AQA Physics A
Unit 1: Particles, Quantum
Phenomena and Electricity
Dr K. A. Newson
Head of Physics TGSG 2008
Overview of New AS Physics Course
Unit 1
Unit 2
Unit 3 (ISA)
75 minute
exam.
75 minute
exam.
Internally
assessed
practical (½)
% of total AS
level marks
40
40
20
% of total A
level marks
20
20
10
How is it
examined?
Physics A Grading System
Unit 1
Unit 2
Unit 3 (ISA)
AS Qualification
Marks
Grade
Grade Marks Grade Total Marks
96 -120
A
A
48 - 60
A
240 - 300
A
84 - 95
B
B
42 - 47
B
210 - 239
B
72 - 83
C
C
36 - 41
C
180 - 209
C
60 - 71
D
D
30 - 35
D
150 - 179
D
48 - 59
E
E
24 - 29
E
120 - 149
E
0 - 47
U
U
0 - 23
U
0 - 119
U
Grade
New A* Grade


New A* grade is for complete A level, it is not
awarded for the AS course alone.
To gain an A* grade students must both:


Gain at least 270 (90%) Uniform marks on the A2
Gain at least 480 (80%) Uniform marks overall on
the A level course.
Overview of Unit 1






1 Matter and Radiation
2 Quarks and Leptons
3 Quantum Phenomena
4 Electric Current
5 Direct Current Circuits
6 Alternating Currents
Matter and Radiation
1.1 Inside the atom
1.1 Inside the atom
Aims of lesson



Understand the structure of the atom and nucleus,
Develop your knowledge of the constituent particles that
make up the atom/nucleus
Know what is meant by an Isotope and how to represent them
Prior knowledge



Every atom consists of a small positively charged nucleus
surrounded by negatively charged electrons;
The nucleus consists of neutrons and protons which are
roughly equal in mass;
Protons and electrons have equal but opposite charges,
neutrons are uncharged.
Large and small

Physics deals with the smallest sub-nuclear particle to the
entirety of the Universe. Physicists need to use a vast range of
values, to make this easier prefixes are used:

For example, a high energy particle collider may accelerate
protons to energies of 3000000000eV, its far simpler and less
prone to error to simply say 3GeV.
In this case G means Giga which means 109 eV.
The full list of prefixes is given to you as a handout.

1.1 Inside the atom
The basic particles...
Proton
positive
charge
Neutron
no
charge
Electron
negative
charge
Describing the atom
Atomic Number
- the number of protons
3
Neutron Number
- the number of neutrons
4
Mass (nucleon) Number
- the number of nucleons
+
=
7
Properties of protons,
neutrons and electrons
Property →
Particle ↓
Mass
(relative to p)
Charge
(relative to p)
Proton
p
1.67 x10-27 kg
1 amu (u)
+ 1.6 x 10-19C Nucleon (part
of nucleus)
(+1e)
Neutron
N
1.67 x10-27 kg
1 amu (u)
Electron
e
9.1 x10-31 kg
1/1800 amu
Description
0
Nucleon
- 1.6 x 10-19C
(-1e)
Found in a
cloud around
the nucleus
Isotopes
Isotopes are atoms of the same element i.e. that have the same
number of protons, but have different numbers of neutron.
Examples of Isotopes
For example, there are 3 isotopes of hydrogen:
1
1)
H (hydrogen)
1
2)
2
1
H (AKA deuterium)
3
H
(AKA
tritium)
1
In the case of Tin (Sn) there are 13 isotopes from
3)
112
50
Sn to
125
50
Sn
Nuclear structure
Key words and terminology



Atomic number (Z): the number of protons contained in
the nucleus (AKA proton number).
Mass/Nucleon number (A): the number of protons and
neutrons contained in the nucleus.
Neutron number (N): the number of neutrons contained
in the nucleus.
N=A-Z
Isotope Notation
A quick way of representing isotopes is to use the following
system:
Mass
number (A)
Atomic number
(Z)
U
92
235
Element’ s
chemical
symbol
Specific Charge


The word ‘Specific’ in Physics means ‘per unit mass’,
The specific charge therefore is given by:
Specific charge
(Ckg-1)

=
Charge (C)
Mass (kg)
E.g. the specific charge on a proton is found by:
specific charge
= 1.60 x 10-19 ÷ 1.67 x 10-27
= 9.58 x 107 Ckg-1
Example

a)
b)
Find the specific charge of the following:
Electron of mass 9.11 x 10-31 kg
An Aluminium ion (Al3+) of mass 4.51 x 10-26 kg
Answers
a) 1.76 x 1011 Ckg-1
b) 1.06 x 107 Ckg-1
Nuclear Stability
Aims of lesson


Understand that some isotopes are unstable and examine
the causes of this.
Know about the three types of radioactive decay and what
happens to the nucleus in each case
Prior knowledge


Radioactive substances emit radiation because they are
unstable.
Know about the 3 types of ionising radiation and their
properties e.g. their penetrating abilities.
Nuclear Stability




Protons and neutrons make up the nucleus.
However, the protons are all positively charged;
so how can they form a nucleus when they
should repel one another?
The reason why nuclei do not fall apart arises
from a force which will be new to you…..
This new type of force is called the STRONG
NUCLEAR FORCE (for obvious reasons), it is
also called the Strong interaction.
It is one of the four Fundamental forces.
Fundamental Interactions (forces)
There are four fundamental interactions, and all
forces in nature can be attributed to these four:
1.Gravitational
2.Electromagnetic
3.Weak
nuclear
4.Strong nuclear
Fundamental forces
Force
Relative Range
strength (fm)
Additional Notes
1
10-1
Electromagnetic
10-3
∞
Weak nuclear
10-16
10-3
Gravitational
10-41
∞
Strong nuclear
Force between nucleons
and quarks.
Force between charged
particles
Force controlling beta
decay (see later notes)
Force between masses
Forces in the nucleus
The strong nuclear force holds protons and neutrons
together in the nucleus.
 The strong nuclear force has a very short range
typically about 4 fm (1 fm = 10-15m).
 At distances larger than this it is insignificant and the
repulsive electromagnetic force dominates.
 At very short distances ~0.5 fm the strong nuclear force
becomes repulsive to stop protons and neutrons being
pushed into each other.
The heaviest unstable nuclides
can be alpha emitters
“Strong” force holds
the nucleons
together
“Electromagnetic
(coulomb) force”
forces protons
apart
Strong force must
be greater than
electromagnetic
force for stability
Larger stable nuclides
N>Z
Excess of
neutrons…therefore
Beta minus emitters
Excess of
protons…therefore
Beta plus emitters
Smaller stable nuclides
N=Z
Three types of ionising radiation
all come from the nucleus
Beta
Electrons
(or positrons)
Gamma
Electromagnetic
wave
Alpha
2 protons, 2 neutrons
Useful websites

http://durpdg.dur.ac.uk/lbl/particleadventure/in
dex.html
Electromagnetic waves and
Photons
Electromagnetic Waves
Electromagnetic spectrum
The Electromagnetic spectrum

The Electromagnetic (EM) spectrum is the entire range of
EM waves (radiations).
Properties of EM radiation



All types consist of vibrating magnetic and electric waves
(handout), the two waves are in phase, that is they vibrate
together.
All EM waves are transverse, that is the vibration is at right
angles to the direction of the wave’s motion.
All travel with the same speed (3.00 x 108 ms-1) in a
vacuum.
Radio waves
Microwaves
Infra-red
Visible light
Ultra-violet
X-rays
Gamma-rays
The EM Spectrum
Frequency
Hz
Wavelength
m
Radio
< 400 x 106
~ 1 - 1000
Microwave
~ 2.5 x 109
~ 0.1 – 0.01
Transmitters
Rotating molecules
~10-24 J
~ 1013 - 1014
~ 10-6
Hot objects, LEDs
~ 1 eV (1.6 x 10-19 J)
Type
Infra-red
Origin
Oscillating currents in
aerials
~ (7 – 4) x 10-7 Very hot objects, LEDs
Photon energy
~10-28 – 10-25 J
~2 – 3 eV
Visible
~ 5 x 1014
UV
> 7.5 x 1014
< 4 x 10-7
Extremely hot objects,
discharge tubes,
> 3 eV
X-rays
~ 1018
~ 10-10
Stopping high energy
electrons.
~ 104 eV
Gamma rays
~ 1020
~ 10-12
Nuclear decay
~106 eV
Cosmic rays
>>1020
Very short
From distant parts of
galaxy.
~10 J
Photons

EM waves are emitted in packets called
Photons and energy

Max Planck (1901) concluded that the energy carried
by light and the other types of electromagnetic radiation
existed in discrete packets called Quanta. The energy
E carried by each quantum is given by:
E = hf


Where f is the frequency (in Hz),
h is a constant called the Planck constant. The value of
Planck’s constant is 6.63×10-34 Js.
Photons and energy

In 1905 Einstein proposed that light radiation consists of a
stream of quanta called Photons.
Example:
Q) What is the energy of a photon of red light that has a
wavelength of 650nm?
Answer:
As frequency (f) = speed of light (c) ÷ wavelength (λ)
 f = c/λ, therefore the energy E = hf = hc/λ
Therefore E = (6.63×10-34 × 3×108) ÷ (6.5×10-7)
E = 3.1×10-19 J
Q) What is the energy of violet light (λ = 400nm)
Answer E = 5×10-19 J
Example

1.
2.
What are the photon energies of the following EM
radiations:
X-ray photon of frequency of 2.5 x 1019 Hz
A photon of blue light having a wavelength of 475nm.
Answers
1.
E = hf = (6.63 x 10-34 x 2.5 x 1019) =
E = 1.66 x 10-14 J
2.
E = hf = hc/λ = (6.63 x 10-34 x 3.00 x 108/4.75 x 10-7)
E = 4.19 x 10-19 J
Laser power

A laser beam consists of a stream of monochromatic
(same f and λ) photons.


The power of a laser = energy transferred each second
by the photons.
If the energy of a single photon = hf

Then the laser power = nhf

Where n = number of photons emitted per second
The electron volt

The electron volt (eV) is defined as the energy gained by an
electron when it is moved through a potential difference of
1Volt.
1eV = 1.6 x 10-19J


Sometimes the mass of particles is given in terms of their
mass-energy in units of eV. This arises because of the
relationship between mass and energy (E = mc2). For
example, the mass of an electron (me) is given by:
 Me = 9.1 x 10-31kg = 8.19 x 10-14 J = 0.511 MeV
More about this later.
Rest energy


Einstein said that the mass of a particle when stationary,
its rest mass (m0), has an energy called its rest energy
which is locked up as mass.
The rest energy is given by E = m0c2
Particles and antiparticles

Consider the following decay:
11
6C

11
0 +
5 B + +1 β
This is called beta+ decay, the beta+ particle is usually
known as a positron. It is virtually identical to an electron
other than its charge which is positive. The positron is the
antimatter equivalent of an electron. We call it the
antiparticle of the electron.
Particles and antiparticles
Carl Anderson 1932
Tracks of positrons in a
cloud chamber, identical
to normal beta (β-)
particles but bent in the
opposite direction
indicating the charge on
a positron is opposite to
beta particles.
Particles and antiparticles




The existence of antimatter was predicted by the English
Physicist Paul Dirac in 1928
According to Dirac’s theory for every particle there exists
a corresponding antiparticle.
Such antiparticle has exactly the same rest mass but the
opposite charge (if the particle is charged).
The particle and antiparticle pair annihilate each other if
they meet converting their rest-mass into photons of
gamma radiation.
Annihilation of positron and electron pair

An electron and positron meet and they
release two gamma rays.
E = mc2




When a particle and its anti-particle meet their mass is
converted to radiation energy by the equation E = mc2.
Two photons are produced, this is needed for conservation
of momentum.
We can calculate the frequency of the gamma rays by
equating the photon energy to the mass energy of the
positron-electron pair.
See next slide.
E = mc2 and annihilation



The mass-energy of particles is sometimes expressed in
units of eV
The energy of each gamma photon = hf
If each electron has a mass of 0.511 MeV

Then 0.511MeV = 9.1 x 10-31Kg so equating the
energies: mec2 = hf therefore f is given:

f = mec2 /h = 1.23 x 1020Hz

(so λ = (c/f) = 2.42 x 10-12m
Pair production


Pair production is basically the reverse of
annihilation.
In pair production a photon creates an
electron and a positron and disappears in the
process.
Pair production
Incident photon
(gamma ray)
Pair production
The electron and
positron follow curved
trajectories in a
magnetic field. How can
we tell from the tracks
that they have opposite
charges?
Energy and Pair production


The energy of each gamma photon = hf
If each electron or positron has a mass of 0.511 MeV

Then the energy of the energy of the gamma photon
must be equal to 2 x 0.511 MeV.

I.e. the photon must have an energy of 1.022MeV,
this is equal to 1.08 x 10-13J

f = E /h = 2.46 x 1020Hz

So λ = (c/f) = 1.21 x 10-12m
Particle interactions
Interaction of particles




An interaction is the process by which two particles exert
forces on each other.
In other words momentum is transferred between the
particles (if no other forces act).
For example, as two electrons approach one another
they repel each other and move away from each other.
The Physicist Richard Feynman found that the force
between the two particles is caused by the exchange of
an exchange particle in this case a virtual photon. The
term virtual means they cannot be detected directly and
if we did detect them this would stop the force from
acting.
Feynman Diagrams



Feynman represented the interaction of particles my means of
a Feynman diagram.
For example, the electromagnetic interaction between two
electrons.
The paths do not show the paths of the particles. The virtual
photon (γ) is represented by a wave.
_
_
Future
e
e
γ
Time
_
e
Past
Exchange particle
is a virtual photon γ
_
e
Feynman diagrams
_
e
Future
γ
Time
_
e
Past
_
e
Exchange particle
is a virtual photon γ
_
e
Analogy

See analogy with skaters on page 13 of text book
Weak nuclear force (weak interaction)



The electromagnetic force results from the exchange of
a virtual photon between the two charged particles.
What about beta decay? What force is at work?
The force responsible for beta decay is the weak nuclear
force (interaction).
Beta +/- decay


At GCSE you given a simplified version of what goes on
in beta decay.
In reality there is an addition particle emitted with the
beta +/- particle. This particle is called the neutrino (ν)
_
(and antineutrino ν).
11
6C
11
0 +
5 B + +1 β +
14
6C
_
11
0 7 N + -1 β + ν
ν
Beta plus decay
Beta minus decay
Neutrinos


Neutrinos are very low mass neutral particles which do
not interact very much other particles.
Neutrinos and antineutrinos interact by means of the
weak interaction (see later notes).
Weak interaction and W bosons
The exchange particle involved with the weak interaction
is called the W boson.
W bosons have:
 Non-zero rest mass (unlike photons)
 Very short range (~ 0.001fm)
 Are charged, W+ is positively charged whilst W- are
negatively charged.

Feynman diagrams for B- decay



Note that in the Feynman diagrams charge is conserved.
In β− decay, the weak interaction converts a neutron (n)
−) and an
into a proton (p)
while
emitting
an
electron
(β
_
antineutrino (ν).
Recall that a proton consists of uud quarks, a neutron
consists of udd quarks. So β− decay is due to the
conversion of a down quark (d) to an up quark (u) by
emission of a W− boson; the W− boson subsequently
decays into an electron and an antineutrino.
Feynman diagrams for B- decay
_
ν
Feynman diagrams for B+ decay

In β+ decay, the weak interaction converts a proton (p)
into a neutron (n) while emitting a positron (β+) and an
neutrino (ν).

Recall that a neutron consists of udd quarks, a proton
consists of uud quarks. So β+ decay is due to the
conversion of an up quark (u) to a down quark (d) by
emission of a W+ boson; the W+ boson subsequently
decays into an positron and an neutrino.
Feynman diagrams for B- decay
t
n
_
ν
udd
W+
udu
P
Neutrino interactions



Neutrinos rarely interact with other particles:
However a neutrino can interact with a neutron making it
turn into a proton by means of the weak interaction. The
process also releases a β− particle.
In equation form:
1
0
n+ ν
W
-
1
1
p+ β
0
-1
Feynman diagrams for neutronneutrino interaction
p
Future
WTime
n
Past
e
_
_
(b)
Exchange particle
is a W- boson
n
Antineutrino interactions



Antineutrinos interact in a similar way:
An antineutrino can interact with a proton making it turn into
a neutron by means of the weak interaction. The process
also releases a β+ particle.
In equation form:
1
1
_
p+ ν
W
+
1
0
n+ β
0
+1
Feynman diagrams for protonantineutrino interaction
n
Future
+
e
W+
Past
(b )
Exchange particle
is a W+ boson
Time
p
+
_
n
Another weak interaction
Electron capture


A proton in a proton rich nucleus may turn into a neutron
by interacting through the weak interaction with an inner
shell electron form outside the nucleus.
The W+ boson turns the electron into a neutrino.
1
p
+
1
0
-1
e-
W
+
1
n
+
ν
0
Feynman diagram for electron
capture
n
Future
n
W+
Time
p
Past
e
_
Summary of Weak interactions






The exchange particle is the W+ or W- boson
W+ or W- bosons have a non-zero mass
W+ or W- bosons are charged
When a proton interacts with the weak interaction the W+
boson is the exchange particle.
When a neutron interacts with the weak interaction the
W- boson is the exchange particle.
Charge is always conserved in all weak interactions
Building blocks of nature



Fundamental particles are particles which are believed to
have no substructure, that is particles which cannot be split
into smaller particles.
Hadrons are particles composed of fundamental particles
known as quarks.
For each type of quark there exists an antiquark
The quark family

There are thought to be 6 types (or flavours) of quarks,
arranged in 3 groups called generations:
Quark type
1st generation
2nd generation
3rd generation
= Up (u)*
Charge (p =1)
+⅔
Down (d)*
-⅓
= Strange (s)*
+⅔
Charm (c)
-⅓
= Top (t)
Bottom (b)
* Only need to worry about these quarks.
+⅔
-⅓
More about quarks



For each type of quark there exists an anti-quark. The
charge on an anti-quark is the opposite to its quark
equivalent.
For example a up quark (u) has
a charge of +2/3,
_
therefore an anti-up quark ( u ) has a charge of -2/3.
Protons are _made up of 3 quarks (2u and 1d). Therefore
antiprotons
p are made up of 2anti-up and 1 anti-down
___
quark ( uud ).
Particle families

We can categories all particles in the following manner:
All particles
Matter and antimatter
Hadrons
Leptons
Baryons
Mesons
(3 quarks)
(quark and antiquark)
Hadrons



Particles (and antiparticles) that interact by the strong
interaction (and electromagnetic interaction if charged)
are called hadrons. They decay through the weak
interaction with the exception of protons which are stable.
They are composed of quarks (and antiquarks).
Hadrons may be further divided into two categories:
Baryons and Mesons.
Baryons



Baryons all consist of a combination of 3 quarks
(antibaryons consist of 3 antiquarks).
The quarks are bound together by the strong interaction. It
is impossible to isolate a solitary quark.
Baryons include the familiar protons and neutrons, but
also include more exotic particles…see table in hand out.
Mesons


All known mesons are believed to consist of a
quark- antiquark pair.
There are approximately 140 types of quark.
For example a pion p- consists of a down
quark and an anti-up quark (du).

See handout for table of examples.
Leptons




Leptons are a family of particles that do not interact
through the strong interaction, they include electrons and
neutrinos, see full list in hand out.
Leptons do interact through the weak interaction and, if
charged, through the electromagnetic interaction.
There are 3 types of lepton (and the antiparticle), each
type has its own associated neutrino
Leptons are thought to be fundamental, that is they
cannot be broken down into simpler particles.
Leptons
τ
Name
Symbol Charge (p)
Electron/(positron)
e- / e +
-1 / +1
Electron
ν
ν
0
e
e
neutrino/antineutrino
Muon/(antimuon)
μ- / μ+
-1 / +1
Muon
ν
ν
0
μ
μ
neutrino/antineutrino
- / τ+
-1 / +1
τ
Tauon/(antitauon)*
Tau*
neutrino/antineutrino
ντ ν-τ
0
Mass (MeV)
0.511
>0
105.5
>0
1777
>0
Particle Collisions



Many of the exotic particles are studied by colliding beams
of particles and antiparticles into each other.
It is important that the conservation of mass-energy is
applied:
The total mass-energy of particles and antiparticles does
not change after the collision. That is:
Total energy =
Rest energy of
particles
+ Kinetic energy of
particles
Particle Collisions

Therefore by applying the conservation of energy
(mass):
Rest energy of =
Total energy
products
before collision
_
Kinetic energy
of products
Particle collisions

Knowing the rest energy of the products is this gives
information as to the possible identity of the products. For
example: if we have a collision between a proton and an
antiproton each with a rest energy of 1GeV and if each has
a kinetic energy of 2GeV, the total energy is given by:
Total energy before collision (Etot)
Etot = Total rest energy + Total kinetic energy
= (2 + 2) + (1 + 1) = 6 GeV

Therefore the resulting collision could produce a range of
products as long as the conservation rules are applied and
the products total energy does not exceed 6 GeV.
Lepton number conservation




Leptons are assigned a lepton number = +1
Antileptons are assigned a lepton number = -1
Non-leptons are assigned a lepton number = 0
In any interaction the total lepton number is conserved.
Example of lepton number
conservation 1

Consider the two interactions:
νe + n
Lepton
number =
Lepton
number =
→
p +
+1
0
0
νe
and
+ n
_
+1
0
→
X
p
0
e-1
+
 Allowed
Because lepton
number is conserved
e+
 Not Allowed
-1
Because lepton
number is not
conserved
Example of lepton number
conservation 2

Muon decay: which of the following decays is allowed?
Use the lepton number conservation rule to find out.
μ-
→
e-
+
or
μ-
→
e-
+
_
νe + νμ
_
_
νe + νμ
Example of lepton number
conservation

Consider the two interactions:
μ- → eLepton
number =
+1
+1
νe
+
νμ 
+1
-1
+1
and
_
_
e
μ- →
X
Lepton
number =
+
_
+1
+
νe
-1
+
Because lepton
number is conserved
νμ 
-1
Allowed
Not Allowed
Because lepton
number is not
conserved
Lepton branches
Consider the muon decay below:
μ


→
?
e-
+
νe
+
_
νμ
The above change satisfies both the rules regarding the
conservation of charge and lepton number. But this change is
never seen. Why?
Remember there 3 branches (families) of lepton, each branch
has the particle together with its neutrino and their antiparticles.
The conservation rule must be applied to each lepton branch in
any interaction. Therefore, a muon must produce muon
neutrino and not an antineutrino. So you must check that each
family has its lepton number conserved.
K mesons (Kaons)


Strange particles were discovered in experiments in which
protons (or neutrons) and pions (π mesons) collide with each
other.
They were called strange particles because they decayed into
through the weak interaction to give either:
1. π mesons, or
2. π mesons and protons
1. Particles that decayed to give only π mesons were called K
mesons.
2. Particles that decayed to give π mesons and protons were
sigma (Σ) particles. See diagram
K mesons (Kaons)
π meson
K meson
π meson (green) collides
with proton (red)
π meson
π meson
Σ (sigma)
particle
proton
Quarks and Antiquarks
1
π- + p → K+ + Σ-
Was observed
2
π+ + n → K+ + Σ0
Was observed
3
π- + n → K0 + Σ-
Was observed
4
π- + n → K- + Σ0
Was not observed
Conservation of strangeness!!





Why were some reactions observed but not others?
Remember that one of the six flavours of quark is the
strange quark.
To explain why some reactions are not observed it is
necessary to introduce another conservation rule: the
conservation of strangeness number (S)
Strange quarks have a strangeness number of -1,
strange antiquarks have a strangeness number of +1, all
other quarks have a strangeness number of zero.
In any strong interaction strangeness is always
conserved.
Strangeness in the quark family
Quarks
Up
Charge
(p=1)
Strangeness
S
Antiquarks
Down Strange
Up
Down Strange
_
d
_
s
u
d
s
_
u
+⅔
-⅓
-⅓
-⅔
+⅓
+⅓
0
0
-1
0
0
+1
Conservation of strangeness!!
1
π- + p → K+ + Σ-
Was observed
2
π+ + n → K+ + Σ0
Was observed
3
π- + n → K0 + Σ-
Was observed
4
π- + n → K- + Σ0
Was not observed
Explanation of interactions

Using the conservation of strangeness it is easy to see
why some reactions were observed and others were not.

For example, in reaction 1 below:
π- + p → K+ + Σ

The π- and p both have S = 0, the K+ has S = +1 and the
Σ- has S = -1, so strangeness is conserved.
However, consider reaction 4 below:
π- + n → K- + Σ0
The π- and p both have S = 0, the K- has S = -1 and the Σ0
has S = -1, so strangeness is conserved.
Conservation of baryon number
A further conservation rule covers baryons.
 Baryon numbers are assigned as follows:
+1 for baryons
-1 for antibaryons
0 for leptons and mesons.
(another way to consider this is that quarks have a baryon
number of 1/3 and antiquarks have a baryon number = -1/3)

In any reaction the total baryon number is always conserved
Example of baryon number conservation

Consider the reaction:

This reaction is observed because it obeys the conservation
of baryon number rule. I.e. p = 1, p = -1, and π+ and π- both
= 0, so both sides = 0.
However, consider the reaction:

_
p + p → π+ + π-
_
p + p → p + π+
This reaction is not observed …..Why?
In terms of baryon numbers, the left side has a total of 0, but
the right side has a baryon number totalling 0. What other
conservation rule is disobeyed?
 Answer: Charge
Chapter 3: Quantum phenomena
Atomic Spectra

Electrons in an atom can only have only certain specific
energies. These energies are called the electron
energy levels. An energy level diagram shows the
possible energies for an electron in a given atom.
For example: the energy level diagram for hydrogen
0 eV
Ionisation level
-0.8 eV
Second excited state
-3.4 eV
First excited state
Energy (eV)
-1.5 eV
-13.6 eV
Ground state
Energy changes




Absorption of energy causes an electron to go a higher
energy level.
The atom gains energy.
Emission of energy causes the electron to fall to a lower
energy level.
The atom loses energy.



Ground state and excitation of electrons
Normally electrons occupy the lowest energy levels
available. The atom and electrons are said to be the
ground state.
The electron can absorb energy; this causes it to
move to a higher energy level. The atom is said to
be in an excited state.
A precise amount of energy is needed to excite an
electron from one level to another. The energy is
equal to the difference in energy between the two
levels.
Example
if an electron in a hydrogen
atom is excited from the
ground state to the first
excited state it must absorb
energy (ΔE) equal to E2 – E1
ΔE
-3.4 eV
First excited
state
-13.6 eV
Ground
state
= -3.4 - -13.6 = 10.2 eV
Emission contd.




An electron in an excited state must lose energy for
it to move back down to the ground state. The
energy must again be exactly equal to the difference
in energy between the two levels.
Example: in the above electron energy level for
hydrogen, how much energy (ΔE) must an electron
in the 2nd excited state lose for it to fall directly back
to the ground state?
ΔE = energy of the excited state – energy of the
ground state
ΔE = -1.5 – (-13.6) = 12.1 eV
Emission

Under usual
conditions, an
excited electron will
only temporarily
remain excited.
Excited electrons
soon return to the
ground state, this
may be directly (A)
or via another
energy level (B).
A
B
-1.5 eV
2nd Excited
state (E3)
-3.4 eV
1st Excited
state (E2)
-13.6 eV
Ground state (E1)
Ionisation energy
Energy (eV)
0 eV
Ground state
(-13.6 eV for H)
e.g. for hydrogen the ionisation energy = 13.6eV
Points regarding ionisation


Electrons in the highest energy level are defined as
having zero energy. Consequently, all other energy
levels are negative.
Free electrons have only kinetic energy and their total
energy is always positive.
Allowable changes and electron
transition


The energy levels for an atom are fixed and
are unique for that particular chemical
element.
All atoms of the same element have the
same energy levels.
Electron excitation


Electrons can be excited to higher energy levels
by many different ways:
Usually by the absorption of electromagnetic
radiation
Also by:


collision with other particles (e- impact)
heating
Important point
The energy (ΔE) needed to excite an electron to a higher energy
level is exactly the difference in energy between the two levels,
i.e.
ΔE = E2 – E1
E2
Excited
state
E1
Initial
state
Important point

If the excitation is caused by the absorption of a photon,
the photon energy (hf) must be equal to the difference in
energy between the two energy levels,
i.e.

hf = ΔE = E2 – E1
If the excitation is caused by electron impact, the electrons
kinetic energy does not have to match ΔE as any surplus
kinetic energy is retained by the colliding electron.
i.e.
KEinitial = ΔE + KEfinal
Emission of energy


When an electron falls to a lower energy level it
rids itself of the surplus energy (ΔE) by emitting
a photon of EM radiation.
The value of ΔE is defined as being the exact
difference between the energies of the two
energy levels (E1 and E2).
Therefore:

ΔE = (E1 - E2) = hf
Ionisation

If an electron can gain enough energy to
reach the highest energy level, it can leave
the atom (is free). The atom is said to be
ionised.
The ionisation energy is defined as: the energy
required to completely remove an electron (in
its ground state) from an atom.
Emission Spectra



Because the atoms of each element have a
characteristic set of energy levels, they will emit a set of
unique characteristic frequencies.
Analysis of the frequency (or λ) of the light emitted can
be used to identify the element.
The group of frequencies of radiation emitted are called
a line spectrum. Each element has its own unique
spectrum.
The atomic hydrogen spectrum
Types of spectra
Quantum Physics Part 2
The Photoelectric effect
The Photoelectric effect


Free electrons are held in a metal by the electrostatic
attraction of the positively charged nuclei.
If an electron is to be removed from the surface of a metal
it must be given energy.
Photoelectric emission is the release of electrons from
the surface of a solid (metal) when electromagnetic
radiation is incident on the surface.
Photoemission
Incident EM
radiation
Released
electrons
(photoelectrons)
Metal
The work function (Φ)
The minimum
amount of energy to
remove an electron
from a substance is
called the work
function (symbol Φ).
Different materials
have different work
functions:
Material
Caesium
Potassium
Lithium
Zinc
Work function (J)
3.43 × 10-19
3.6 × 10-19
3.71 × 10-19
6.6 × 10-19
Threshold frequency

To remove the electron from the material, the photon energy
(hf) must be at least equal to the work function.

The lowest frequency at which the photon energy is sufficient
to remove an electron is called the threshold frequency
(fmin).

i.e.
hfmin = Φ
Threshold frequency

If the frequency of the EM radiation is less than the work
function no electron is released. The condition for photoelectron
emission is:
hf ≥ Φ

Note: in calculations sometimes a maximum wavelength that
corresponds to the threshold frequency is given, this is called
the cut-off wavelength (λo).
λo = c ÷ fmin
Einstein’s Photoelectric equation

For photoelectric emission, the photon energy must be
at least equal to the work function of the material.

What happens to the excess energy if the photon
energy is greater than the work function?

The answer is that the excess energy is taken by the
released electron as kinetic energy (Ek).

Conservation of energy gives Einstein’s equation:
Einstein’s Photoelectric equation
Max kinetic energy
of electron
photon
Work
=
function
energy
So
Ek = hf – Φ
Example
3.0 eV
Photon
arrives
Electron emitted
with kinetic
energy of 0.7 eV
Lithium metal Φ = 2.3 eV
Example 2
Light of frequency 6.7 × 1014 Hz shines onto clean Caesium
metal. What is the maximum kinetic energy and speed of the
electrons emitted?
The work function of caesium metal is 3.43 × 10-19J and
Planck’s constant is 6.63× 10-34 Js.
Electron emitted
with kinetic
energy = ?
Photon
frequency =
6.7 × 1014 Hz
Caesium Φ = 3.43 × 10-19 J
Answer
Using Einstein’s equation gives
 Ek = hf – Φ
 Ek = (6.63× 10-34 × 6.7 × 1014) – 3.43 × 10-19
 Ek = 1.01 × 10-19 J


As Ek = ½mv2, this gives v2 = 2Ek/m
therefore v = 4.71× 105 ms-1
Photoelectric effect experiments

A simple photocell may be used to determine
a value for the Planck constant (h).
Apparatus:
Light of known
frequency f
_
pA
R
(
E
Photoelectric
cell
V
+
Method






The potentiometer R is set so that the voltmeter reads zero.
Light is directed onto the emitting electrode (E).
If frequency f ≥ fo photoelectrons are emitted.
The release of the photoelectrons is detected by the current
flowing through the picoammeter.
The voltage across the photocell is increased until the
current decreases to zero. This voltage value is called the
stopping potential (Vs).
The stopping potential is recorded for a range of frequencies.
The photocell
The work done in moving a charge q
through a potential difference V is
given by:
 Work done = qV
 The energy needed to reach the
collector electrode (c) = qV
_
+
c
Determination of Planck’s constant

The energy needed to reach the collector electrode = qV

Where q = electron charge (1.6 ×10-19C)

Where V is the voltage across the photocell

V is increased until the electrons cannot reach the collector.

The stopping potential Vs is the voltage when electrons fail
to reach the collector.

At this stage: KE of electron = qVs
(1)
Determination of Planck’s constant

From Einstein’s photoelectric equation:
Electron KE = ½mv2 = hf – Φ
From equation (1) :
qVs = hf – Φ

Therefore:



Vs
Φ
hf
=
–
q
q
and
Vs
=
hc
qλ
–
Φ
q
Determination of Planck’s constant
Using the equations:
Vs
hf
=
q
–
Φ
q
or
Vs
=
hc
qλ
–
Φ
Q
Allows the value of Planck’s constant to be determined. This is
achieved by recording the stopping potential and either the
corresponding frequency or wavelength.
Determination of Planck’s constant
Stopping
potential (Vs)
If the stopping potential Vs (Y-axis) is plotted against the frequency f (or
1/λ) (X-axis), the graph can be used to determine Planck’s constant h.
Gradient of graph = h/q
fo
Φ/q
Photon
frequency (Hz)
Note, if the X-axis is 1/λ,
the gradient is hc/q
Saturation and photocurrents




The photoelectrons are released from the surface with a
range of kinetic energies.
If the collector electrode has a negative potential not all the
electrons will reach the collector (their kinetic energy < qV).
However, if the collector electrode potential is made positive,
the photocurrent increases (as more electrons are collected).
At the saturation potential ALL the electrons are collected.
Saturation and photocurrents
+
c
Photocurrent
-
Vs
_
Saturation
0
+
Voltage
Frequency, current and intensity

The intensity (brightness) of the light determines the number
of photons that are incident on the surface. The greater the
intensity, the higher the photocurrent (providing that f ≥ fo).
Photocurrent
Saturation
bright
dim
Vs
0
Voltage V
Frequency, current and intensity (II)

If red and blue lights of equal intensity are used, the
resulting graph would be:
Photo-current
Red
light
Blue
light
Vs
Voltage V
The blue light has a
higher (-)Vs as it has a
higher photon energy.
Frequency, current and intensity (II)


The intensity of the two radiations is the same. This means
that both the red and blue lights deliver the same amount of
energy per second.
However, each red light photon is lower in energy than the
blue light so there are a greater number of red light
photons. The greater number of red light photons causes
the photocurrent to be higher for the red light source.
Summary on the Photoelectric effect
You should know that:





Energy has to be supplied to remove electrons from a metal.
Surface electrons are the easiest to remove (and need lowest energy).
Different metals require differing amounts of energy to release their
surface electrons, this energy is called the work function (Ф).
A photon can release a surface electron only if it has sufficient energy
(hf ≥ Ф).
Photon energy is determined by frequency f, so there is a minimum
frequency called the threshold frequency fo that can release electrons
(hfmin = Ф).
Duality: particle or wave?



One of the greatest debates in Physics: what is light?
Is light a wave or a particle?
Isaac Newton : light consists of a stream of particles
called corpuscles.
Christian Huygens 1655: Light is a wave that
propagates through an invisible medium called the
luminiferous ether.
Thomas Young 1802: Double slit experiment,
explained by the principle of wave superposition.
Duality: particle or wave?





Augustine Fresnel 1815: Explained diffraction.
Interference of light waves through oil films.
Leon Focault 1850: measured speed of light in air
and water.
James Clerk Maxwell 1865: light is an EM wave.
Heinrich Hertz 1887: light shares properties with
radio waves.
Albert Einstein 1905: Photoelectric effect; cannot
be explained if light is a wave, but can be explained
if light acts as a particle.
Duality: particle or wave?


Contemporary ideas: light can be thought of as being
neither a true wave nor particle. However, light does show
wave-like and particle-like behaviour but never both at the
same time.
So the nature of light is essentially schizophrenic, but what
about other particles………?????
Matter waves

Prince Louis de Broglie
(1924) established an
equation for the
relationship between a
particle’s momentum and
its wavelength.
Or in symbol form:
Planck’s constant
Wavelength =
momentum
h
h
λ =
=
mv
p
Matter waves

Electrons are a form of matter, so these waves are
called matter waves; the wavelength of matter waves
is sometimes called the de Broglie wavelength.

de Broglie concluded that everyday objects are too
heavy to observe their wave properties, i.e. their
wavelengths are too small to be observed.
Example:

Calculate the wavelength of a Physics student of mass
50kg running at a speed of 5ms-1 to catch her bus.
λ =
h
mv
6.63 X 10-34
=
=
50 X 5
2.65×10-36 m
The wave character of the student is so immensely smaller
than the student, that there are no experiments we can
perform that can probe her wave character. Hence, we
never observe students diffracting.
Example 2
• When the mass of the object is very small, the wave
properties can be detected experimentally.
 For example, calculate the wavelength of an electron which
has a velocity of 1×106 ms-1
h
6.63 × 10-34
-10 m
λ =
=
=
7.29
×10
9.1× 10-31 x 1×106
mv
• The above electron has a similar wavelength to x-rays.
• In 1927 the diffraction of electrons was observed
experimentally by C. J. Davisson.
Wave-Particle duality



Most properties of light, such as diffraction, interference and
polarisation can be explained by light being considered as a
wave.
Wave theory cannot explain the photoelectric effect.
However, photoelectric effect can be explained by light
being considered as a particle.
When electrons are fired at graphite or metal targets, a
diffraction pattern is produced. Electron diffraction is used
as a probe of matter in Physics and as an analytical
technique in Chemistry.
Electron diffraction
X-Ray diffraction
Summary of conservation
rules
In any interaction charge is always conserved
Additionally, in
