= 4 r 2 = 6.09 1 0 1 0 m2
= 5800 K
F
6,0007,500
yellow-white
If temperature
G
5,2006,000
yellow
then total power radiated = A T 4
K
3,7005,200
orange
M
2,4003,700
red
Spectral classes do not need to be mentioned but are used in
many text books.
Summary
If we know the distance to a star we can analyse the light from
the star and work out:
the chemical composition (by analysing the absorption spectrum)
= 5.67 1 0 - 8 6.09 1 0 1 8
(5800 4 )
= 3.9 1 0 2 6 W
The radius of the star r is linked to its surface area, A, using
the equation A = 4 r 2 .
the luminosity (using measurements of the brightness and
the distance away)
the surface area of the star (using the luminosity, the surface
temperature and the StefanBoltzmann law) .
the surface temperature (using a measurement of m a x and
Wiens law see page 90)
O p t i O n D A s t r O p h ys i c s
195
nsss
Stellar typeS anD black holeS
The source o energy or our Sun is the usion o hydrogen into helium. This is also true or many other stars.
There are however, other types o object that are known to exist in the Universe.
Type of object
Description
Red giant stars
As the name suggests, these stars are large in size and red in colour. Since they are red, they are
comparatively cool. They turn out to be one o the later possible stages or a star. The source o energy is
the usion o some elements other than hydrogen. Red supergiants are even larger.
White dwarf stars
As the name suggests, these stars are small in size and white in colour. Since they are white, they are
comparatively hot. They turn out to be one o the fnal stages or some smaller mass stars. Fusion is no
longer taking place, and a white dwar is just a hot remnant that is cooling down. Eventually it will cease
to give out light when it becomes sufciently cold. It is then known as a brown dwarf.
Cepheid variables
These are stars that are a little unstable. They are observed to have a regular variation in brightness and
hence luminosity. This is thought to be due to an oscillation in the size o the star. They are quite rare but
are very useul as there is a link between the period o brightness variation and their average luminosity.
This means that astronomers can use them to help calculate the distance to some galaxies.
Neutron stars
Neutron stars are the post-supernova remnants o some larger mass stars. The gravitational pressure has
orced a total collapse and the mass o a neutron star is not composed o atoms it is essentially composed
o neutrons. The density o a neutron star is enormous. Rotating neutron stars have been identifed as
pulsars.
Black holes
Black holes are the post-supernova remnant o larger mass stars. There is no known mechanism to stop
the gravitational collapse. The result is an object whose escape velocity is greater than the speed o light.
See page 1 50.
main Sequence StarS
The general name or the creation o
nuclei o dierent elements as a result
o fssion reactions is nucleosynthesis.
Details o how this overall reaction
takes place in the Sun do not need to
be recalled by SL candidates, but HL
candidates do need this inormation.
One process is known as the proton
proton cycle or pp cycle.
1
1H
step 1
+ 11 H 21 H + 01 e + + 00
e+
p
In order or any o these reactions to take
place, two positively charged particles
(hydrogen or helium nuclei) need to
come close enough or interactions to
take place. Obviously they will repel one
another.
This means that they must be at a high
temperature.
I a large cloud o hydrogen is hot
enough, then these nuclear reactions
can take place spontaneously. The power
radiated by the star is balanced by the
power released in these reactions the
temperature is eectively constant.
The star remains a stable size because
the outward pressure o the radiation is
balanced by the inward gravitational pull.
But how did the cloud o gas get to be
at a high temperature in the frst place?
As the cloud comes together, the loss
o gravitational potential energy must
mean an increase in kinetic energy and
hence temperature. In simple terms the
gas molecules speed up as they all in
towards the centre to orm a proto-star.
Once ignition has taken place, the star
can remain stable or billions o years.
See page 205 or more details.
n
p
p
step 2
n
p
2
1H
+ 11 H 32 He + 00
Fg
n p
p
p
step 3
p pn
p pn
Fg
3
2 He
Fg
collapse of cloud
under gravity gives
molecular KE
+ 32 He 42 He + 2 11 p
n p
pn
p
p
the proton-proton cycle (p-p cycle)
196
Fg
O p t i O n D A s t r O p h ys i c s
cloud of gas
With sucient KE,
nuclear reactions
can take place.
t hzsprss d
hr Diagram
The point of classifying the various types of stars is to see if any patterns exist. A useful way of making this comparison is the
HertzsprungRussell diagram. Each dot on the diagram represents a different star. The following axes are used to position
the dot.
The vertical axis is the luminosity of the star as compared with the luminosity of the Sun. It should be noted that the scale is
logarithmic.
The horizontal axis a scale of decreasing temperature. Once again, the scale is not a linear one. (It is also the spectral class of
the star OBAFGKM)
The result of such a plot is shown below.
10 6
10 2
our Sun
10 0
luminosity/L.
10 4
0
00
0
3
4
50
0
6
7
00
0
50
00
10
25
50
00
00
0
0
0
10 -2
10 -4
surface
temperature/K
A large number of stars fall on a line that (roughly) goes from top left to bottom right. This line is known as the main sequence and
stars that are on it are known as main sequence stars. Our Sun is a main sequence star. These stars are normal stable stars the only
difference between them is their mass. They are fusing hydrogen to helium. The stars that are not on the main sequence can also be
broadly put into categories.
In addition to the broad regions, lines of constant radius can be added to show the size of stars in comparison to our Suns radius.
These are lines going from top left to bottom right.
10 2
10 s
ol a r
1 so
ola r
10 - 2
10 3
red su pergia n ts
r ra d
ii
ra d i i
l a r ra
0 .1 s
sola
sol a
r ra d
ii
red gia n ts
instability
strip
d ius
ra d i u
sol a
s
r ra d
m a in
seq u en ce
Su n
ius
0
3
00
0
6
00
0
00
10
50
00
0
wh ite
d wa rfs
eective
temperature/K
maSS-luminoSity relation for main Sequence StarS
For stars on the main sequence, there is a correlation between the star' s mass, M, and its luminosity, L. Stars that are brighter on
the main sequence (i.e. higher up) are more massive and the relationship is:
L M3 . 5
O p t i O n D A s t r O p h ys i c s
197
cd vbs
principleS
mathematicS
Very small parallax angles can be measured using
satellite observations (e.g. Gaia mission) but even
these measurement are limited to stars that are about
1 00 kpc away. The essential difculty is that when we
observe the light rom a very distant star, we do not
know the dierence between a bright source that is
ar away and a dimmer source that is closer. This is the
principal problem in the experimental determination
o astronomical distances to other galaxies.
The process o estimating the distance to a galaxy (in which the
individual stars can be imaged) might be as ollows:
A Cepheid variable star is quite a rare type o star.
Its outer layers undergo a periodic compression and
contraction and this produces a periodic variation in
its luminosity.
A Cepheid variable star undergoes
periodic compressions and
contractions.
increased
luminosity
lower
luminosity
Use the luminosityperiod relationship or Cepheids to estimate the
average luminosity.
apparent brightness
Use the average luminosity, the average brightness and the inverse
square law to estimate the distance to the star.
O
1
2
3
4 5
6
7
8
10 6
10 5
10 4
10 3
10 2
10
1
1
2
5
10 20 50 100 period / days
General luminosityperiod graph
These stars are useul to astronomers because the
period o this variation in luminosity turns out to
be related to the average absolute magnitude o the
Cepheid. Thus the luminosity o a Cepheid can be
calculated by observing the variations in brightness.
example
A Cepheid variable star has a period o 1 0.0 days and apparent peak brightness o 6.34 1 0 - 1 1 W m- 2
The luminosity o the Sun is 3.8 1 0 2 6 W. Calculate the distance to the Cepheid variable in pc.
Using the luminosityperiod graph (above)
peak luminosity = 1 0 3 . 7 L s u n = 501 2 3.8 1 0 2 6 = 1 .90 1 0 3 0 W
L=b
4r2
____
L
r = ____
4b
_________________
1 .90 1 0 3 0
= ____________________
4 6.34 1 0 - 1 1
= 4.88 1 0 1 9 m
4.88 1 0 1 9 pc
= ___________
3.08 1 0 1 6
= 1 590 pc
198
O p t i O n D A s t r O p h ys i c s
9 10 11 time/days
Variation o apparent magnitude or a particular Cepheid variable
peak luminosity/L SUN
When we observe another galaxy, all o the stars
in that galaxy are approximately the same distance
away rom the Earth. What we really need is a light
source o known luminosity in the galaxy. I we had
this then we could make comparisons with the other
stars and judge their luminosities. In other words we
need a standard candle that is a star o known
luminosity. Cepheid variable stars provide such a
standard candle.
Locate a Cepheid variable in the galaxy.
Measure the variation in brightness over a given period o time.
rd g ss
after the main Sequence
The massluminosity relation (page 1 97) can be used to compare
the amount o time dierent mass stars take beore the hydrogen
uel is used. Consider a star that is 1 0 times more massive than our
Sun. This means that the luminosity o the larger star will be (1 0) 3 . 5
= 3,1 62 times more luminous that our Sun. Since the source o
this luminosity is the mass o hydrogen in the star, then the larger
star eectively has 1 0 times more uel but is using the uel at
more than 3000 times the rate. The more massive star will fnish
1
its uel in ___
o the time. A star that has more mass exists or a
300
shorter amount o time.
I it has sufcient mass, a red giant can continue to use
higher and higher elements and the process o nucleosynthesis
can continue.
newly formed red giant star
helium-burning
shell
A star cannot continue in its main sequence state orever. It is
using hydrogen into helium and at some point hydrogen in
the core will become rare. The usion reactions will happen
less oten. This means that the star is no longer in equilibrium
and the gravitational orce will, once again, cause the core to
collapse.
This collapse increases the temperature o the core still
urther and helium usion is now possible. The net result is or
the star to increase massively in size this expansion means
that the outer layers are cooler. It becomes a red giant star.
red
giant star
dormant hydrogenburning shell
400 million km
carbonoxygen core
core of star
nucleosynthesis
old, high-mass red giant star
700 million km
hydrogen-burning shell
helium-burning shell
carbon-burning shell
neon-burning shell
oxygen-burning shell
silicon-burning shell
iron core
star runs out
of hydrogen
collapses
further
helium fusion possible
due to increased
temperature
expansion
This process o usion as a source o energy must come to an
end with the nucleosynthesis o iron. The iron nucleus has one
o the greatest binding energies per nucleon o all nuclei. In
other words the usion o iron to orm a higher mass nucleus
would need to take in energy rather than release energy. The
star cannot continue to shine. What happens next is outlined
on the ollowing page.
O p t i O n D A s t r O p h ys i c s
199
S v
Page 199 showed that the red giant phase or a star must eventually
come to an end. There are essentially two possible routes with
dierent fnal states. The route that is ollowed depends on
the initial mass o the star and thus the mass o the remnant
that the red giant star leaves behind: with no urther nuclear
reactions taking place gravitational orces continue the
collapse o the remnant. An important critical mass is called
the Chandrasekhar limit and it is equal to approximately
1 .4 times the mass o our Sun. Below this limit a process
called electron degeneracy pressure prevents the urther
collapse o the remnant.
h r Diagram interpretation
All o the possible evolutionary paths or stars that have been
described here can be represented on a H R diagram. A
common mistake in examinations is or candidates to imply
that a star somehow moves along the line that represents the
main sequence. It does not. Once ormed it stays at a stable
luminosity and spectral class i.e. it is represented by one
fxed point in the H R diagram.
evolution of a low-mass star
luminosity
poSSible fateS for a Star (after reD giant
phaSeS)
I a star has a mass less than 4 Solar masses, its remnant
will be less than 1 .4 Solar masses and so it is below the
Chandrasekhar limit. In this case the red giant orms a
planetary nebula and becomes a white dwarf which
ultimately becomes invisible. The name planetary nebula is
another term that could cause conusion. The ejected material
would not be planets in the same sense as the planets in our
Solar System.
ejection of
planetary nebula
to white dwarf
ma
in s
equ
red giant
phase
en c
e
surface temperature
planetary nebula
luminosity
evolution of a high-mass star
low-mass star
(e.g. type G)
to black hole / neutron star
white dwarf
red giant
red giant
phase
I a star is greater than 4 Solar masses, its remnant will
have a mass greater than 1 .4 Solar masses. It is above the
Chandrasekhar limit and electron degeneracy pressure is not
sufcient to prevent collapse. In this case the red supergiant
experiences a supernova. It then becomes a neutron star or
collapses to a black hole. The fnal state again depends on mass.
ma i
n se
que
n ce
surface temperature
pulSarS anD quaSarS
larger-mass
star (e.g. type
A, B, O)
very large-mass
supernova
red supergiant
black hole
large-mass
supernova
neutron star
A neutron star is stable due to neutron degeneracy pressure.
It should be emphasized that white dwars and neutron stars
do not have a source o energy to uel their radiation. They
must be losing temperature all the time. The act that these
stars can still exist or many millions o years shows that the
temperatures and masses involved are enormous. The largest
mass a neutron star can have is called the Oppenheimer
Volkoff limit and is 23 Solar masses. Remnants above this
limit will orm black holes.
200
O p t i O n D A s t r O p h ys i c s
Pulsars are cosmic sources o very weak radio wave energy
that pulsate at a very rapid and precise requency. These have
now been theoretically linked to rotating neutron stars. A
rotating neutron star would be expected to emit an intense
beam o radio waves in a specifc direction. As a result o
the stars rotation, this beam moves around and causes the
pulsation that we receive on Earth.
Quasi-stellar objects or quasars appear to be point-like sources
o light and radio waves that are very ar away. Their red
shits are very large indeed, which places them at the limits
o our observations o the Universe. I they are indeed at
this distance they must be emitting a great deal o power or
their size (approximately 1 0 4 0 W! ) . The process by which this
energy is released is not well understood, but some theoretical
models have been developed that rely on the existence o
super-massive black holes. The energy radiated is as a result o
whole stars alling into the black hole.
t b b dl
expanSion of the uniVerSe
I a galaxy is moving away rom
the Earth, the light rom it will
be red shited. The surprising
act is that light rom almost all
galaxies shows red shits
almost all o them are moving
away rom us. The Universe is
expanding.
At frst sight, this expansion
seems to suggest that we are in
the middle o the Universe, but
this is a mistake. We only seem
to be in the middle because it
was we who worked out the
velocities o the other galaxies.
I we imagine being in a
dierent galaxy, we would get
exactly the same picture o the
Universe.
vA
the uniVerSe in the
paSt the big bang
vD
A
B
D
us (at rest)
C
vC
vB
As far as we are concerned, most galaxies
are moving away from us.
vA
This point, the creation o the Universe,
is known as the Big Bang. It pictures all
the matter in the Universe being crushed
together (very high density) and being very
hot indeed. Since the Big Bang, the Universe
has been expanding which means that, on
average, the temperature and density o the
Universe have been decreasing. The rate o
expansion would be expected to decrease
as a result o the gravitational attraction
between all the masses in the Universe.
vD
vus
D
A
vC
B (at rest) C
Any galaxy would see all the other galaxies
moving away from it.
A good way to picture this expansion is to think o the Universe as a sheet o
rubber stretching o into the distance. The galaxies are placed on this huge
sheet. I the tension in the rubber is increased, everything on the sheet moves
away rom everything else.
As the section of rubber sheet expands, everything moves away from everything else.
A urther piece o evidence or the Big
Bang model came with the discovery o
the Cosmic microwave background
(CMB) radiation by Penzias and Wilson.
They discovered that microwave radiation
was coming towards us rom all directions
in space. The strange thing was that the
radiation was the same in all directions
(isotopic) and did not seem to be linked
to a source. Further analysis showed that
this radiation was a very good match
to theoretical black-body radiation
produced by an extremely cold object a
temperature o just 2.73 K.
This is in perect agreement with the
predictions o Big Bang. There are two
ways o understanding this.
1.
All objects give out electromagnetic
radiation. The requencies can be
predicted using the theoretical
model o black-body radiation. The
background radiation is the radiation
rom the Universe itsel which has
now cooled down to an average
temperature o 2.73 K.
Intensity per unit wavelength range
coSmic microwaVe
backgrounD raDiation
I the Universe is currently expanding, at
some time in the past all the galaxies would
have been closer together. I we examine
the current expansion in detail we fnd that
all the matter in the observable universe
would have been together at the SAME
point approximately 1 5 billion years ago.
Note that this model does not attempt to
explain how the Universe was created, or by
Whom. All it does is analyse what happened
ater this creation took place. The best way
to imagine the expansion is to think o the
expansion o space itsel rather than the
galaxies expanding into a void. The Big Bang
was the creation o space and time. Einsteins
theory o relativity links the measurements
o space and time so properly we need to
imagine the Big Bang as the creation o space
and time. It does not make sense to ask
about what happened beore the Big Bang,
because the notion o beore and ater (i.e.
time itsel) was created in the Big Bang.
2.
Some time ater the Big Bang,
radiation became able to travel through
the Universe (see page 21 0 or details).
It has been travelling towards
us all this time. During this time
the Universe has expanded this
means that the wavelength o this
radiation will have increased (space
has stretched) . See page 21 0 or
anisotropies in the CMB.
peak wavelength ~1.1 mm (microwaves)
individual data
points for background
microwave radiation
theoretical
line for 2.73 K
black-body
radiation
wavelength
O p t i O n D A s t r O p h ys i c s
201
g
DiStributionS of galaxieS
Galaxies are not distributed randomly throughout space. They tend to be ound
clustered together. For example, in the region o the Milky Way there are twenty
or so galaxies in less than 2.5 million light years.
The Virgo galactic cluster (50 million light years away rom us) has over 1 ,000 galaxies
in a region 7 million light years across. On an even larger scale, the galactic clusters
are grouped into huge superclusters o galaxies. In general, these superclusters oten
involve galaxies arranged together in joined flaments (or bands) that are arranged as
though randomly throughout empty space.
motion of galaxieS
As has been seen on page 201 it is a surprising observational act that the vast
majority o galaxies are moving away rom us. The general trend is that the more
distant galaxies are moving away at a greater speed as the Universe expands. This
does not, however, mean that we are at the centre o the Universe this would be
observed wherever we are located in the Universe.
As explained on page 201 , a good way to imagine this expansion is to think o space
itsel expanding. It is the expansion o space (as opposed to the motion o the galaxies
through space) that results in the galaxies relative velocities. In this model, the red shit
o light can be thought o as the expansion o the wavelength due to the stretching
o space.
time and expansion of Universe
light wave as emitted from a distant galaxy
light wave when it
arrives at Earth
mathematicS
Examle
I a star or a galaxy moves away rom us, then the wavelength
o the light will be altered as predicted by the Doppler eect (see
page 1 02) . I a galaxy is going away rom the Earth, the speed
o the galaxy with respect to an observer on the Earth can be
calculated rom the red shit o the light rom the galaxy. As long
as the velocity is small when compared with the velocity o light,
a simplifed red shit equation can be used.
A characteristic absorption line oten seen in stars is due to
ionized helium. It occurs at 468.6 nm. I the spectrum o a star
has this line at a measured wavelength o 499.3 nm, what is the
recession speed o the star?
v
Z = ___
__
c
0
Where
0
= 6.55 1 0 2
v = 6.55 1 0 - 2 3 1 0 8 m s - 1
= 1 .97 1 0 7 m s - 1
= change in wavelength o observed light (positive i
wavelength is increased)
0 = wavelength o light emitted
v = relative velocity o source o light
c = speed o light
Z = red shit.
202
(499.3 468.6)
Z = ___
= _______________
468.6
O p t i O n D A s t r O p h ys i c s
hs w d s s
experimental obSerVationS
hiStory of the uniVerSe
Although the uncertainties are large, the general trend or
galaxies is that the recessional velocity is proportional to the
distance away rom Earth. This is Hubbles law.
I a galaxy is at a distance x, then Hubbles law predicts its
velocity to be H0 x. I it has been travelling at this constant
speed since the beginning o the Universe, then the time that
has elapsed can be calculated rom
recessional velocity / km s -1
10 000
distance
Time = ________
speed
x
= ____
H0 x
8 000
6 000
1
= ___
H0
4 000
This is an upper limit or the age o the Universe. The
gravitational attraction between galaxies predicts that the
speed o recession decreases all the time.
2 000
0
0
20 40 60 80 100 120 140
distance / Mpc
size of R
observable
universe
Mathematically this is expressed as
v d
or
v = H0 d
T 1
H0
where H0 is a constant known as the Hubble constant. The
uncertainties in the data mean that the value o H0 is not
known to any degree o precision. The SI units o the Hubble
constant are s - 1 , but the unit o km s - 1 Mpc- 1 is oten used.
1
H0
time
now
the coSmic Scale factor (R)
Page 202 shows how the Doppler red shit equation, = ___
_vc , can be used to calculate the recessional velocity, v, o certain
0
galaxies. This equation can only be used when v c or in other words, the recessional velocity, v, has to be small in comparison to
the speed o light, c. There are however plenty o objects in the night sky (e.g. quasars) or which the observed red shit, z, is greater
than 1 .0. This implies that their speed o recession is greater than the speed o light. In these situations it is helpul to consider a
quantity called the cosmic scale actor (R) .
As introduced on page 201 , the expansion o the Universe is best pictured as the expansion o space itsel. The expansion o the
Universe means that a measurement undertaken at some time in the distant past, or example the wavelength o light emitted by an
object 1 0 million years ago, will be stretched and will be recorded as a larger value when measured now. All measurement will be
stretched over time and this can be considered as a rescaling o the Universe (the Universe getting bigger) .
The cosmic scale actor, R, is a way o quantiying the expansion that has taken place. In the above example, i the wavelength
was emitted 1 0 million years ago with wavelength 0 when the scale actor was R0 , the wavelength measured today would have
increased by to a larger value ( = 0 + ) . This is because the cosmic scale actor has increased by R (to the larger value
R
. The ratio o the measured wavelengths, __
, is equal to the
R = R0 + R) . All measurements will have increased by the ratio, __
R
R
__
ratio o the cosmic scale actors, R , so the red shit ratio, z is given by:
0
0
0
- 0
= _____
- 1 = ___
R -1
z = ____
= ___
R0
0
0
0
R - 1
or z = ___
R0
R
So a measured red shit o 4 means that __
= 5. I we consider R to be the present size o the observable Universe, then the light
R
must have been emitted when the Universe was one fth o its current size.
0
O p t i O n D A s t r O p h ys i c s
203
t vs
SupernoVae anD the accelerating uniVerSe
Supernovae are catastrophic explosions that can occur in the development o some stars (see page 200) . Supernovae are rare events
(the last one to occur in our galaxy took place in 1 604) but the large number o stars in the Universe means that many have been
observed. An observer on the Earth sees a rapid increase in brightness (hence the word nova = new star) which then diminishes
over a period o some weeks or months. Huge amounts o radiated energy are emitted in a short period o time and, at its peak, the
apparent brightness o a single supernova oten exceeds many local stars or individual galaxies.
Supernovae have been categorized into two dierent main types (see page 207 or more details) according to a spectral analysis
o the light that they emit. The light rom a type II supernova indicates the presence o hydrogen (rom the absorption spectra)
whereas there is no hydrogen in a type I supernova. There are urther subdivisions o these types (Ia, Ib, etc.) based on dierent
aspects o the light spectrum.
Type Ia supernovae are explosions involving white dwar stars. When these events take place, the amount o energy released can
be predicted accurately and these supernovae can be used as standard candles. By comparing the known luminosity o a type Ia
supernova and its apparent brightness as observed in a given galaxy, a distance measurement to that galaxy can be calcuated. This
technique can be used with galaxies up to approximately 1 ,000 Mpc away.
The expanding Universe (which is consistent with the Big Bang model) means that that the cosmic scale actor, R, is increasing. As
a result o gravitational attraction, we might expect the rate at which R increases to be slowing down. Analysis o a large number
o type Ia supernovae has, however, provided strong evidence that not only is the cosmic scale actor, R, increasing but the rate
at which it increases is getting larger as time passes. In other words the expansion o the Universe is accelerating. The evidence
rom type Ia supernovae identies this eect rom a time when the universe was approximately __23 o its current size. Note that this
acceleration is dierent to the very rapid period o expansion o the early Universe which is called infation.
The mechanisms that cause an accelerating Universe are not ully understood but must involve an outward accelerating orce to
counteract the inward gravitational pull. There must also be a source o energy which has been given the name dark energy
(see page 21 2) .
dark energy
accelerated expansion
dark ages
development of
galaxies
ination
1 st stars
13.7 billion years
Source: NASA/WMAP Science Team
204
O p t i O n D A s t r O p h ys i c s
HL
n s Js
the JeanS criterion
nuclear fuSion
As seen on page 1 96, stars orm out o interstellar clouds o
hydrogen, helium and other materials. Such clouds can exist
in stable equilibrium or many years until an external event
(e.g. a collision with another cloud or the infuence o another
incident such as a supernova) starts the collapse. At any given
point in time, the total energy associated with the gas cloud
can be thought o as a combination o:
A star on the main sequence is using hydrogen nuclei to
produce helium nuclei. One process by which this is achieved
is the protonproton chain as outlined on page 1 96. This is
the predominant method or nuclear usion to take place in
small mass stars (up to just above the mass o our Sun) . An
alternative process, called the CNO (carbonnitrogenoxygen)
process takes place at higher temperatures in larger mass stars.
In this reaction, carbon, nitrogen and oxygen are used as
catalysts to aid the usion o protons into helium nuclei. One
possible cycle is shown below:
The negative gravitational potential energy, EP, which
the cloud possesses as a result o its mass and how it is
distributed in space. Important actors are thus the mass
and the density o the cloud.
4
The positive random kinetic energy, EK , that the particles
in the cloud possess. An important actor is thus the
temperature o the cloud.
The cloud will remain gravitationally bound together i
EP + EK < zero. Using this inormation allows us to predict
that the collapse o an interstellar cloud may begin i its mass
is greater than a certain critical mass, MJ . This is the Jeans
criterion. For a given cloud o gas, MJ is dependent on the
clouds density and temperature and the cloud is more likely
to collapse i it has:
1
START
He
1
H
H
12
15 N
C
13 N
15
O
13 C
large mass
14
N
small size
low temperature.
In symbols, the Jeans criterion is that collapse can start i
M > MJ
1
H
proton
neutron
positron
1
H
gamma ray
neutrino
time Spent on the main Sequence
For so long as a star remains on the main sequence, hydrogen burning is the source o energy that allows the star to remain in
hydrostatic equilibrium (see page 1 92) and have a constant luminosity L. A star that exists on the main sequence or a time TM S
must in total radiate an energy E given by:
E = L TM S
This energy release comes rom the nuclear synthesis that has taken place over its lietime. A certain raction f o the mass o the
star M has been converted into energy according to Einsteins amous relationship:
E = f Mc2
L TM S = f Mc2
f Mc2
TM S = _______
L
But the massluminosity relationship applies, L M3 . 5
M
TM S ____
M3 . 5
TM S M- 2 . 5
Thus the higher the mass of a star, the shorter the lifetime that it spends on the main sequence
(
- 2 .5
) (
Time on main sequence for star A
Mass of star A
____________________________
= ____________
Time on main sequence for star B
Mass of star B
Mass of star B
= ____________
Mass of star A
)
2 .5
For example our Sun is expected to have a main sequence lietime o approximate 1 0 1 0 years. How long would a star with 1 00
times its mass be expected to last?
2 .5
1
= 1 0 5 years
Time on MS or 1 00 solar mass star = 1 0 1 0 ____
1 00
( )
O p t i O n D A s t r O p h ys i c s
205
nsss f s
HL
nucleoSyntheSiS o the main Sequence
For so long as a star remains on the main sequence, hydrogen
burning is the source o energy that allows the star to continue
emitting energy whilst remaining in a stable state. More and
more helium exists in the core. A nuclear synthesis involving
helium (helium burning) does release energy (since the binding
energy per nucleon o the products is greater than that o the
reactants) but can only take place at high temperatures.
For high mass stars, the helium burning process can begin
gradually and spread throughout the core whereas in small
mass stars this process starts suddenly. Whatever the mass o
the star, a new equilibrium state is created: the red giant or red
supergiant phase (see page 200) .
A common process by which helium is converted is a series o
nuclear reactions called the triple alpha process in which
carbon is produced.
1.
Two helium nuclei use into a beryllium nucleus (and a
gamma ray) , releasing energy.
4
2
2.
4
2
8
4
He
Be +
The beryllium nucleus uses with another helium nucleus
to produce a carbon nucleus (and a gamma ray) , releasing
energy.
8
4
3.
He +
Be + 42 He
12
6
C+
Some o the carbon produced in the triple alpha process
can go on to use with another helium nucleus to produce
oxygen. Again this process releases energy:
12
6
C + 42 He
16
8
In high and very high mass stars, gravitational contraction means
that the temperature o the core can continue to rise and more
massive nuclei can continue to be produced. These reactions all
involve the release o energy. Typical reactions include:
Production o neon:
12
6
Production o magnesium:
12
6
Production o oxygen:
12
6
C+
12
6
C
20
10
C+
12
6
C
24
12
C+
12
6
4
2
He
Mg +
16
8
C
Ne +
O + 2 42 He
In addition i the temperatures are high enough, neon and
oxygen burning can occur:
20
10
20
10
Production o sulur:
Ne +
Ne +
16
8
O+
4
2
16
8
16
8
He
O
O+
24
12
32
16
4
2
He
Mg +
S+
Many reactions are possible and other heavy nuclei such as
silicon and phosphorus are also produced. Some o these
alternative nuclear reactions also produce neutrons, which can
easily be captured by other nuclei to orm new isotopes. This
process o neutron capture is explored urther below.
In very high mass stars, silicon burning can also take place
which results in the ormation o iron, 52 66 Fe. As explained on
page 1 99, iron has one o the highest binding energies per
nucleon and represents the largest nucleus that can be created
in a usion process that releases energy. Heavier nuclei can be
acquired, but the reactions require an energy input.
O+
nuclear SyntheSiS o heaVy elementS neutron capture
Many o the reactions that take place in the core o stars also
involve the release o neutrons. Since neutrons are without
any charge, it is easy or them to interact with other nuclei that
are present in the star. When a nucleus captures a neutron, the
resulting nucleus is said to be neutron rich. Given enough
time, most o these neutron-rich nuclei would undergo beta
decay. In this process, the neutron changes into a proton,
emitting an electron and an antineutrino:
1
0
n
1
1
A
Z
X+
1
0
p+
n
0
-1
_
+ v
A + 1
Z
X
A + 1
Z + 1
Y+
0
-1
_
+ v
This is known as slow neutron capture or the s-process. The
overall result o the s-process is a new element. Typically the
206
O p t i O n D A s t r O p h ys i c s
s-process takes place during the helium burning stage o a red
giant star. Typically this means that elements that are heavier
than helium but lighter than iron are able to be created.
The alternative process, rapid neutron capture or the
r-process, takes place when the neutrons are present in
such vast numbers that there is not sufcient time or the
neutron-rich nuclei to undergo beta decay beore several more
neutrons are captured. The result is or very heavy nuclei
to be created. Typically the r-process takes place during the
catastrophic explosion that is a supernova. Elements that are
heavier than iron, such as uranium and thorium, can only be
created in this way at very high temperatures and densities.
HL
tys f sv
SupernoVae
Supernovae are among the most gigantic explosions in the Universe (see page 200) . The two categories o supernova are based
on their light curves a plot o how their brightness varies with time and a spectral analysis o the light that they emit. Type I
supernovae quickly reach a maximum brightness (and an equivalent luminosity o 1 0 1 0 Suns) which then gradually decreases over
time. Type II supernovae oten have lower peak luminosities (equivalent to, say, 1 0 9 Suns) .
Type II
Luminosity
Luminosity
Type I
100
200
300
0
days after maximum brightness
0
100
200
300
days after maximum brightness
400
Supernovae types are distinguish by analysis o their light spectra. All type I supernovae do not include the hydrogen spectrum in
the elements identifed and the dierent subdivisions (Ia, Ib and Ic) are based on a more detailed spectral analysis:
Type Ia shows the presence o singly ionized silicon.
Type Ib shows the presence o non-ionized helium.
Type Ic does not show the presence o helium.
All type II supernovae show the presence o hydrogen. The dierent subdivisions (IIP, IIL, IIn and IIb) again depend on the
presence, or not, o dierent elements.
The reasons or these dierences are the dierent mechanisms that are taking place:
Supernova Type Ia
Supernova Type II
Spectra
Does not show hydrogen but does show singly ionized
silicon.
Shows hydrogen.
Cause
White dwar exploding.
Large mass red giant star collapsing.
Context
Binary star system with white dwar and red giant
orbiting each other.
Large star (greater than 8 Solar masses) at the end
o its lietime, using lighter elements up to the
production o iron.
Process
The gravity feld o the white dwar star attracts
material rom the red giant star, thus increasing the
mass o the white dwar.
When the star runs out o uel, the iron centre core
cannot release any urther energy by nuclear usion.
The star collapses under its own gravity orming a
neutron star.
Explosion
The extra mass gained by the white dwar takes the
total mass o the star beyond the Chandrasekhar
limit (1 .4 Solar masses) or a white dwar. Electron
degeneracy pressure is no longer sufcient to halt
the gravitational collapse. Nuclear usion o heavier
elements (up to iron) starts and the resulting sudden
release o energy causes the star to explode with the
matter being distributed throughout space.
Electron degeneracy pressure is not sufcient to halt
the gravitational collapse o the core, but neutron
degeneracy pressure is and the core becomes a
stable and rigid neutron star. The rest o the inalling
material bounces o the core creating a shock wave
moving outwards. This causes all o the outer layers to
be ejected.
O p t i O n D A s t r O p h ys i c s
207
HL
t s d ds
the coSmological principle
The cosmological principle is a pair of assumptions about the
structure of the Universe upon which current models are based.
The two assumptions are that the Universe, providing one only
considers the large scale structures in the Universe, is isotropic
and homogeneous.
An isotropic universe is one that looks the same in every
direction no particular direction is different to any other. From
the perspective of an observer on Earth, this appears to be a
true statement about the large scale structure of the universe,
but the assumption does not only apply to observers on the
Earth. In an isotropic universe all observers, wherever they
are in the universe, are expected to see the same basic random
distribution of galaxies and galaxy clusters as we do on Earth
and this is true in whatever direction they observe.
A homogeneous universe is one where the local distribution
of galaxies and galaxy clusters that exists in one region of the
universe turns out to be the same distribution in all regions of
the universe. Provided one is considering a reasonably large
rotation curVeS mathematical moDelS
The stars in a galaxy rotate around their common centre of
mass. Different models can be used to predict how the speed
varies with distance from the galactic centre.
1.
section of space (e.g. a sphere of radius equal to several hundreds
of Mpc) , then the number of galaxies in that volume of space
will be effectively the same wherever we choose to look in
the universe. Recent discoveries of apparently very large scale
structures in the Universe cause some astrophysicists to question
the validity of the cosmological principle.
Einstein used the cosmological principle to develop a model of
the Universe in which the Universe was static. He did this by
proposing that the gravitational attraction between galaxies
would be balanced by a yet-to-be-discovered cosmological
repulsion. Subsequent analysis of the equations of general
relativity showed that, if the cosmological principle is correct, the
Universe must be non-static. Hubbles observational discovery
of the expansion of the Universe and the existence of CMB
has meant that many physicists now agree that the Universe is
non-static based around the Big Bang model of an expanding
universe. The cosmological principle is also linked to three
possible models for the future of the Universe (see page 21 1 ) .
The star at a given distance r from the centre will orbit in
circular motion because its centripetal force is provided by the
gravitational attraction:
GMm
mv2
_
= _
2
r
r
Near the galactic centre
GM
v2 = _
r
A simple model to explain the different speeds of rotation of stars
near the galactic centre assumes that density of the galaxy near
its centre, , is constant. A star of mass m feels a resultant force
of gravitational attraction in towards the centre. The value of this
resultant force is the same as if the total mass M of all the stars that
are closer to the galactic centre were concentrated in the centre. An
important point to note is that the net effect of all the stars that are
orbiting at radius that is greater than r sums to zero.
speed
density of stars
in galactic centre =
The total mass of stars that orbit closer than of this star, M, is
given by
4 3
r
M = volume density = _
3
4 3
G_
r
4G
3
v2 = _
= _ r2
r
_____
v=
3
_
4G
r
3
i.e. v r
2.
Far away from the galactic centre
Far away from the galactic centre, observations of the number
of visible stars show that the effective density of the galaxy has
reduced so much that individual stars at these distances can
be considered to be freely orbiting the central mass and to be
unaffected by their neighbouring stars. In this situation,
mass of
star, m
radius r
GM
v2 = _
r where M is the mass of the galaxy
__
i.e. v
stars outside r have
overall no net eect
M
r
m
stars inside r have
eect of total mass M
at centre
208
O p t i O n D A s t r O p h ys i c s
_1r
Comparisons with observations of real galaxies show good
agreement with mathematical model (1 ) but no agreement
with mathematical model (2) . The proposed solution is
discussed on page 209.
HL
r vs d d
an initial linear increase in orbital velocity with distance
within the galactic centre
a fat or slightly increasing curve showing a roughly constant
speed o rotation away rom the galactic centre.
Orbital speed (km/s)
rotation curVeS
Galaxies rotate around their centre o mass and the speeds
o this rotation can be calculated or individual stars rom an
analysis o the stars spectra. A rotation curve or a galaxy
show how this orbital speed varies with distance rom the
galactic centre. Most galaxies show:
350
300
250
200
150
100
50
NGC 4378
NGC 3145
NGC 1620
NGC 7664
5
10
15
20
25
distance from centre of galaxy (kpc)
eViDence for Dark matter
As shown above, observed rotation curves or real galaxies
agree with theoretical models within the galactic centre (v r)
but the orbital velocity o stars is not observed to decrease with
distance away rom the centre as would be expected. Instead,
the orbital velocity is roughly constant whatever the radius. I
the orbital velocity v o a star is constant at dierent values o
radius, then
GM
since v2 = _
r
M
_
=
constant
or M r
r
Thus the total mass that is keeping the star orbiting in its galaxy
must be increasing with distance rom the galactic centre. This is
certainly not true o the visible mass (the stars emitting light)
machoS, wimpS anD other theorieS
Astrophysicists are attempting to come up with theories to
explain why there is so much dark matter and what it consists
o. There are a number o possible theories:
The matter could be ound in Massive Astronomical
Compact Halo Objects or MACHOs or short. There is some
evidence that lots o ordinary matter does exist in these
groupings. These can be thought o as low-mass ailed stars
or high-mass planets. They could even be black holes. These
would produce little or no light. Evidence suggests that these
could only account or a small proportion.
that we can see so the suggestion is that there must be dark
matter. In this situation it would have to be concentrated
outside the galactic centre orming a halo around the galaxy.
Further evidence suggests that only a very small amount o this
matter could be imagined to be made up o the protons and
neutrons that constitute ordinary, or baryonic, matter.
Dark matter:
gravitationally attracts ordinary matter
does not emit radiation and cannot be inerred rom its
interactions
is unknown in structure
makes up the majority o the Universe with less than 5% o
the Universe made up o ordinary baryonic matter.
There could be new particles that we do not know about.
These are the Weakly Interacting Massive Particles. Many
experimenters around the world are searching or these
so-called WIMPs.
Perhaps our current theories o gravity are not completely
correct. Some theories try to explain the missing matter as
simply a ailure o our current theories to take everything
into account.
O p t i O n D A s t r O p h ys i c s
209
HL
t s uvs
fluctuationS in cmb
The cosmic microwave background radiation (CMB) is essentially
isotropic (the same in all directions) . This implies that the matter
in the early Universe was uniormly distributed throughout
space with no random temperature variations at all. I this was
precisely the case then the development o the Universe would be
expected to be absolutely identical everywhere and matter would
be uniormly distributed throughout the Universe it would be
without any structure. We know, however, that matter is not
uniormly distributed as it is concentrated into stars and galaxies.
Further analysis o the CMB reveals tiny fuctuations
(anisotropies) in the temperature distribution o the early
Universe in dierent directions. These temperature variations
are typically a ew K compared with the background eective
temperature o 2.73 K. The diagram right is an enhanced
projection which highlights the minor observed variations
in the CMB (with the eects o our own galaxy removed) .
Just like a map includes all the countries o the world, this
projection shows the variation in received CMB rom the
whole Universe.
Variation in CMB as observed by the Wilkinson Microwave
Anisotropy Probe (WMAP)
The minute dierences in temperature imply minor dierences
in densities, which allow structures to be developed as the
Universe expands.
the hiStory of the uniVerSe
We can work backwards and imagine the process that took place soon ater the Big Bang.
Very soon ater the Big Bang, the Universe must have been very hot.
As the Universe expanded it cooled. It had to cool to a certain temperature beore atoms and molecules could be ormed.
The Universe underwent a short period o huge expansion (Infation) that would have taken place rom about 1 0 - 3 6 s ater the
Big Bang to 1 0 - 3 2 s.
Time
1 0- 4 5 s 1 0 - 3 6 s
1 0- 3 6 s 1 0 - 3 2 s
What is happening
Unication o orces
Infation
1 0- 3 2 s 1 0 - 5 s
Quarklepton era
1 0- 5 s 1 0 - 2 s
Hadron era
1 0- 2 s 1 0 3 s
Nucleosynthesis
1 03 s 3 1 0 5
years
3 1 0 5 years
1 0 9 years
Plasma era
(radiation era)
1 0 9 years now
Formation o stars,
galaxies and galactic
clusters
Formation o atoms
Comments
This is the starting point.
A rapid period o expansion the so-called infationary epoch. The reasons or this
rapid expansion are not ully understood.
Matter and antimatter (quarks and leptons) are interacting all the time. There is
slightly more matter than antimatter.
At the beginning o this short period it is just cool enough or hadrons (e.g.
protons and neutrons) to be stable.
During this period some o the protons and neutrons have combined to orm helium
nuclei. The matter that now exists is the small amount that is let over when matter
and antimatter have interacted.
The ormation o light nuclei has now nished and the Universe is in the orm o a
plasma with electrons, protons, neutrons, helium nuclei and photons all interacting.
At the beginning o this period, the Universe has become cool enough or the rst atoms
to exist. Under these conditions, the photons that exist stop having to interact with
the matter. It is these photons that are now being received as part o the background
microwave radiation. The Universe is essentially 75% hydrogen and 25% helium.
Some o the matter can be brought together by gravitational interactions. I this
matter is dense enough and hot enough, nuclear reactions can take place and stars
are ormed.
coSmic Scale factor anD temperature
The expansion o the Universe means that the wavelength o
any radiation that has been emitted in the past will be stretched
over time (see page 202) . Thus the radiation that was emitted
approximately 1 2 billion years ago (shortly ater the Big Bang)
at very short wavelengths is now being received as much longer
microwaves the CMB radiation.
The spectrum o CMB radiation received corresponds to blackbody radiation at a temperature o 2.73 K. The calculation uses
Wiens law to link the peak wavelength, m a x , o the radiation to
the temperature, T, o the black body in kelvins:
2.9 1 0 - 3
m a x = __________
T
1
m a x _
T
210
O p t i O n D A s t r O p h ys i c s
When the radiation was emitted the temperature o the
universe was much hotter, the cosmic scale actor, R, was much
smaller and m a x was also proportionally much smaller.
Since the stretching o the Universe is the cause o the
change in wavelength, then the ratio o cosmic scale actors
at two dierent times must be the same as the ratio o peak
wavelengths so
m a x R
1
1
_
R or T _
T
R
t uvs
future of the uniVerSe (without Dark
energy)
cosmic scale factor
I the Universe is expanding at the moment, what is it going
to do in the uture? As a result o the Big Bang, other galaxies
are moving away rom us. I there were no orces between
the galaxies, then this expansion could be thought o as being
constant.
R
current rate
of expansion
1
now
time
The expansion o the Universe cannot, however, have been
uniorm. The orce o gravity acts between all masses. This
means that i two masses are moving apart rom one another
there is a orce o attraction pulling them back together. This
orce must have slowed the expansion down in the past. What
it is going to do in the uture depends on the current rate o
expansion and the density o matter in the Universe.
cosmic scale factor
HL
R
open
at
closed
now
time
An open Universe is one that continues to expand orever. The
orce o gravity slows the rate o recession o the galaxies down
a little bit but it is not strong enough to bring the expansion to a
halt. This would happen i the density in the Universe were low.
A closed Universe is one that is brought to a stop and then
collapses back on itsel. The orce o gravity is enough to bring
the expansion to an end. This would happen i the density in
the Universe were high.
A fat Universe is the mathematical possibility between open
and closed. The orce o gravity keeps on slowing the expansion
down but it takes an innite time to get to rest. This would
only happen i the Universe were exactly the right density. One
electron-positron pair more, and the gravitational orce would
be a little bit bigger. Just enough to start the contraction and
make the Universe closed.
critical DenSity, c
The theoretical value o density that would create a fat
Universe is called the critical density, c. Its value is not
certain because the current rate o expansion is not easy to
measure. Its order o magnitude is 1 0 - 2 6 kg m- 3 or a ew proton
masses every cubic metre. I this sounds very small remember
that enormous amounts o space exist that contain little or no
mass at all.
The density o the Universe is not an easy quantity to measure.
It is reasonably easy to estimate the mass in a galaxy by
estimating the number o stars and their average mass but the
majority o the mass in the Universe is dark matter.
The value o c can be estimated using Newtonian gravitation.
We consider a galaxy at a distance r away rom an observer with
a recessional velocity o v with respect to the observer.
radius r
total mass in
sphere, M
recessional
velocity =
The total energy ET o the galaxy is the addition o its kinetic
energy EK and gravitational potential energy, EP given by:
ET = EK + EP
1
EK = _
mv2 but Hubbles law gives v = H0 r
2
1
EK = _
m(H0 r) 2
2
average density of universe
inside sphere =
radius r
observer
recessional velocity =
The net eect o all the masses in the Universe outside the
sphere on the galaxy is zero (see page 208 or an analogous
situation) . The galaxy is thus gravitationally attracted in by a
total mass M which acts as though it was located at the observer
as shown (above) .
GMm
4 3
EP = - _
but M = volume density = _
r
r
3
G4r3 m
4Gr2 m
EP = - _ = - _
3r
3
I ET is positive, the galaxy will escape the inward attraction
the universe is open.
I ET is negative, the galaxy will eventually all back in the
universe is closed.
I ET is exactly zero, the galaxy will take an innite time to be
brought to rest the universe is fat. This will occur when the
density o the universe is equal to the critical density c.
4Gr2 cm
1
_
m(H0 r) 2 = _
2
3
2
8Gr cm
mH0 2 r2 = _
3
2
3H0
c = _
8G
O p t i O n D A s t r O p h ys i c s
211
HL
D
coSmic DenSity parameter
The cosmic density parameter, 0 is the ratio o the average
density o matter and energy in the Universe, , to the critical
density, c
0 = ___
C
I 0 > 1 , the universe is closed.
I 0 < 1 , the universe is open.
I 0 = 1 , the universe is fat.
Dark energy
Gravitational attraction between masses means that the rate
o expansion o the Universe would be expected to decrease
with time. Measurements using type Ia supernovae as standard
candles have provided strong evidence that the expansion
has not, in act, been slowing down over time (see page 204) .
Observations currently indicate that the Universes rate o
expansion has been increasing.
Dark matter is hypothesized to explain the missing matter
that must exist within galaxies or the known laws o
gravitational attraction to be able to explain a galaxys rate
o rotation. Dark matter adds to the attractive force of
gravity acting within galaxies implying more unseen
mass than had been previously expected, hence the name
dark mass.
Currently there is no single accepted explanation or this
observation and, o course, it is possible that our theories o
gravity and general relativity need to be modied. Perhaps we
are on the brink o discovery o new physics. Whatever the
cause, the reason or the Universes accelerating expansion has
been given the general name dark energy.
The observation that expansion o the Universe is
accelerating means that then there must be a orce that is
counteracting the attractive orce o gravity. Dark energy
opposes the attractive force of gravity between
galaxies. The resulting increase in energy implies an unseen
source o energy, hence the name dark energy.
Dark energy and dark matter are two dierent concepts. In
both cases experimental evidence implies their existence but
physicists have yet to agree a theoretical basis that explains the
existence o either concept.
effect of Dark energy on the coSmic Scale factor
The existence o dark energy counteracts the attractive orce o gravity. This will cause the cosmic scale actor to increase over time.
The graph below compares how a fat Universe is predicted to develop with and without dark energy.
cosmic scale
factor, R
at Universe with dark energy
(accelerating expansion)
at Universe without dark energy
(approaches maximum size)
now
212
O p t i O n D A s t r O p h ys i c s
time
HL
asss s
aStrophySicS reSearch
Much o the current undamental research that is being
undertaken in astrophysics involves close international
collaboration and the sharing o resources. Scientists can be
proud o their record o international collaboration. For example,
at the time that the previous edition o this book was being
published, the Cassini spacecrat had been in orbit around Saturn
or several years sending inormation about the planet back to
Earth and is currently (201 4) continuing to produce data.
The CassiniHuygens spacecrat was unded jointly by ESA (the
European Space Agency) , NASA (the National Aeronautics and
Space Administration o the United States o America) and ASI
(Agenzia Spaziale Italiana the Italian Space Agency) . As well
as general inormation about Saturn, an important ocus o the
mission was a moon o Saturn called Titan. The Huygens probe
was released and sent back inormation as it descended towards
the surace. The inormation discovered is shared among the
entire scientic community. Many current projects, or example
the Dark Energy Survey (involving more than 1 20 scientists or
23 institutions worldwide) , continue this process.
All countries have a limited budget available or the scientic
research that they can undertake. There are arguments both or,
and against, investing signicant resources into researching the
nature o the Universe.
It is one o the most undamental, interesting and important
areas or humankind as a whole and it thereore deserves to
be properly researched.
All undamental research will give rise to technology that
may eventually improve the quality o lie or many people.
Lie on Earth will, at some time in the distant uture, become
an impossibility. I humankinds descendents are to exist in
this uture, we must be able to travel to distant stars and
colonize new planets.
Arguments against:
The money could be more useully spent providing ood,
shelter and medical care to the many millions o people who
are suering rom hunger, homelessness and disease around
the world.
I money is to be allocated on research, it is much more
worthwhile to invest limited resources into medical research.
This oers the immediate possibility o saving lives and
improving the quality o lie or some suerers.
It is better to und a great deal o small diverse research
rather than concentrating all unding into one expensive
area. Sending a rocket into space is expensive, thus unding
space research should not be a priority.
Is the inormation gained really worth the cost?
Future research, such as the Euclid mission to map the
geometry o the dark Universe continues to be planned.
Arguments or:
Understanding the nature o the Universe sheds light on
undamental philosophical questions like:
Why are we here?
Is there (intelligent) lie elsewhere in the Universe?
current obSerVationS
Three recent scientic experiments that have studied the CMB
in detail have together added a great deal to our understanding
o the Universe. Particular experiments o note include:
calculated the Hubble constant to be 67.1 5 km s - 1 Mpc- 1
NASAs Cosmic Background Explorer (COBE)
showed the Universe to be fat, 0 = 1
NASAs Wilkinson Microwave Anisotropy Probe (WMAP)
calculated the Universe to be composed o 4.6% atoms,
23% dark matter and 71 .4% dark energy.
ESAs Planck space observatory.
Together these experiments have:
mapped the anisotropies o the CMB in great detail and with
precision
discovered that the rst generation o stars to shine did so
200 million years ater the Big Bang, much earlier than
many scientists had previously expected
calculated the age o the Universe as 1 3.75 0.1 4 billion
years old
showed that their results were consistent with the Big Bang
and specic infation theories
In summary, current scientic evidence suggests that, when dark
matter and dark energy are taken into consideration, the Universe:
is fat
has a density that is, within experimental error, very close to
the critical density
has an accelerating expansion
is composed mainly o dark matter and dark energy.
O p t i O n D A s t r O p h ys i c s
213
ib questons astrophyscs
This question is about determining some properties o the star
Wol 359.
3.
a) The star Wol 359 has a parallax angle o 0.41 9 seconds.
(i) Describe how this parallax angle is measured.
[4]
(ii) Calculate the distance in light-years rom Earth
to Wol 359.
[2]
(iii) State why the method o parallax can only be
used or stars at a distance o less than a ew
hundred parsecs rom Earth.
[1 ]
b) The ratio
a) The spectrum o light rom the Sun is shown below.
relative
intensity
1.
0.4
[4]
0.2
0
0
Show that the ratio
luminosity o Wol 359
_____________________
is 8.9 1 0 4 . (1 ly = 6.3 1 0 4 AU)
luminosity o the Sun
c) The surace temperature o Wol 359 is 2800 K and
its luminosity is 3.5 1 0 2 3 W. Calculate the radius o
Wol 359.
4.
B
A
1.0 10 3
luminosity L/L s
1.0 10 1
a)
3.0 10 4 1.2 10 4
3.0 10 3
surface temperature T/K
5.
(i) Draw a circle around the stars that are red giants.
Label this circle R.
[1 ]
(ii) Draw a circle around the stars that are white
dwars. Label this circle W.
(iii) Draw a line through the stars that are main
sequence stars.
b) Explain, without doing any calculation, how
astronomers can deduce that star B has a larger
diameter than star A.
[2]
[2]
a) Explain how Hubbles law supports the Big Bang
model o the Universe.
[2]
b) Outline one other piece o evidence or the model,
saying how it supports the Big Bang.
[3]
c) The Andromeda galaxy is a relatively close galaxy,
about 700 kpc rom the Milky Way, whereas the
Virgo nebula is 2.3 Mpc away. I Virgo is moving
away at 1 200 km s 1 , show that Hubbles law
predicts that Andromeda should be moving away
at roughly 400 km s 1 .
[1 ]
Apparent brightness o the Sun
Apparent brightness o star A
Mean distance o Sun rom Earth
1 pc
=
=
=
=
1 .4
4.9
1 .0
2.1
[3]
A quasar has a redshit o 6.4. Calculate the ratio o the
current size o the universe to its size when the quasar
emitted the light that is being detected.
[3]
HL
6.
[1 ]
d) Explain why the distance o star A rom Earth cannot be
determined by the method o stellar parallax.
[1 ]
Explain the ollowing:
a) Why more massive stars have shorter lietimes
[2]
b) The jeans criterion
[2]
c) How elements heavier than iron are produced by stars [2]
[3]
1 0 3 W m2
1 0 9 W m2
AU
1 0 5 AU
[4]
i B Q u E s t i O n s A s t r O p h ys i c s
e) I light o wavelength 500 nm is emitted rom
Andromeda, what would be the wavelength observed
rom Earth?
[1 ]
d) How type 1 a supernovae can be used as standard candles [2]
c) Using the ollowing data and inormation rom the
HR diagram, show that star A is at a distance o about
800 pc rom Earth.
214
(i) The elements present in its outer layers.
(ii) Its speed relative to the Earth.
d) Andromeda is in act moving towards the Milky Way,
with a speed o about 1 00 km s 1 . How can this
discrepancy rom the prediction, in both magnitude
and direction, be explained?
[3]
1.0 10 -1
1.0 10 -3
[2]
b) Outline how the ollowing quantities can, in principle,
be determined rom the spectrum o a star.
[2]
The diagram below shows the grid o an HR diagram,
on which the positions o selected stars are shown.
(L S = luminosity o the Sun.)
500 1000 1500 2000 2500 3000
wavelength / nm
Use this spectrum to estimate the surace temperature
o the Sun.
d) By reerence to the data in (c) , suggest why Wol 359 is
neither a white dwar nor a red giant.
[2]
1.0 10 5
0.8
0.6
apparent brightness o Wol 359
_____________________________
is 3.7 1 0 1 5 .
apparent brightness o the Sun
2.
1.0
7.
e) The signifcance o observed anisotropies in the
Cosmic Microwave background
[2]
) The signifcance o the critical density o universe
[2]
g) The evidence or dark matter
[2]
h) What is meant by dark energy
[2]
Calculate the critical density or o the universe using the
Hubble constant o 71 km s - 1 Mpc- 1
[3]
17 a P P e n d i x
gp
Plotting graPhs axes and best fit
All the ints have been ltted crrectly.
The reasn r ltting a grah in the frst lace is that it allws
us t identiy trends. T be recise, it allws us a visual way
reresenting the variatin ne quantity with resect t
anther. When ltting grahs, yu need t make sure that all
the llwing ints have been remembered:
Errr bars are included i arriate.
The grah shuld have a title. Smetimes they als need a key.
The scales the axes shuld be suitable there shuld nt,
curse, be any sudden r uneven jums in the numbers.
The inclusin the rigin has been thught abut. Mst
grahs shuld have the rigin included it is rare r a grah
t be imrved by this being missed ut. I in dubt include it.
Yu can always draw a secnd grah withut it i necessary.
The fnal grah shuld, i ssible, cver mre than hal the
aer in either directin.
The axes are labelled with bth the quantity (e.g. current)
AND the units (e.g. ams) .
The ints are clear. Vertical and hrizntal lines t make
crsses are better than 45 degree crsses r dts.
A best-ft trend line is added. This line NEVER just jins the
dts it is there t shw the verall trend.
I the best-ft line is a curve, this has been drawn as a single
smth line.
I the best-ft line is a straight line, this has been added WITH
A RULER.
As a general rule, there shuld be rughly the same number
ints abve the line as belw the line.
Check that the ints are randmly abve and belw the
line. Smetimes ele try t ft a best-ft straight line
t ints that shuld be reresented by a gentle curve. I
this was dne then ints belw the line wuld be at the
beginning the curve and all the ints abve the line
wuld be at the end, r vice versa.
Any ints that d nt agree with the best-ft line have been
identifed.
Measuring intercePt, gradient and area
under the graPh
The gradient a curve at any articular int is the gradient
the tangent t the curve at that int.
Grahs can be used t analyse the data. This is articularly easy
r straight-line grahs, thugh many the same rinciles can
be used r curves as well. Three things are articularly useul:
the intercept, the gradient and the area under the graph.
y
1. Intercept
In general, a grah can intercet (cut) either axis any number
times. A straight-line grah can nly cut each axis nce and ten
it is the y-intercept that has articular imrtance. (Smetimes
the y-intercet is reerred t as simly the intercet.) I a
grah has an intercet zer it ges thrugh the rigin.
Proportional nte that tw quantities are rrtinal i the
grah is a straight line THAT pASSES THRoUGH THE oRIGIN.
Smetimes a grah has t be cntinued n (utside the range
the readings) in rder r the intercet t be und. This rcess is
knwn as extrapolation. The rcess assuming that the trend
line alies between tw ints is knwn as interpolation.
extrapolation
y
P
y
x
x
x
x
at point P on the curve,
gradient = y
x
gradient of straight
line = y
x
3. Area under a graph
The area under a straight-line grah is the rduct
multilying the average quantity n the y-axis by the quantity
n the x-axis. This des nt always reresent a useul hysical
quantity. When wrking ut the area under the grah:
I the grah cnsists straight-line sectins, the area can be
wrked ut by dividing the shae u int simle shaes.
I the grah is a curve, the area can be calculated by cunting
the squares and wrking ut what ne square reresents.
y-intercept
The extrapolation of the
graph continues the
trend line.
y
The line is interpolated
between the points.
The units r the area under the grah are the units n the
y-axis multilied by the units n the x-axis.
I the mathematical equatin the line is knwn, the area
the grah can be calculated using a rcess called integration.
y
y
2. Gradient
The gradient a straight-line grah is the increase in the y-axis
value divided by the increase in the x-axis value.
The llwing ints shuld be remembered:
A straight-line grah has a cnstant gradient.
The triangle used t calculate the gradient shuld be as large
as ssible.
x
area under graph
x
area under graph
The gradient has units. They are the units n the y-axis
divided by the units n the x-axis.
only i the x-axis is a measurement time des the gradient
reresent the RATE at which the quantity n the y-axis
increases.
APPEN D I X
215
gp y d dm p
equation of a straight-line graPh
All straight-line graphs can be described using one general
equation
You should be able to see that the physics equation has exactly
the same orm as the mathematical equation. The order has
been changed below so as to emphasize the link.
v = u + at
m and c are both constants they have one fxed value.
c represents the intercept on the y-axis (the value y takes
when x = 0)
m is the gradient o the graph.
In some situations, a direct plot o the measured variable will
give a straight line. In some other situations we have to choose
careully what to plot in order to get a straight line. In either
case, once we have a straight line, we then use the gradient and
the intercept to calculate other values.
For example, a simple experiment might measure the velocity o
a trolley as it rolls down a slope. The equation that describes the
motion is v = u + at where u is the initial velocity o the object. In
this situation v and t are our variables, a and u are the constants.
y = c + mx
By comparing these two equations, you should be able to see
that i we plot the velocity on the y-axis and the time on the
x-axis we are going to get a straight-line graph.
velocity v / m s -1
y = mx + c
y and x are the two variables (to match with the y-axis and
the x-axis) .
20
15
10
gradient = 20 = 4 m s -2
5
5
1
intercept = 0
2
3
4
5
time t / s
The comparison also works or the constants.
c (the y-intercept) must be equal to the initial velocity u
m (the gradient) must be equal to the acceleration a
Identiy which symbols represent variables and which
symbols represent constants.
The symbols that correspond to x and y must be variables and
the symbols that correspond to m and c must be constants.
I you take a variable reading and square it (or cube, square
root, reciprocal etc.) the result is still a variable and you
could choose to plot this on one o the axes.
You can plot any mathematical combination o your original
readings on one axis this is still a variable.
Sometimes the physical quantities involved use the symbols m
(e.g. mass) or c (e.g. speed o light) . Be careul not to conuse
these with the symbols or gradient or intercept.
Example 1
Example 2
I an object is placed in ront o a lens we get an image. The
image distance v is related to the object distance u and the ocal
length o the lens f by the ollowing equation.
1 +_
1 =_
1
_
u
v
f
There are many possible ways to rearrange this in order to get
it into straightline orm. You should check that all these are
algebraically the same.
uv
v + u = _
f
The gravitational orce F that acts on an object at a distance r
away rom the centre o a planet is given by the equation
GMm where M is the mass o the planet and the
F = _
r2
m is mass o the object.
I we plot orce against distance we get a curve (graph 1 ) .
GMm
We can restate the equation as F = ____
+ 0 and i we plot F
r
1
on the y-axis and __
on
the
x-axis
we
will get a straight-line
r
(graph 2) .
or
v
v
_
_
u = f - 1
or
1 = _
1 - _
1
_
u
v
f
v
u
With a little rearrangement we can oten end up with the
physics equation in the same orm as the mathematical
equation o a straight line. Important points include
In this example the graph tells us that the trolley must
have started rom rest (intercept zero) and it had a constant
acceleration o 4.0 m s - 2 .
(v + u)
choosing what to Plot to get a
straight line
gradient = 1
f
gradient = 1
f
uv
intercept = 0
v
intercept = 1
1
A
2
F
force F
2
A
gradient = GMm
B
C
distance r
C
B
intercept = 0
216
APPEN D I X
I
r2
1
u
2
intercept = 1
f
gradient = 1
1
v
gpc y mc fuc
Then lg (a) = b
[t be abslutely recise lg1 0 (a) = b]
Mst calculatrs have a lg buttn
n them. But we dnt have t use 1 0
as the base. We can use any number
that we like. Fr examle we culd
use 2.0, 563.2, 1 7.5, 42 r even
2.71 8281 828459045235360287471 4.
Fr cmlex reasns this last number
IS the mst useful number t use! It
is given the symbl e and lgarithms
t this base are called natural
logarithms. The symbl fr natural
lgarithms is ln (x) . This is als n mst
calculatrs.
If p = e q
Then ln (p) = q
When an exerimental situatin invlves
a wer law it is ften nly ssible t
transfrm it int straight-line frm by
taking lgs. Fr examle, the time erid
f a simle endulum, T, is related t its
length, l, by the fllwing equatin.
A lt f the variables will give a curve,
but it is nt clear frm this curve what
the values f k and p wrk ut t be. on
t f this, if we d nt knw what the
value f p is, we can nt calculate the
values t lt a straight-line grah.
The int f lgarithms is that they can
be used t exress sme relatinshis
(articularly wer laws and
exnentials) in straight-line frm. This
means that we will be ltting grahs
with lgarithmic scales.
6
7
8
9
10
11
A lgarithmic scale increases by the
same rati all the time.
10
1
0
10
10
1
10
2
1 00
10
Bth the gravity frce and the electrstatic
frce are inverse-square relatinshis. This
means that the frce (distance aart) - 2 .
The same technique can be used t
generate a straight-line grah.
l / metres
force =
k
(distance apart) 2
ln (T) = ln (k l p )
3
1 000
distance apart
log (force)
The trick is t take lgs f bth sides f
the equatin. The equatins belw have
used natural lgarithms, but wuld wrk
fr all lgarithms whatever the base.
intercept = log (k)
gradient = -2
ln (T) = ln (k) + ln (l p )
log (distance apart)
ln (T) = ln (k) + p ln (l)
This is nw in the same frm as the
equatin fr a straight line
y = c + mx
A nrmal scale increases by the same
amunt each time.
4 5
In l
plt f ln (time erid) versus
ln (length) gives a straight-line grah
Fr examle, the cunt rate R at any
given time t is given by the equatin
Inverse square relatinshi direct lt
and lg-lg lt
R
( )
1 = - ln (c)
ln _
c
These rules have been exressed fr
natural lgarithms, but they wrk fr
all lgarithms whatever the base.
3
intercept = ln (k)
Time erid versus length fr a simle
endulum
ln (cn ) = n ln (c)
2
gradient = p
k and p are cnstants.
ln (c d) = ln (c) + ln (d)
1
The gradient will be equal t p
The intercet will be equal t
ln (k) [s k = e ( in te rce t) ]
T = k lp
The werful nature f lgarithms
means that we have the fllwing rules
ln (c d) = ln (c) - ln (d)
Thus if we lt ln (T) n the y-axis and
ln (l) n the x-axis we will get a straightline grah.
In T
If a = 1 0 b
Power laws and logs
(log log)
force
logs base ten and base
Mathematically,
T / seconds
hl
R0
R = R0 e - t
R = R0 e -t
R0 and are cnstants.
If we take lgs, we get
ln (R) = ln (R0 e - t)
Natural lgarithms are very imrtant
because many natural rcesses are
exnential. Radiactive decay is an
imrtant examle. In this case, nce
again the taking f lgarithms allws
the equatin t be cmared with the
equatin fr a straight line.
ln (R) = ln (R0 ) - t [ln (e) = 1 ]
ln (R) = ln (R0 ) - t ln (e)
This can be cmared with the equatin
fr a straight-line grah
t
ln (R)
exPonentials and logs
(log linear)
ln (R) = ln (R0 ) + ln (e - t)
intercept = ln (R0 )
gradient = -
y = c + mx
t
Thus if we lt ln (R) n the y-axis and t
n the x-axis, we will get a straight line.
Gradient = -
Intercet = ln (R0 )
APPEN D I X
217
Answers
Topic 1 (Page 8): Measurements and uncertainties
1. (a) (i) 0.5 acceleration down the slope (a) (iv) 0.36 ms - 2
2. C 3. D 4. D 5. (b) 2.4 0.1 s (c) 2.6 0.2 ms - 2
6. (b) (i) -3; (b) (ii) 2.6 1 0 - 4 Nm- 3
Topic 9 (Page 104): Wave phenomena
;
1. B 2. (b) 27.5 m s - 1 3. (a) 0.2; 4. (a) (i) zero; (ii) or _
2
-1 0
(iii) zero; (b) 1 1 0 nm; 5. (b) (i) 1 .5 1 0 m;
(d) (ii) 5.0 1 0 1 9 m s - 2
Topic 2 (Page 24): Mechanics
1. C 2. D 3. B 4. B 5. (a) 520 N; (b) (i) 1 .2 MJ; (b) (ii) 270 W
6. (a) equal; (b) let; (c) 20 km hr- 1 ; (e) car driver; () No
7. (c) 3.50 N
Topic 10 (Page 111): Fields
1. A 2. C 3. C 4. (a) (i) 1 .9 1 0 1 1 J; (a) (ii) 7.7 km hr- 1 (a) (iii)
2.2 1 0 1 2 J; (c) 2.6 hr 5. (b) (i) 2.5 m s - 2
Topic 3 (Page 32): Thermal Physics
1. B 2. D 3. D 4. D 5. (a) (i) length = 20 m, depth = 2 m,
width = 5m, temp = 25 C; (a) (ii) $464; (b) (i) 84 days
6. (a) (i) 7.8 J K mol- 1
Topic 4 (Page 50): Waves
1. C 2. C 3. (a) longitudinal (b) (i) 0.5 m; (ii) 0.5 mm;
(iii) 330 m s-1 4. (c) (i) 2.0 Hz; (ii) 1.25 (1.3) cm; () (i) 4.73 10-7 m;
(ii) 0.510 mm 5. 45
Topic 5 (Page 64): Electricity and magnetism
1. C 2. A 3. (c) (ii) 7.2 1 0 1 5 m s - 2 (c) (iv) 1 00 v 4. (d) B;
(e) (i) Equal; (ii) approx. 0.4A; (iii) lamp A will have greater
power dissipation;
Topic 6 (Page 68): Circular motion and gravitation
M;
1. A 2. A 3. C 4. (a) (ii) No; (b) 1 .4 m s - 1 5. (b) (i) g = G _
R2
M
_
27
24
(b) (ii) 1 .9 1 0 kg 6. (a) g = G 2 ; (b) 6.0 1 0 kg;
R
Topic 7 (Page 81): Atomic nuclear and particle physics
1. B 2. D 3. A 4. D 5. B 6. (a) (i) uud; (ii) electron is
undamental; (iii) 3 quarks or 3 antiquarks; (iv) a quark and an
__ 0
2
26
24
4
] 8. (b) (i) 1 H + 1 2 Mg 1 1 Na + 2 He
antiquark; 7. uu [
12
12
9. (a) 6 C 7 N + -01 + ; (b) (ii) 1 1 600 years; 10. (a) (i) 3;
(b) (i) 1 .72 1 0 1 9 11. A
Topic 8 (Page 94): Energy production
1. (c) 1 5 MW (d) (i) 20% 4. (a) 1 000 MW;
(b) 1 200 MW; (c) 1 7% ; (d) 43 kg s - 1 5. (c) 1 .8 MW
Topic 11 (Page 120): Electromagnetic induction
1. D 3. B 4. B 5. D 6. (b) 0.7 v 7. (a) 7.2 1 0 - 4 C; (b) 2.9
1 0 - 3 s; (c) (ii) 5CR = 3.5 s; (c) (iii) No
Topic 12 (Page 130): Quantum and nuclear
1. C 2. B 3. (b) ln R & t; (c) Yes; (e) 0.375 hr- 1 ; (h) 1 .85 hr
5. (b) (i) 6.9 1 0 - 3 4 Js; (b) (ii) 3.3 1 0 - 1 9 J 6. 4.5 1 0 4 Bq
Option A (Page 151): relativity
2. (a) (i) 1 .40c; (a) (ii) 0.95c; (c) 6.0 1 0 1 9 J 3. a) 2 yrs;
b) 4 yrs ; c) x = 5 ly; d) 0.5 c 4. (c) ront; (d) T:1 00 m, S:87 m;
(e) T:75m, S:87 m; 5. (a) (i) zero; (a) (ii) 2.7 m0 c2 ; (b) (i) 0.923 c;
(b) (ii) 2.4 m0 c2 ; (b) (iii) 3.6 m0 c2 ; (c) agree.
Option B (Page 170): Engineering physics
2. a) 0.95 N m; b) 25.2 J; c) 1 3.4 N 3. (a) No; (b) Equal; (c)
300 J; (d) -500 J; (e) 500 J; () 1 50 J; (g) 1 6% 4. (b) 990K; (c)
(i) 1 ; (c) (ii) 2 & 3; (c) (iii) 3; 6. Laminar (R=1 200) 7. (a) 2Hz;
(b) 21 mW
Option C (Page 189): Imaging
1. (a) 1 4 cm behind mirror, virtual, upright, magnifed ( 2) ;
(b) 24 cm behind diverging lens, real, inverted, magnifed ( 3) ;
(c) 4.5 cm behind second lens, real, upright & diminished (0.25)
2. (d) upside down; (e) 60 cm; 4. (a) 1 0 dB; (b) 0.5 mW
5. (a) 1 20 MHz; (b) (iii) d = 38mm, l = 1 30mm
6. (b) (ii) 4 mm;
(b) (iii) 9.3 mm; (b) (iv) 1 8.6 mm
Option D (Page 214): Astrophysics
1. (ii) d = 7.78 ly, (c) r = 8.9 1 0 7 m; 3. (a) 5800 K
4. (e) 499.83 nm 5. 1 4% o current size 7. 9.5 1 0 - 2 7 kg m- 3
Origin of individual questions
The questions detailed below are all taken rom past IB examination papers and are all IB.
Topic 1: Measurement and uncertainties
1 N99S2(S2) 2 M98H1 (5) 3 N98H1 (5) 4 M99H1 (3)
5 M98SpH2(A2) 6 N98H2(A1 )
Topic 8: Energy production
1 NO1 S3(C1 ) 2 M99S3(C1 ) 3 M98SpS3(C3)
5 M98S3(C2) 6 M1 22H2(B2.1 )
Topic 2: Mechanics
1 M98S1 (2) 2 M98S1 (4) 3 M98S1 (8)
6 N00H2(B2) 7 M091 S2(A2)
Topic 9: Wave phenomena
1 N01 H1 (24) 2 N98H2(A5) 3 M1 1 1 H3(G3)
5 M1 01 S2(A2)
Topic 3: Thermal Physics
1 N99H1 (1 5) 2 N99H1 (1 6) 3 N99H1 (1 7)
6 M1 1 2H2(A5) 7 M091 S2(A2)
Topic 4: Waves
1 M01 H1 (1 4) 2 N1 0H1 (1 5)
3 N03S2(A3)
Topic 10: Fields
3 N1 0H1 (24) 4 N98H2(B4)
5 M98 Sp2(B2)
4 5 N04H2(B4.1 )
Topic 5: Electricity and magnetism
4 N03 HL2 Q2.2
Topic 6: Circular motion and gravitation
1 N1 0S1 (7) 2 M1 1 1 H1 (4) 3 M1 01 S1 (8) 4 M1 1 1 2(A5)
5 M08 SpS3(A3) 6 N05H2(B2.1 ) part question sections (d)
to (g)
Topic 7: Atomic, nuclear and particle physics
1 N98S1 (29) 2 M99S1 (29) 3 M99S1 (30 4 M98SpS1 (29)
5 M98SpS1 (30) 8 M98S2(A3) 9 M99S2(A3) 10 M99H2(B4)
11 M1 22H1 (32)
218
An s we r s to q u e s ti o n s
4 M98SpS3(C2)
5 N09 HL3 G4
5 N01 H2(A3)
Topic 11: Electromagnetic induction
1 N00H1 (31 ) 3 M98H1 (33) 4 M1 1 2H1 (24)
6 N98H2(A4)
5 N99H1 (34)
Topic 12: Quantum and nuclear physics
1 N1 0H1 (34) 2 M01 H1 (35) 3 N00H2(A1 )
Option A relativity
2 M1 1 1 H3(2) 4 M00H3(G1 )
5 N01 H3(G2)
6 M092H3(3)
Option B Engineering physics
3 N01 H2(B2) 4 N98H2(A2)
Option C imaging
2 N00H3(H1 ) 5 M03H3(D2)
Option D Astrophysics
1 M1 01 H3(E1 ) 2 M1 1 1 H3(E2)
3 N01 H3(F2)
4 N98H3(F2)
Index
Page numbers in italics reer to question sections.
A
absolute magnitude 1 94
absolute uncertainties 5
absolute zero 29
absorption spectra 69
greenhouse gases 92
acceleration 9, 1 4, 1 08
acceleration-time graphs 1 0
acceleration, velocity and displacement
during simple harmonic motion
[SHM] 34, 95
equations o uniorm angular
acceleration 1 52
experiment to determine ree-all
acceleration 1 3
uniormly accelerated motion 1 1
achromatic doublets 1 78
acoustic impedance 1 87
addition 5
air resistance 1 6
albedo 90
alpha radiation 72
alpha decay 72, 1 27
alternating current 54
coil rotating in a magnetic eld
ac generator 1 1 4
diode bridges 1 1 5
losses in the transmission o power
115
RMS values 1 1 4
transormer operation 1 1 4
transmission o electrical power 1 1 5
ammeters 56
amperes 63
ampliers 1 83
amplitude 39, 48
angular impulse 1 57
angular magnication 1 77
angular momentum 1 57, 1 65
conservation o angular momentum
1 57
angular motion 1 24
angular size 1 77
anisotropies 21 0
antimatter 73
antineutrinos 1 29
antinodes 48
antiparticles 78
apparent magnitude 1 94
Archimedes principle 1 64
assumptions 3
asteroids 1 90
astronomical units (AUs) 1 91 , 1 93, 1 94
astrophysics 1 90, 21 4
accelerating Universe 204
astrophysics research 21 3
Big Bang model 201
Cepheid variables 1 98
cosmological principle and
mathematical models 208
current observations 21 3
dark energy 21 2
uture o the Universe 21 1
galactic motion 202
HertzsprungRussell diagram 1 97
history o the Universe 21 0
Hubbles law and cosmic scale actor
203
luminosity 1 94
nature o stars 1 92
nuclear usion the Jeans criterion
205
nucleosynthesis 1 96
nucleosynthesis o the main sequence
206
objects in the Universe 1 901
red giant stars 1 99
rotation curves and dark matter 209
stellar evolution 200
stellar parallax 1 93
stellar spectra 1 95
types o supernovae 207
atomic clock requency shits 1 48
atomic energy levels 1 29
atomic physics 69, 77, 81
atomic model 77
evidence or atomic model 77
explanation o atomic spectra 69
structure o matter 77
atomic spectra 69, 1 23
atoms 26
attenuation 1 83
attenuation coecient 1 85
mass attenuation coecient 1 85
Avogadro constant 30
B
background radiation 73
background count 73
Balmer series 1 23
barium meals 1 86
barrel distortion 1 78
baryonic matter 209
baryons 78, 79
base units 2
batteries 60
becquerels 73
Bernoulli eect 1 65
applications o the Bernoulli equation
1 66
Bernoulli equation 1 65
beta radiation 72
beta decay 72, 1 29
Big Bang model 201
binding energy 75
binding energy per nucleon 76
binoculars 45
bireringence 41
black-body radiation 90, 1 94
black holes 1 50, 1 96, 200
Schwarzchild radius 1 50
blue shit 1 95
Bohr model o the atom 1 24
boundary conditions 49
Boyles law 31
Brackett series 1 23
Brewsters law 41
buoyancy 1 64
C
cancer 72
capacitance 1 1 7
capacitors 1 1 6
capacitor [RC] discharge circuits 1 1 8
capacitor charging circuits 1 1 9
capacitors in series and parallel 1 1 7
energy stored in charged capacitor
119
carbon dioxide 92
carbon xation 92
Carnot cycles 1 63
Carnot engine 1 63
Carnot theorem 1 63
cells 60
Celsius scale 25
Cepheid variables 1 96
mathematics 1 98
principles 1 98
chain reaction 76
Chandrasekhar limit 200
charge 51
charge capacity 60
Coulombs law 51
current 54
particle acceleration and electric
charge 1 45
Charless law 31
chemical energy 22
chlorofuorocarbons (CFCs) 92
circuits 55
capacitor [RC] discharge circuits 1 1 8
capacitor charging circuits 1 1 9
investigating diode-bridge rectication
circuit experimentally 1 1 6
parallel circuits 56
potential divider circuits 57
rectication and smoothing circuits
116
sensor circuits 57
series circuits 56
circular motion 657, 68
angular velocity and time period 66
circular motion in a vertical plane 66
examples 65
mathematics o circular motion 65
mechanics o circular motion 65
Newtons law o universal gravitation
67, 68
radians 66
collisions 23
colour 79
comets 1 90
communications 1 84
coaxial cables 1 84
optical bres 1 84
wire pairs 1 84
complex numbers 1 25
composite particles 78
compression 1 4, 1 6
I n dex
219
compression waves 35
concentration o solutions 42
conduction 89
conduction electrons 54
conductors 51
conjugate quantities 1 26
conservation o energy 22
constant pressure 29, 1 60
constant temperature 29
constant volume 29
constellations 1 901
constructive intererence 40, 47
continuity equation 1 65
continuous spectrum 69
continuous waves 35
convection 89
conventional current 54
converging lenses 1 72, 1 73
convex lenses 1 73
Copenhagen interpretation 1 25
cosmic density parameter 21 2
cosmic microwave background (CMB)
radiation 73
fuctuations in CMB 21 0
cosmic scale actor 203
cosmic scale actor and temperature
21 0
eect o dark energy on the cosmic
scale actor 21 2
cosmological principle 208
Coulombs law 51
coulombs 53
couples 1 54
critical angle 45
critical density 21 1
CT (computed tomography) scans 1 86
current 54
D
damping 1 68
dark energy 204, 21 2
eect o dark energy on the cosmic
scale actor 21 2
dark matter 209
gravity 21 2
MACHOs, WIMPs and other theories
209
Davisson and Germer experiment 1 22
De Broglie hypothesis 1 22
deormation 1 4
derived units 2
destructive intererence 40, 47
dielectric material 1 1 7
diraction 46
basic observations 46, 97
Davisson and Germer experiment 1 22
diraction and resolution 1 01
diraction grating 99
electron diraction experiment 1 22
examples o diraction 46
explanation 97
multiple-slit diraction 99
practical signicance o diraction 46
resolvance o diraction gratings 1 01
single-slit diraction with white light
97
uses o diraction gratings 99
diode bridges 1 1 5
220
I n dex
investigating diode-bridge rectication
circuit experimentally 1 1 6
direct current 54
discharge characteristics 60
dispersion 1 83
acceleration, velocity and displacement
during simple harmonic motion
[SHM] 34,
95
diverging lenses 1 75
denitions and important rays 1 75
images created by diverging lens 1 75
division 5
Doppler eect 1 02
Doppler broadening 1 03
examples and applications 1 03
mathematics o the Doppler eect 1 02
moving observer 1 02
moving source 1 02
drag 1 6
drit speed equation 54
drit velocity 54
dwar planets 1 90
dynamic riction 20
E
Earth 1 90
day 1 91
year 1 91
earthquake waves 35
eddy currents 1 1 5
eciency 22
Einstein model o light 1 21
elastic collisions 23
elastic potential energy 22
electric elds 52, 61
comparison between electric and
gravitational elds 1 1 0
comparison between electric and
magnetic elds 1 32
energy dierence in an electric eld
53
representation o electric elds 52
electric potential dierence 53, 1 09
electric potential energy 53, 1 09
electrical conduction in a metal 54
electrical energy 22
electrical meters 56
electricity 51 60, 64
electric charge and Coulombs law 51
electric circuits 55
electric current 54
electric elds 52
electric potential energy and electric
potential dierence 53, 1 09
example o use o Kirchos laws 59
internal resistance and cells 60
potential divider circuits and sensors
57
resistivity 58
resistors in series and parallels 56
electromagnetic orce 71
electromagnetic induction 1 1 21 9, 1 20
alternating current 1 1 41 5
capacitance 1 1 7
capacitor charge 1 1 9
capacitor discharge 1 1 8
induced electromotive orce (em) 1 1 2
Lenzs law and Faradays law 1 1 3
rectication and smoothing circuits
116
electromagnetic waves 37, 89
electromotive orce (em) 60
induced em 1 1 2
production o induced em by relative
motion 1 1 2
transormer-induced em 1 1 3
electron degeneracy pressure 200
electrons 77
nuclear scattering experiment
involving electrons 1 28
orbital 1 25
electronvolts 53
electrostatic orce 1 4, 1 6, 51
electrostatic potential energy 22
elementary particles 78
emission spectra 69
emissivity 90
energies, range o 1
energy 22
concepts o energy and work 22
conservation o energy 22
energy fow or stars 1 92
energy types 22
mass and energy 1 43
relativistic momentum and energy
1 44
wave energy 48
energy degradation 1 62
energy production 8293, 94
electrical power production 82
energy conversions 82
ossil uel power production 84
global warming 93
greenhouse eect 92
hydroelectric power 87
new and developing technologies 88
nuclear power 856
primary energy sources 83
radiation 90
secondary energy sources 88
solar power 87, 91
thermal energy transer 89
wind power and other technologies
88
energy sources 83
comparison o energy sources 83
non-renewable energy sources 83
renewable energy sources 83
specic energy and energy density 83
energy transer 35, 89
energy transormations 22
wind power 88
engineering physics 1 52, 1 70
Bernoulli examples 1 66
equilibrium examples 1 55
rst law o thermodynamics 1 61
fuids at rest 1 64
fuids in motion Bernoulli eect 1 65
orced oscillations and resonance
1 689
heat engines and heat pumps 1 63
Newtons second law moment o
inertia 1 56
rotational dynamics 1 57
second law o thermodynamics and
entropy 1 62
solving rotational problems 1 58
thermodynamic systems and concepts
1 59
translational and rotational
equilibrium 1 54
translational and rotational motion
1 523
viscosity 1 67
work done by an ideal gas 1 60
entropy 1 62
equilibrium 1 6
equilibrium examples 1 55
hydrostatic equilibrium 1 64
translational and rotational
equilibrium 1 54
equipotentials 1 06
equipotential suraces 1 06
examples o equipotentials 1 06
relationship to eld lines 1 06
equivalence principle 1 46
bending o light 1 46
errors 4
error bars 6
estimation 3
evaporation 89
exchange bosons 78
exchange particles 78
excited state 72
exponential processes 73, 1 29
F
ar point 1 77
Faradays law 1 1 3
application o Faradays law to moving
and rotating coils 1 1 3
Feynman diagrams 80
bre optics 1 823
elds 1 05, 1 1 1
describing elds 1 05
electric and gravitational elds
compared 1 1 0
electric potential energy and potential
1 09
equipotentials 1 06
eld lines 1 05, 1 06
gravitational potential energy and
potential 1 07
orbital motion 1 08
potential [gravitational and electric]
1 05
potential and eld strength 1 09
propagation 1 32
uniorm elds 1 1 0
rst harmonic 49
ssion 76
fuid riction 1 6
fuid resistance 1 3
fuids at rest 1 64
buoyancy and Archimedes principle
1 64
denitions o density and pressure
1 64
hydrostatic equilibrium 1 64
Pascals principle 1 64
variation o fuid pressure 1 64
fuids in motion 1 656
Bernoulli eect 1 656
ideal fuid 1 65
laminar fow, streamlines and the
continuity equation 1 65
fux density 1 1 2
fux linkage 1 1 3
fux losses 1 1 5
orces 1 4
couples 1 54
dierent types o orces 1 4, 1 6
orces as vectors 1 4
undamental orces 71 , 78
magnitude 1 05
measuring orces 1 4
moment o orce (torque) 1 54
particles that experience and mediate
the undamental orces 71
ossil uels 84
advantages and disadvantages 84
eciency o ossil uel power stations
84
energy transormations 84
ractional uncertainties 5
rames o reerence 9, 1 31
inertial rame o reerence 1 33, 1 42
ree-body diagrams 1 4
ree-all 1 1
experiment to determine ree-all
acceleration 1 3
requency 48
atomic clock requency shits 1 48
driving requency 1 68
Larmor requency 1 88
natural requency and resonance 1 68
threshold requency 1 21
ultrasound 1 87
riction 1 4, 1 6
coecient o riction 20
static and dynamic actors aecting
riction 20
undamental units 2
usion 28, 76
nuclear usion 205
G
galaxies 1 91
distributions o galaxies 202
experimental observations 203
motion o galaxies 202
rotation curves 208, 209
Galilean transormations 1 31
ailure o Galilean transormation
equations 1 31
gamma radiation 72
gases 25, 26
equation o state 30
experimental investigations 29
gas laws 2930
greenhouse gases 92
ideal gases and real gases 30
molecular model o an ideal gas 31
gauge bosons 78
Geiger counters 73
general relativity 1 46
applications o general relativity to the
universe as a whole 1 49
geometric optics 43
geometry o mirrors and lenses 1 76
global positioning systems 1 48
global warming 93
evidence or 93
mechanisms 93
possible causes 93
gluons 79
GM tubes 73
graphs 1 0
acceleration-time graphs 1 0
choosing what to plot to get a straight
line 21 6
displacement-time graphs 1 0
equation o a straight-line graph 21 6
exponentials and logs 21 7
graphical representation o uncertainty
4
intererence o waves 40
logs base ten and base 21 7
measuring intercept, gradient and area
under the graph 21 5
plotting graphs axes and best t 21 5
power laws and logs 21 7
rotational motion 1 58
simple harmonic motion [SHM] 34
velocity-time graphs 1 0
gravitational elds 1 1
comparison between electric and
gravitational elds 1 1 0
gravitational eld strength 67
gravitational orce 1 4, 1 6
gravitational lensing 1 48
gravitational potential 1 07
escape speed 1 07
gravitational potential energy 22, 1 07
gravitational potential gradient 1 08
gravitational red shit 1 47
evidence to support gravitational red
shit 1 48
gravity 71
centre o gravity 1 55
dark energy 21 2
dark matter 21 2
eect o gravity on spacetime 1 49
greenhouse eect 92, 93
greenhouse gases 92
H
hadrons 78
hal-lie 74
example 74
investigating hal-lie experimentally
74
simulation 74
hal-value thickness 1 85
harmonics 49
heat 26, 1 59
heat engines 1 63
heat fow 25
heat pumps 1 63
methods o measuring heat capacities
and specic heat capacities 27
phases o matter and latent heat 28
specic heat capacity 27
Heisenberg uncertainty principle 1 26
estimates rom uncertainty principle
1 26
HertzsprungRussell diagram 1 97
interpretation 200
Higgs bosons 78, 79
I n dex
221
Hubble constant 203
Hubbles law 203
Huygens principle 97
hydraulic systems 1 64
hydroelectric power 87
advantages and disadvantages 87
hydrogen spectrum 1 23
hydrostatic equilibrium 1 64, 1 92
hysteresis 1 1 5
I
ideal gases 30
ideal gas laws 29, 30
ideal gas processes 1 61
kinetic model o an ideal gas 31
work done during expansion at
constant pressure 1 60
image ormation 1 71
image ormation in convex lenses 1 73
image ormation in mirrors 1 76
real and virtual images 1 71
stick in water 1 71
wave model o image ormation 1 72
imaging 1 71 88, 1 89
aberrations 1 78
astronomical refecting telescopes 1 80
channels o communication 1 84
compound microscope and
astronomical telescope 1 79
converging and diverging mirrors 1 76
converging lenses 1 72
dispersion, attenuation and noise in
optical bres 1 83
diverging lenses 1 75
bre optics 1 82
image ormation 1 71
image ormation in convex lenses 1 73
radio telescopes 1 81
simple magniying glass 1 77
thin lens equation 1 74
ultrasonic imaging 1 878
X-ray imaging techniques 1 86
X-rays 1 85
impulse 23
incompressibility 1 65
inelastic collisions 23
inra-red 89
insulators 51 , 89
intensity 39, 90, 1 85, 1 88
intererence o waves 40
thin parallel lms 1 00
two-source intererence 47, 98
intermolecular orces 26
internal energy 22, 26, 1 59
internal energy o an ideal monatomic
gas 1 59
internal resistance 60
determining internal resistance
experimentally 60
invariant quantities 1 36
inverse square law o radiation 39
isotopes 70
J
Jeans criterion 205
K
Kelvin scale 25
222
I n dex
kilograms 1 7
kinetic theory 26
kinetic energy 22, 33
kinetic riction 20
Kirchos circuit laws 55
example o use o Kirchos laws 59
L
laminar fow 1 65
lamination 1 1 5
latent heat 28
methods o measuring latent heat 28
length contraction 1 38
calculation o time dilation and length
contraction 1 40
derivation o length contraction rom
Lorentz transormation 1 38
lengths, range o 1
lenses 1 72
centre o curvature 1 72
chromatic aberration 1 78
ocal length 1 72, 1 75
ocal point 1 72, 1 75
geometry o mirrors and lenses 1 76
linear magnication 1 72, 1 74
power 1 72
principal axis 1 72
spherical aberration 1 78
thin lens equation 1 74
Lenzs law 1 1 3
leptons 78
lepton amily number 78
lit 1 4, 1 6
light
bending o light 1 46
bending o star light 1 48
circularly polarized light 41
light clock 1 37
light curves 207
light energy 22
light gates 1 1
light waves 35, 40
light years (lys) 1 91
light-dependent resistors (LDRs) 57
monochromatic light 47
partially plane-polarized light 41
plane-polarized light 41
polarized light 41
speed o light 1 32
unpolarized light 41
waveparticle duality 1 22
liquid-crystal displays [LCDs] 42
liquids 26
logarithmic unctions 21 7
natural logarithms 21 7
longitudinal waves 35
longitudinal sound waves in a pipe 49
Lorentz transormations 1 34
derivation o eect rom Lorentz
transormation 1 37
derivation o length contraction rom
Lorentz transormation 1 38
Lorentz actor 1 34
Lorentz transormation example 1 34
luminosity 1 94
Lyman series 1 23
M
MACHOs (Massive Astronomical Compact
Halo Objects) 209
magnetic elds 61
comparison between electric and
magnetic elds 1 32
magnetic eld in a solenoid 63
straight wire 63
two parallel wires 63
magnetic orce 1 4, 1 6, 64
examples o the magnetic eld due to
currents 62
magnetic eld lines 61
magnetic orce on a current 62
magnetic orce on a moving charge 62
magniying glass 1 77
angular magnication 1 77
angular size 1 77
near and ar point 1 77
magnitude 1 , 1 05, 1 94
orders o magnitude 1 , 3
Maluss law 41
mass 1 9
centre o mass 1 52
mass and energy 1 43
mass deect 75
point masses 67
range o masses 1
unied mass units 75
material dispersion 1 83
mathematics 5
Cepheid variables 1 98
Doppler eect 1 02
exponential decay 1 29
gravitational red shit 1 47
motion o galaxies 202
parabolic motion 1 2
stellar parallax 1 93
two-source intererence 47
wind power 88
matter structure 77
matter waves 1 22
Maxwells equations 1 32
mean position 33
measurement 1 7, 8
mechanics 923, 24
energy and power 22
equilibrium 1 6
fuid resistance and ree-all 1 3
orces and ree-body diagrams 1 4
mass and weight 1 9
momentum and impulse 23
motion 91 2
Newtons rst law o motion 1 5
Newtons second law o motion 1 7
Newtons third law o motion 1 8
solid riction 20
work 21
mesons 78, 79
metabolic pathways 72
meteorites 1 90
methane 92
micrometers 58
microscopes
compound microscopes 1 79
scanning tunnelling microscopes 1 27
travelling microscopes 98
microscopic vs macroscopic 26
conduction 89
ideal gases 30
Milky Way galaxy 1 91
Minkowski diagrams 1 3942
mirrors 1 76
geometry o mirrors and lenses 1 76
molar gas constant 30
molar mass 30
mole 30
molecules 26
moment o inertia 1 56
example 1 56
moments o inertia or dierent objects
1 56
momentum 1 7
conservation o momentum 23
equations 1 44
linear momentum and impulse 23
relativistic momentum and energy
1 44
units 1 44
use o momentum in Newtons second
law 23
motion 9
equations o uniorm motion 1 1
equilibrium 1 6
example o equation o uniorm
motion 1 0
alling objects 1 1
fuid resistance and ree-all 1 3
orces and ree-body diagrams 1 4
rames o reerence 9, 1 31
graphical representation 1 0
instantaneous vs average 9
Newtons rst law o motion 1 5
Newtons second law o motion 1 7
Newtons third law o motion 1 8
practical calculations o uniormly
accelerated motion 1 1
projectile motion 1 2
uniormly accelerated motion 1 1
multiplication 5
muons 78
muon experiment 1 38
N
near point 1 77
nebulae 1 90
negative temperature coecient (NTC)
57
neutrinos 1 29
neutrons 77
neutron capture 206
Newtons rst law o motion 1 5
law o universal gravitation 67, 68
second law o motion 1 7
third law o motion 1 8
nitrous oxide 92
NMR (Nuclear Magnetic Resonance) 1 88
comparison between ultrasound and
NMR 1 88
nodes 48
noise 1 83
normal reaction 1 4, 1 6
nuclear energy 22
nuclear energy levels 1 29
nuclear usion 205
nuclear physics 706, 81 , 1 30
ssion and usion 76, 205
hal-lie 74
nuclear energy levels and radioactive
decay 1 29
nuclear reactions 75
nuclear stability 70
nucleus 1 28
nuclide notation 70
radioactivity 723
strong nuclear orce 71
weak nuclear orce 71
nuclear power 856
advantages and disadvantages 85
enrichment and reprocessing 86
usion reactors 86
health and saety risk 86
moderator, control rods and heat
exchanger 85
nuclear weapons 86
thermal meltdown 86
nuclear reactions 75
articial transmutations 75
mass deect and binding energy 75
unied mass units 75
units 75
worked examples 75
nucleosynthesis 1 96
nuclear synthesis o heavy elements
neutron capture 206
nucleosynthesis o the main sequence
206
nucleus 1 28
deviations rom Rutherord scattering
in high energy experiments 1 28
nuclear radii and nuclear densities
1 24
nuclear scattering experiment
involving electrons 1 28
size 1 28
nuclides 70
O
Ohms law 55
ohmic and non-ohmic devices 55
OppenheimerVolko limit 200
optic bre 1 82
attenuation 1 83
capacity 1 83
communications 1 84
material dispersion 1 83
noise, ampliers and reshapers 1 83
types o optic bre 1 82
waveguide dispersion 1 83
optically active substances 41
orbital motion 1 08
energy o an orbiting satellite 1 08
gravitational potential gradient 1 08
weightlessness 1 08
oscillations 33
damping 1 68
natural requency and resonance 1 68
phase o orced oscillations 1 69
Q actor and damping 1 68
undamped oscillations 96
output ripple 1 1 6
ozone 92
ozone layer 92
P
pair production and pair annihilation 1 23
parabolas 1 2
parallax 1 93
parsecs (pcs) 1 91 , 1 93
particle acceleration and electric charge
1 45
particle physics 7880, 81
classication o particles 78
conservation laws 78
exchange particles 78
Feynman diagrams 80
leptons 78
particles that experience and mediate
the undamental orces 71
quantum chromodynamics 79
quarks 79
standard model 78, 79
Pascals principle 1 64
Paschen series 1 23
path dierence 47
percentage uncertainties 5
periscopes 45
permittivity 1 1 7
Pund series 1 23
phase 48
in phase 33, 40
out o phase 33, 40
phase dierence 40
photoelectric eect 1 21 , 1 85
example 1 21
stopping potential experiment 1 21
photons 1 45
photovoltaic cells 87
piezoelectric crystals 1 87
pion decay 1 31 , 1 45
Plancks constant 1 24
plane o vibration 41
planetary nebula 200
planetary systems 1 90
planets 1 90
polarization 41
concentration o solutions 42
urther examples 42
liquid-crystal displays [LCDs] 42
optically active substances 41
polarizing angle 41
polaroid sunglasses 42
stress analysis 42
positive eedback 93
potential [electric or gravitational] 1 05,
1 07, 1 09
equipotentials 1 06
potential and eld strength 1 09
potential barrier 1 27
potential dierence [electric and
gravitational] 1 05, 1 09
potential due to more than one charge
1 09
potential energy store 33
potential inside a charged sphere 1 09
potentiometers 57
PoundRebkaSnider experiment 1 48
power 22, 82
power dissipation 55
powers 5
precession 1 88
prexes 2
I n dex
223
pressure law 31
primary cells 60
prismatic refectors 45
projectile motion 1 2
horizontal component 1 2
mathematics o parabolic motion 1 2
vertical component 1 2
proper length 1 36
proper time 1 36
protonproton (pp) cycle 1 96
protons 77
pulsars 1 96, 200
pumped storage 87
Q
Q actor and damping 1 68
quality 1 85
quantities 1 , 2
quantized energy 69
quantum chromodynamics 79
quantum physics 1 21 7, 1 30
atomic spectra and atomic energy
states 1 23
Bohr model o the atom 1 24
Heisenberg uncertainty principle and
loss o determinism 1 26
matter waves 1 22
photoelectric eect 1 21
Schrdinger model o the atom 1 25
tunnelling, potential barrier and
actors aecting tunnelling
probability 1 27
quarks 78, 79
quark connement 79
quasars 200
R
r-process 206
radians 66
radiant energy 22
radiation 89
black-body radiation 90, 1 94
cosmic microwave background (CMB)
radiation 73
equilibrium and emissivity 90
intensity 90
isotropic radiation 201 , 21 0
radioactive decay 72
mathematics o exponential decay
1 29
nature o alpha, beta and gamma
decay 72
random decay 73
radioactivity 723
background radiation 73
eects o radiation 72
ionizing properties 72
properties o alpha, beta and gamma
radiations 72
radiation saety 72
random errors 4
rays 35, 39
ray diagrams 43, 1 71 , 1 73
real gases 30
rectication 1 1 5, 1 1 6
red shit 1 03, 1 95
gravitational red shit 1 47
refection 43
224
I n dex
law o refection 43
refection and transmission 43
refection o two-dimensional plane
waves 43
total internal refection and critical
angle 45
types o refection 43
reraction 445
double reraction 41
methods or determining reractive
index experimentally 45
reraction o plane waves 44
reractive index and Snells law 44
total internal refection and critical
angle 45
relativity 1 31 50, 1 51
black holes 1 50
curvature o spacetime 1 49
equivalence principle 1 46
general relativity 1 46, 1 49
gravitational red shit 1 47
invariant quantities 1 36
length contraction and evidence to
support special relativity 1 38
Lorentz transormations 1 34
mass and energy 1 43
Maxwells equations 1 32
reerence rames 1 31
relativistic mechanics examples 1 45
relativistic momentum and energy
1 44
spacetime diagrams [Minkowski
diagrams} 1 3942
special relativity 1 33
supporting evidence 1 48
time dilation 1 37
twin paradox 1 401
velocity addition 1 35
relaxation time 1 88
reshapers 1 83
resistance 55
investigating resistance 58
resistivity 58
resistors in parallels 56
resistors in series 56
resolution 1 01
resonance 1 68, 1 88
examples o resonance 1 69
phase o orced oscillations 1 69
resonance tubes 49
rest mass 1 36
restoring orce 33
Reynolds number 1 67
root mean square (RMS) 1 1 4
rotation curves 209
mathematical models 208
rotational equilibrium 1 54
rotational motion 1 52
bicycle wheel 1 52
energy o rotational motion 1 57
problem solving and graphical work
1 58
relationship between linear and
rotational quantities 1 53
summary comparison o equations o
linear and rotational motion 1 58
Rydberg constant 1 23
Rydberg ormula 1 23
S
s-process 206
Sankey diagram 82
scalars 7
scattering 1 28, 1 85
Schrdinger model o the atom 1 25
Schwarzchild radius 1 50
scientic notation 3
secondary cells 60
recharging secondary cells 60
sensors 57
sensor circuits 57
SI units 2, 1 44
signicant gures 3
signicant gures in uncertainties 4
simple harmonic motion [SHM] 33
acceleration, velocity and displacement
during SHM 34, 95
energy changes during SHM 34, 96
equation 95
identication o SHM 95
mass between two springs 33
two examples o SHM 95
simultaneity 1 33
singularity 1 50
situation diagrams 65
smoothing circuits 1 1 6
Snells law 44
solar energy 22
solar power 87, 91
advantages and disadvantages 87
solar constant 91
Solar System 1 90
solid riction 20
solids 26
sound waves 35, 40
investigating speed o sound
experimentally 38
spacetime
curvature o spacetime 1 49
spacetime interval 1 36
spacetime diagrams 1 39
calculation o time dilation and length
contraction 1 40
examples 1 39, 1 40
representing more than one inertial
rame on same spacetime diagram
1 42
twin paradox 1 401
special relativity 1 33
length contraction and evidence to
support special relativity 1 38
postulates o special relativity 1 33
simultaneity 1 33
specic heat capacity 27
spectral linewidth 1 26
spectrometers 99
standing waves 48
stars 1 901
bending o star light 1 48
binary stars 1 92
brown dwar stars 1 96
classication o stars 1 95
energy fow or stars 1 92
luminosity and apparent brightness
1 94
main sequence stars 1 96
mass-luminosity relation or main
sequence stars 1 97
neutron stars 1 96, 200
red giant stars 1 96, 1 99, 200
red supergiant stars 1 96
stellar clusters 1 91
stellar evolution 200
stellar parallax 1 93
stellar types and black holes 1 96
time spent on the main sequence 205
white dwar stars 1 96, 200
static riction 20
steady fow 1 65
Stean-Boltzmann law 90, 1 95
stellar spectra 1 95
absorption lines 1 95
Stokes law 1 67
streamlines 1 65
stress analysis 42
strobe photography 1 1
strong interactions 79
strong nuclear orce 71
subtraction 5
Sun 1 91
equilibrium 1 92
supernovae 200
supernovae and the accelerating
Universe 204
types o supernovae 207
superposition 40
superposition o wave pulses 40
systematic errors 4
T
tangential stress 1 67
telescopes
array telescopes 1 81
astronomical refecting telescopes 1 80
astronomical telescopes 1 79
cassegrain mounting 1 80
comparative perormance o Earthbound and satellite-borne
telescopes 1 81
comparison o refecting and reracting
telescopes 1 80
Newtonian mounting 1 80
radio intererometry telescopes 1 81
single dish radio telescopes 1 81
temperature 25, 26
cosmic scale actor and temperature
21 0
temperature dierence 27
tension 1 4, 1 6
thermal capacity 27
thermal energy 22
conduction 89
convection 89
radiation 89
thermal energy transer 89
thermal equilibrium 25, 90
thermal physics 25, 32
gas laws 2930
heat and internal energy 26
molecular model o an ideal gas 31
phases [states] o matter and latent
heat 28
specic heat capacity 27
thermistors 57
thermodynamics 1 59
rst law o thermodynamics 1 61
heat 1 59
internal energy 1 59
internal energy o an ideal monatomic
gas 1 59
second law o thermodynamics 1 62
surroundings 1 59
thermodynamic system 1 59
work 1 59
thin lens equation 1 74
real is positive 1 74
thin parallel lms 1 00
applications 1 00
conditions or intererence patterns
1 00
phase changes 1 00
ticker timers 1 1
time 201
time constant 1 1 8
velocity-time graphs 1 0
time dilation 1 37
calculation o time dilation and length
contraction 1 40
derivation o eect rom rst principles
1 37
derivation o eect rom Lorentz
transormation 1 37
time period 66
isochronous time period 33
range o times 1
tomography 1 86
torque 1 54
totally inelastic collisions 23
transormer operation 1 1 4
resistance o the windings (joule
heating) 1 1 5
step-up and step-down transormers
114
turns ratio 1 1 4
transient oscillations 1 68
translational equilibrium 1 6, 1 54
translational motion 1 52
bicycle wheel 1 52
relationship between linear and
rotational quantities 1 53
transverse waves 35
transverse waves on a string 49
travelling waves 35, 48
trigonometry 7
tubes o fow 1 65
tunnelling 1 27
alpha decay 1 27
scanning tunnelling microscopes 1 27
turbulent fow 1 65
Reynolds number 1 67
two-source intererence 47
double-slit intererence 98
Youngs double-slit experiment 47, 98
U
ultrasound 1 87
A- and B-scans 1 87
acoustic impedance 1 87
choice o requency 1 87
comparison between ultrasound and
NMR 1 88
piezoelectric crystals 1 87
relative intensity levels 1 88
uncertainties 47, 8
error bars 6
estimating the uncertainty range 4
graphical representation o uncertainty
4
Heisenberg uncertainty principle 1 26
mathematical representation o
uncertainties 5
signicant gures in uncertainties 4
uncertainty in intercepts 6
uncertainty in slopes 6
uniormly accelerated motion 1 1
universal gravitation 67, 68
Universe 1 91
Big Bang 201
closed Universe 21 1
cosmic scale actor 203
expansion o the Universe 201
fat Universe 21 1
uture o the Universe 21 1
history o the Universe 21 0
history o the Universe 203
open Universe 21 1
supernovae and the accelerating
Universe 204
upthrust 1 4, 1 6
V
vaporization 28
vectors 7
addition/subtraction o vectors 7
components o vectors 7
orces as vectors 1 4
representing vectors 7
vector diagrams 65
velocity 9
acceleration, velocity and displacement
during simple harmonic motion
[SHM] 34,
95
angular velocity and time period 66
change in velocity 1 4
Galilean equation 1 35
relative velocities 9, 1 401
velocity addition 1 35
velocity gradient 1 67
velocity-time graphs 1 0
wave equations 36
vernier callipers 58
Very Long Baseline Intererometry 1 81
viscosity 1 67
non-viscosity 1 65
voltmeters 56
volts 53
W
water 92
water ripples 35, 40
waveparticle duality 1 22
wave phenomena 951 03, 1 04
diraction 97
Doppler eect 1 023
multiple-slit diraction 99
resolution 1 01
simple harmonic motion [SHM] 956
two-source intererence 98
waveunction 1 25
I n dex
225
waveguide dispersion 1 83
wavelength 48
waves 3349, 50
boundary conditions 49
crests and troughs 35
diraction 46
electromagnetic spectrum 37, 89
graphs o simple harmonic motion 34
intensity 39
investigating speed o sound
experimentally 38
nature and production o standing
[stationary] waves 48
oscillations 33
polarization 41 2
refection 43
reraction 445
superposition 40
travelling waves 35
two-source intererence o waves 47
wave characteristics 36
wave energy 48
wave equations 36
wave model o image ormation 1 72
wave pulses 35
waveronts 35, 39
waves along a stretched rope 35
weak interactions 79
weak nuclear orce 71
weight 1 6, 1 9
weightlessness 1 08
Wiens law 90
WIMPs (Weakly Interacting Massive
Particles) 209
wind power 88
advantages and disadvantages 88
work 21 , 1 59
concepts o energy and work 22
denition 21
examples 21
heat and work 26
pV diagrams and work done 1 60
when is work done? 21
work done by an ideal gas 1 60
work unction 1 21
X
X-rays 1 85
basic X-ray display techniques 1 85
imaging techniques 1 86
intensity, quality and attenuation 1 85
Y
Youngs double-slit experiment 47, 98
226
I n dex
OXFORD IB STUDY GUIDES
Physics
2014 edition
o r T h e I B d I p lo m a
Author
Ti Kik
Csy suting t pysics Cus Bk, tis cnsiv stuy
gui efectively reinorces t ky cncts t tst sybus t Sl
n hl (fst xin 2016) . pck wit ti assessment guidance,
it suts t igst civnt in xs.
Oxord IB study guides build unrivalled assessment potential.
Yu cn tust t t:
Comprehensively cv t sybus, tcing IB scifctins
rinc all t ky tics in cncis, us-iny t, cementing
understanding
Efectively prepare stunts ssssnt wit visin sut n
exam strategies
Us c n stigtw ngug t support EAL learners
mti is snt in cncis cunks,
helping students ocus
Suting Cus Bk,
v wit t IB
978 0 19 839213 2
digtic t bks wn
cnging cncts, building
understanding
1
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