Name ………………………………………………………
Advancing Physics A2
Chapter 17
Probing deep into matter
Student Notes
John Mascall
January 2010
The King’s School, Ely
Assessable learning outcomes for Ch 17
Candidates should demonstrate evidence of:
1. knowledge and understanding of phenomena, concepts and relationships by describing and
explaining:
(i) use of particle accelerators to generate high-energy beams of particles for scattering
(details of the construction of accelerators are not required);
(ii) evidence from scattering for a small massive nucleus within the atom;
(iii) evidence of discrete energy levels in atoms
(for example, from collisions with electrons or from line spectra);
(iv) a simple model of the atom as the quantum behaviour of electrons in a confined space;
(v) simple picture of the internal structure of protons and neutrons;
(vi) relativistic calculations for particles travelling at very high speed, eg in particle accelerators
or cosmic rays;
2. scientific communication and comprehension of the language and representations of physics,
by making appropriate use of the terms:
(i) energy level, scattering, nucleus, proton, neutron, nucleon, electron, positron, quark,
gluon, neutrino, hadron, lepton, antiparticle;
by sketching and interpreting:
(ii) paths of scattered particles;
(iii) electron standing waves in simple models of an atom;
3. quantitative and mathematical skills, knowledge and understanding by making calculations and
estimates involving:
(i) motion of a charged particle in magnetic field using F = qvB;
(ii) kinetic and potential energy of a scattered charged particle;
(iii) Erest = mc2 and relativistic factor γ = Etotal / Erest
A Revision Checklist for Chapter 17 can be found on the Advancing Physics
CD-ROM.
Page | 2
Section 17.1
Creation and annihilation
Learning outcomes
●
Particles can be created and annihilated in matter–antimatter pairs.
●
Electrons and positrons have opposite electric charge and lepton number.
●
Electromagnetic interactions arise from the exchange of photons.
●
Electrons are fermions, which obey the Pauli exclusion principle.
●
Leptons are particles like the electron, neutrino and their antiparticles.
●
Total energy, momentum and electric charge are conserved in interactions.
A survey
In this opening section we give a broad brush account of the particle physicist's view of the
Universe. We start with the student's book (pp 189-191), which discusses the role of
electron/positron annihilation in the PET scanner – antimatter in everyday life. A very
useful programme that enables you to see how diagnosis can be made with a PET
scanner can be found at http://www.insidestory.iop.org/. In addition, Section 9 of the
Classroom Video ‘Radioisotopes at Work’ is a 2 minute clip showing PET scanning in
action together with the related theory.
Making PET scans
A pair of gamma rays are emitted in opposite
directions as a result of electron–positr on
annihilation inside the patient
Scintillator: captures a gamma-ray
photon and emits lower-energy
photons into photomultiplier tubes
Display Material 20O
OHT 'Making PET scans'
sig nal p ro cessing

Photom ultiplier:
incoming photons create
a cascade of electrons,
giving an electrical pulse
output
One pair of detectors will
respond almost
simultaneously. This near
coincidence shows that
the two gamma-ray
photons came from a
common source. The tiny
time difference between
the two signals is then
used to work out where
they came from along the
line between the
detectors.

A nn ih ilatio n of po sitron s f ro m o xygen -15
Oxygen-15 is an unstable isotope. It is
us ed to ‘label’ a m olecule like glucos e
that is used in body metabolism .
+
e + It dec ay s by  decay,
emitting a positron (and a
neutr ino).
T he pos itron (e + ) c ollides alm ost im mediately with
an electron (e – ) in on e of the s urrounding a tom s.
e+
e–
T he elec tron and positron annihilate, c reating a pair
of gam ma-ray photons that travel in opposite
direc tions. PET sc anners detect these gamm a-ray
photons.


Scintillators are
arranged in a grid
on the inside
sur face of the
scanner. In any
short period of time
many detectors will
res pond to
gamma-ray
photons from
many different
annihilations inside
the body. A
com puter
produces a sliceby- slice map of
activity in the brain.
A positron and electron annihilate each other, emitting a pair of gammaray photons travelling in opposite directions. The detection of many of
these leads to a map of brain activity being built up
Page | 3
Display Material 30O
OHT 'Conserved quantities in electron–positron annihilation'
Conserved quantities
e–
Simplify:
assume head-on
collision with
equal speeds
You should appreciate that the
following quantities are conserved
during electron-positron
annihilation:
e+


Energy is conserved
total energy before
total energy after
=
= kinetic energy of particles
+ rest energy of particles
*this includes rest energy given by
Erest = mc2
energy after is
energy of gamma
photons
= 2  0.511 MeV
minimum value of energy
before is rest energy:
= 2 mc 2 = 2  0.511 MeV
Energy*
Momentum
Charge
Momentum is conserved
total linear momentum before
e–
=
total linear momentum after

e+
same mass; equal and
opposite velocities

energy E,
momentum p = E/c
photons identical,
momentums opposite
total momentum before = 0
total momentum = 0
Electric charge is conserved
total charge before
charge
(–e) + (+e) = 0
=
total charge after
charge
0 +0= 0
Energy, momentum and electric charge are always conserved in
electron–positron annihilation
Questions - use the data sheet to find the values needed.
1
Use the formula E = mc2 to show that the rest energy for an electron is 0.511 MeV.
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Page | 4
2
Calculate the wavelength of a photon that has an energy of 0.511 MeV (this is the
the minimum energy available for one photon from the annihilation described
above). You should use a copy of the electromagnetic spectrum to check that
the wavelength obtained is consistent with that of a gamma ray.
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3
(a) Use the formula p = E/c to calculate the momentum of each of the gamma ray
photons that would be emitted as a result of the annihilation of a stationary electron
and a stationary positron. You should understand why this calculation is of
theoretical interest only!
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(b) Suggest how your answer would differ if the electron and positron were to collide
head-on with equal speeds as suggested in the diagram above.
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Pair creation is shown on page 191 and is discussed on page 192 (see also the diagram
below). More images of both pair creation and annihilation can be seen on Display
Material 10S OHT 'Annihilation and pair production: bubble chamber pictures'
With pair creation, an electron and
positron are produced from a single
gamma ray under certain conditions.
Page | 5
Question
(a) Why do the electron and positron move in spiral paths?
…………………………………………………………………………………………………………
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(b) Why do the electron and positron spiral in opposite directions?
…………………………………………………………………………………………………………
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Pair creation is less simple than it might seem. Imagine the process of annihilation above
in reverse. Two identical gamma rays would come together to create an electron-positron
pair. This should happen because energy, momentum and charge are all conserved.
However, it is rather unlikely! Pair creation could occur with a single photon of twice the
energy of each of the photons considered - this would conserve energy and charge - but
linear momentum would not be conserved. Momentum can be conserved if the gamma ray
photon passes close to the nucleus of an atom so that the nucleus recoils to conserve
momentum. In this case the gamma ray must have an energy equal to at least 2mc2 which
is the rest energy of the electron and positron.
Question
Calculate the minimum energy of a single gamma ray that is able to create an electronpositron pair by pair creation.
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Page | 6
Display Material 40O
OHT 'Pair creation and annihilation'
Pair annihilation and creation
Annihilation


gamma energy
= 2  0.511 MeV plus
kinetic energy of electrons
e–
e+
Creation?
e–
e+
extremely rare
(cannot bring two
identical photons
together)


e–
e+
Pair creation
gamma energy
= 2  0.511 MeV (minimum)
nucleus carries away
momentum, to conserve
momentum and energy
close to
nucleus

Note that charge is conserved in all three situations described by the diagram.
You should find out more about the bubble chamber and how to interpret the photographs
obtained by working through Activity 10S Software Based 'Bubble chamber photographs'.
A taste of real particle physics work can be gained by working through Activity 50S
Software-based ‘Identifying particles using Lancaster Particle Physics software’. The
software required can be downloaded from http://lppp.lancs.ac.uk.
Page | 7
Particles and fields
Here we present an overview, showing how conservation laws shape our understanding of the Universe at
the deepest level. Little of the detail appears in the specification so it would be possible to omit this section
having read the relevant part of the student’s book.
You will see on pages 192-193 that the discovery of processes in which particles are created and destroyed
led to a completely new way of looking at fields and matter.
Chapters 15 and 16 deal with the older classical model of electric and magnetic fields and chapter 11 deals
with the classical view of the gravitational field. In classical electromagnetism, charged particles are the
source of the electric and magnetic fields (called the electromagnetic field). These fields exert forces on
other charged particles. The particles of matter and the fields created by the charged particles are
completely different things.
According to the more modern quantum theory, the electromagnetic field is a creator and destroyer of
photons. Photons are created at one point and travel to another point where they are absorbed. In the
process of travelling they ‘try all paths’ (see Chapter 7). It is the interaction of the field with the charges on
the particles that causes the photons to be created and destroyed. Charge is then thought of as being
something that is proportional to the quantum amplitude to create or destroy a photon – it represents the
strength of the interaction between an electron and a photon.
According to quantum theory, forces arise between a pair of charges as a result of the charges exchanging
photons which carry energy and momentum. Whether the force is attractive or repulsive depends on charge
because the sign of the charge determines the phase of the quantum amplitude to create or destroy a
photon. The end result is determined by adding up all the amplitudes taking account of phase, so the result
may be attraction or repulsion. This idea is represented visually in the diagram below.
Display Material 50O
OHT 'Quantum fields create and destroy particles'
Quantum fields create and destroy particles
cannot happen
cannot happen
Electromagnetic quantum
field creates a photon
charge e
photon carries
away energy
and momentum
Electromagnetic quantum field
destroys a photon
photon delivers
energy and
momentum
charge e
Virtual photons
The photon appears and then
vanishes within the interaction so it
can never be detected. These
particles are merely exchanged –
that means they are emitted by one
particle and absorbed by another.
However, their effects have an
influence on the probability of the
process.
energy and momentum not
both conserved
combined process can happen
virtual
(unobservable)
photon
charge e
charge e
photon exchanged:
charges exchange energy and
momentum
energy and momentum conserved overall
Forces between charges arise from the exchange of momentum through
exchange of virtual photons
Page | 8
Remember that a basic rule of quantum theory is to ‘try all
paths’. In the diagrams 70O and 80O below, each picture
shows one way for a given process to happen – each is like
a possible path for that process. For each of the two
diagrams the quantum amplitude (phasor arrow) for that set
of possible paths can be written down. By adding these
quantum amplitudes over all places and times, taking
account of phase, it is possible to arrive at a total amplitude
for the phasor arrow for each of the two processes. The
probability of each process is then found by squaring the
resultant amplitude.
Display Material 70O
Feynman diagrams
Feynman diagrams are used here to give a
visual impression of the event. They were
devised to give a means of keeping track of all
the possible ways particles can interact. Each
diagram pictures one way for a given process
to happen. You can ignore any reference to
Feynman diagrams if you wish as, whilst they
appear in the textbook, they do not appear in
the specification. If you wish to understand
Feynman diagrams you might like to consult
QED (stands for Quantum Electrodynamics) by
Richard P Feynman published by Penguin
Books in 1990. The title goes on to describe the
book as ‘The strange story of light and matter’.
It is certainly that!
One simple point you will pick up in Chapter 3
is that the y axis represents time and the x axis
represents space.
OHT 'Feynman diagrams show possibilities to be combined'
Feynman diagrams show possibilities to be combined
Propagation and interaction of a pair of electrons
Electrons at A and B arrive at C and D
no photon exchange
C
D
A
A
B
electrons just travel A to C and
B to D
Electrons are
identical
so there is no way
to tell these apart:
D
C
add the phasor
arrows for both
diagrams
B
electrons just travel A to D
and B to C
one photon exchange
D
D
C
C
B
A
electrons exchange a photon
A
B
No way to tell
these apart:
add the phasor
arrows for both
diagrams
electrons exchange a
photon
.... plus diagrams with more photons...
For each diagram, add phasor arrows for all possible space-time locations A, B, C, D.
Add total phasor arrows for each type of diagram.
Quantum rule: Try all possible ways to interact
Page | 9
Display Material 80O
OHT 'Ways for an electron to scatter a photon'
Ways for an electron to scatter a photon
In each diagram one electron and one photon come in, and one electron and one photon go out.
All diagrams represent the same process.
interaction with virtual pair
creation and annihilation
Simple electron–photon interactions
e–
e–
B
B
e–
A
e+
A
e–
Electron:
absorbs photon at A
travels to B
emits photon at B
A
e–
B
plus other
more
complex
diagrams
e–
Electron:
emits photon at A
travels to B
absorbs photon at B
photon creates e –, e+ pair at B
e + goes to A
e – and e + annihilate at A,
emitting photon
Phasor arrows for all diagrams are added up to find total amplitude and phase for scattering
Fermions and bosons
If two particles such as electrons or protons arrive at the same point in space-time, their
quantum amplitudes subtract – that means they add up with opposite phase. The phasor
arrows point in opposite directions leading to a total amplitude of zero leading to a situation
where the two particles can never be at the same place at the same time. Particles like
these are called fermions. Chemists will have come across the idea when thinking about
building up the electronic structure of atoms other than hydrogen. In that case two
electrons can occupy the same space if they have different ‘spin’ because they are said to
have different quantum states as a result. This idea leads to the Pauli exclusion principle
that no two particles, electrons in atoms or protons and neutrons in nuclei, ever share the
same quantum state.
Photons do exactly the opposite. The amplitudes of two identical photons at the same
point in space-time add up with the same phase. The total amplitude doubles if they have
identical amplitudes to start with. This principle is used to good effect in a laser where
many photons in the same state join together. Particle that behave this way are known as
bosons. All particles whose exchange gives rise to forces, such as photons and gluons,
are bosons.
The link with ‘spin’ is that fermions are said to have a ‘spin’ of ½ [in reality this is
½ (h/2)] and bosons are said to have a ‘spin’ of 1 [in reality this is (h/2)].
Page | 10
OHT 'Identical particles – bosons and fermions
Display Material 90O
Identical particle s in the same state
Identical particles in the same state
Photons (bosons)
Electrons (fermions)
Photons are indistinguishable.
These possibilities cannot be distinguished.
Add the phasors for the two diagrams.
C
Electrons are indistinguishable.
These possibilities cannot be distinguished.
Add the phasors for the two diagrams.
D
C
X
D
Y
A
X
B
photon X goes to D
photon Y goes to C
photon X goes to C
photon Y goes to D
D
C
X
Y
A
B
C
Y
A
D
X
Y
B
A
B
electron X goes to C
electron Y goes to D
electron X goes to D
electron Y goes to C
bring points C and D together
bring points C and D together
E
E
Exchanging particles
reverses phase.
Adding phasors gives
zero amplitude.
Phasors for the two
diagrams become
the same.
Adding phasors
gives 2  amplitude
4  intensity.
A
A
+
B
B
+
=0
=
There is zero probability for two electrons to be at the same space-time
point (be in same state)
Photons in a given state increase the chance of others joining them in
the same state
This is the Pauli exclusion principle
Beta decay and conservation laws
The positrons used in the PET scans come from a form of beta decay in which a positive
electron is emitted instead of the more usual negative electron. This is because the
nucleus has too many protons for the number of neutrons.
If the number of neutrons is too many for the number protons a negative electron is
emitted as when the strontium-90 used in the laboratory decays. A first attempt at
describing this decay would be as follows:
90
90
Sr
38
0
Y +
39
e
-1
The lower numbers show that charge is conserved. Remember that protons have a charge
of +1 and the electron emitted has a charge of -1. A neutron has changed into a proton so
the total number of protons and neutrons is unchanged. It is convenient to invent a new
term for the family of particles that includes the proton and the neutron. We call members
of this family baryons. That means that baryon number is conserved in this interaction.
Now that we know about antiparticles, we can appreciate that protons and neutrons will
have their own antiparticles – these will have a baryon number of -1.
Page | 11
Baryon
proton
neutron
antiproton
antineutron
Baryon number
+1
+1
-1
-1
The equation above would suggest that the beta particles emitted would have a fixed
energy – this would equal the difference between the rest energies of the strontium and
yttrium nuclei. The fixed energy should be 0.546 MeV but experimental observation
suggests that the beta particle energy varies from zero up to the maximum of the expected
value of 0.546 MeV. Where does the excess energy go?
Pauli suggested that a third particle is emitted to conserve energy. This new particle must
also conserve momentum and charge. To conserve charge, this new particle must have
zero charge. This new particle is now called an antineutrino. To see why we have
introduced an antiparticle as opposed to a particle, we must appreciate that the new
particle and the electron produced come from the same family – we call them both
leptons. Lepton number must also be conserved and this works if we give the electron a
lepton number of 1 (it’s a particle) and the new particle a lepton number of -1 (it’s an
antiparticle).
Lepton
electron
neutrino
positron (antielectron)
antineutrino
Lepton number
+1 (lepton)
+1 (lepton)
-1 (antilepton)
-1 (antilepton)
Note that the electron and positron are regarded as antiparticles because they have
opposite charge whereas the neutrino and antineutrino are regarded as antiparticles
because they have opposite lepton number.
For many years it has been thought that the antineutrino has a rest energy of zero so it
travels at the speed of light rather like a photon, and has a momentum p = E/c. Because
the antineutrino interacts extremely weakly with matter it took 26 years to detect it.
The full equation for the beta decay of strontium-90 is now:
90
90
Sr
38
0
Y +
39
0 __
e +
-1
0
Questions
1
Check that the conservation rules apply to this equation by completing the table
below.
Conservation rule
baryon number
charge
lepton number
Left-hand side
Right-hand side
Page | 12
2
Complete the corresponding equation for positron emission given below.
…
15
0
….. +
O
…
8
0
e +
…
Display Material 100O
0
OHT 'Beta decay of strontium-90'
Missing energy in beta decay
Beta decay of strontium-90
rest
energy
strontium-90
expected energy 0.546 MeV
–
0.546 MeV
yttrium-90
38 protons
52 neutrons
39 protons
51 neutrons
Energy spectrum of beta decay of strontium-90
fraction of beta
particles per
unit energy
range
0
energy/MeV
0.546
Beta decay of strontium-90, including antineutrino emission
rest
energy
strontium-90
–
energy E
0.546 MeV

energy 0.546 MeV – E
yttrium-90
Neutrons carry away the ‘missing’ energy
The weak interaction
The change in the nature of particles that occurs in radioactive beta decay is caused by
the weak interaction. This works through the exchange of bosons in the same way as
electromagnetism. However, the bosons are not photons but Z0, W + and W -. These are
massive particles. You are not required to know any detail.
Page | 13
Section 17.2 Scattering and scale
Learning outcomes
●
Scattering experiments reveal the structures of atoms, nuclei and nucleons; the
smaller the scale the greater the energy needed.
●
Atoms have tiny dense positively charged nuclei, made of protons and neutrons
packed together at high density.
●
Quarks with fractional electric charges combine in threes to form neutrons, protons
and other particles.
●
The strong ‘colour’ force between quarks is carried by gluons, which, like photons,
are bosons.
●
At high energies, scattering experiments create a large number of new kinds of
particle.
Historical scattering
Looking at small objects inaccessible to optical radiation requires different techniques, one
of which is Rutherford scattering. Here we introduce Rutherford scattering in its
established historical context.
Students wishing to refresh their GCSE knowledge may wish to revisit the Sunflower
Physics software. Those who have not seen alpha particle tracks in a cloud chamber
would benefit from seeing a quick demonstration at this stage. The apparatus used by
Geiger and Marsden was not many times larger than the cloud chamber!
Display Material 110O
OHT 'Alpha particle scattering experiment'
Rutherford’s scattering experiment
lead block to select
narrow beam of alpha
particles
radium source of
alpha particles
thin gold
foil
scattered alpha
particles
alpha particle
beam
microscope to view zinc
sulphide screen and count
alpha particles
vary angle of
scattering
observed
zinc sulphide screen;
tiny dots of light where
struck by alpha particle
Page | 14
Under the direction of Rutherford, Geiger and Marsden fired alphas particles from a natural
radioactive source at a thin gold foil. The pattern of the scattering led to Rutherford’s
model of the atom with a small, dense, positively charged nucleus surrounded by
distant orbiting electrons. They already knew that the alpha particles were helium nuclei
with a charge of +2e but they had no idea of the size of the alpha particles or of the nature
of the gold nuclei that were the target. They knew that the alpha particles had an energy of
about 5 MeV which is large enough to remove electrons from the gold atoms so they
expected the alpha particles top go right through the gold foil with very little deflection.
A qualitative description of the results is given in the table below. Use current knowledge
of the structure of the atom to complete the table by adding the conclusions that could be
drawn from the observations.
Observation
A very small number of alphas particles
were deflected right back from the foil so
they must have been scattered through a
very large angle.
Conclusion
Most of the alpha particles passed through
the foil with very little deflection.
The diagrams below show the mechanism for the scattering as we now understand it
together with some tests that were carried out to provide further evidence for the
Rutherford model of the atom.
Notes:
Page | 15
OHT 'Rutherford’s picture of alpha particle scattering'
Display Material 120O
Rutherford’s picture of alpha scattering
N umber scattered decreas es with angle
10 5
10 4
nucleus paths of scattered
alpha particles
10 3
R utherford’s
predic tion for a
s mall, m ass ive
c harged nuc leus
10 2
10 1
10 0
0
30 60 90 120 150 180
s c attering angle/degree
F o r calculatio n s
fo rce F =
a lpha particle
s c attered
2Z e 2
4 0d 2
charge +2e

s cattering angle
d
aiming error b
As sump tion s:
a lpha pa rtic le is the He nuc leus, cha rge + 2e
gold nucleus h as charge + Z e, and is much m ore
m as siv e than alpha particles
s c attering forc e is inv erse sq uare el ectrical
repuls ion
gold nuc leus
c harge + Ze
e qual force F b ut
nuc leus is mas siv e,
s o little r ecoil
TEST:
Are slowed -down alp ha particles
scat tered m ore?
TEST:
Does using nuclei of smaller charge
scatter alpha particles less?
Z
Z
reduce alpha
energy with
absorber
r eplace foil by
metal of smaller
atomic number
less energetic alpha
particle turned
around further from
the nucleus
lower speed
Z
alpha particle gets closer
to nucleus of smaller
charge and is deflected
less
smaller nucleus with less
charge, e.g. aluminium
Careful investigation of alpha scattering supported the nuclear model of
the atom
The path of the alpha particle should be understood both in terms of the forces acting and
in terms of the transfer between kinetic and potential energy.
The scattering will be modelled using a 1/r gravitational hill.
The following software activities provide a dynamic view of scattering.
Activity 80S Software Based 'Probes scattered by a target'
using File 40L Launchable File 'Scattered probes'
Activity 90S Software Based 'Many probes scattered by a target'
using File 50L Launchable File 'Many probes scattered'
Students who enjoy a challenge might like to try their hand at Activity 70E Experiment
‘Probing arrangements’.
Here the north-seeking poles of magnets are used to model repulsive interactions.
Page | 16
If the special case of a head-on collision is considered, an upper limit can be estimated for
the size of the nucleus. This was done by Rutherford and his team.
Display Material 130O
OHT 'Distance of closest approach'
The formula for electrical
potential energy of two charges is
Ep = q1q2
40r
or just
Ep = k q1q2
r
where k ≈ 9 × 109 VC-1m-1.
You should know:
● Diameter of atom ≈ 10-10 m
● Diameter of nucleus ≈ 10-14 m
● The nucleus is about 10 000 times smaller than the atom.
● Most of the atom is empty space. If the nucleus were the size of a small marble,
the atom would be about the size of a football pitch.
Page | 17
Question
Re-work the calculation above using the formula for electrical potential energy given in the
text box above together with the value of k supplied. The values of q1 and q2 can be
deduced from the diagram.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
Changing probe, smaller scale
Here we extend the scattering idea to smaller scales, an example of fruitful recycling of a
single idea. You should be aware of the decrease in scale, together with the seemingly
paradoxical fact that probing smaller and smaller things requires larger and larger
accelerators and higher energies. Geiger and Marsden's apparatus fitted on a table top; a
detector at CERN is a big as a small house.
A brief look at the next activity shows the rapid increase in particle energy that has been
available for scattering experiments over the last century. Note that the energy goes up by
a factor of 10 every decade. The Y-axis is logarithmic of course.
1019
Planck scale
1012
GUT scale
108
highest energy
cosmic rays
104
LHC
Tevatron
CERN
post-war
accelerators
1
Lawrence
Rutherford
Thomson
0.001
1900
1950
2000
2050
2100
2150
year
You can plot your own graph using Activity 110S Software Based 'The Livingston curve'.
The data is available on File 70T Spreadsheet Data Table 'Energies of accelerators'
The costs of funding the accelerator programmes can be explored using Activity 120P
Presentation 'The funding of particle physics' if you wish.
Page | 18
You should be aware of the basic facts about the nucleus given below (pp 200-201).
You should know that the proton
and neutron have approximately
the same mass. Neutrons have no
charge whereas protons have a
charge +1e.
The number of negative electrons
equals the number of protons in
the nucleus in a neutral atom. The
number of electrons in the neutral
atom determines its chemical
properties.
Electron scattering can be used for scattering experiments just like alpha particles.
Electrons are charged (negative in this case) but they are not affected by the strong
nuclear force in the way that alpha particles are (pp 201-202). That means that electrons
can be used to glean close-up detail of the nucleus without them being affected by nonelectrical forces as would be the case with alpha particles. This is Rutherford scattering on
a much more energetic scale; this type of scattering enables the charge distribution inside
the nucleus to be mapped.
In addition, electrons, like photons, can be thought of as having a wavelength given by the
de Broglie relation:
λ=h
p
where p is the momentum and h is the Planck constant.
Page | 19
Electrons are diffracted by the small spherical nucleus in the same way that light is
diffracted when it passes through a small hole. The diffraction pattern consists of a series
of concentric rings. The first minimum of the diffraction pattern occurs when sin  = 1.22λ/d
where d is the diameter of the circular nucleus. In chapter 6 we used the equation in the
form sin = λ/d because the waves were passing through a slit and not a circular hole. The
diffraction pattern is superimposed on the Rutherford scattering curve as both effects are
present. The diameter of the nucleus can be found from measurements of the diffraction
part of the curve. Note that diffraction effects are present with conventional alpha particle
scattering but they are hard to see. You should be able to suggest why that is the case.
…………………………………………………………………………………………………………
The experimental set-up and results for electron scattering are shown below (see p 186).
Meaningful results are only obtained from nuclear scattering when the wavelength is very
small. That means that the electrons must be accelerated to several hundred MeV which is
much larger than the rest energy, mc2, of 0.5 MeV. From the equation E2 = p2c2 + m2c4 you
should see that a good approximation for the momentum is:
p≈E
c
That means that λ = h ≈ hc
p E
Page | 20
Questions
1
Show that 100 MeV electrons have a de Broglie wavelength comparable with
nuclear dimensions. Use the data sheet for relevant information not supplied.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
2
Identify the angle at which the first minimum for electron diffraction occurs in the
graph shown above and use this to calculate the size of the nucleus. Note that the
experiment was carried out with 100 MeV electrons.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
Page | 21
Measurements of nuclear radius enable nuclear volume to be calculated using the formula
V = 4 r3 where r is the nuclear radius.
3
A graph of volume against nucleon (mass) number is quite a good straight line indicating
that the nuclear density is approximately the same for all nuclei (pp 202-203).
Display Material 140O
OHT 'Density of nuclear matter'
Density of nuclear m atter
Volum e of nucleus increases linearly with number of nucleons
Electron scattering measures
radius r of nucleus
calculate v olume =
4
3
r 3
1500
1 97
1000
122
59
500
28
16
0
Sr
Si
C
He
4
1
88
Sb
O
12
0
Co
Au
H
50
100
150
number of nucleons
200
Estimate fro m g rap h:
100 nucleons in
volume 700  10 –45 m 3
Data:
mass per nucleon u
= 1.7  10 –2 7 kg
volum e per nucleon
= 7  10 –45 m 3
Calcu late d ensit y:
density =
density =
mass
volum e
1.7  10– 27 kg
–
7  10 45 m 3
= 2.4  1017 kg m – 3
Density of ‘nuclear matter ’ is roughly 2  101 7 kg m –3
A matchbox full of nuclear matter would have a mass of five billion
tonnes
Matter of this density is only found on Earth in the nucleus of every atom. However, larger
bodies called neutron stars (Chapter 11) are thought to have the same density; neutron
stars are estimated to be a few kilometres in diameter. Neutron stars spin rapidly and emit
pulses of radio waves as they spin. These were first detected via the pulsating radio waves
received on Earth (Jocelyn Bell Burnell and Anthony Hewish) and are thought to have
been formed when a massive star (see supernovae in Chapter 12) ran out of nuclear fuel,
collapsed, and exploded leaving behind a core made of neutrons.
Page | 22
Question
Calculate the mass of a neutron star having a radius of 10 km.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
Since the volume of a nucleus is
proportional to nucleon number
A we can show that the nuclear
radius r = r0A1/3. The value of r0
is effectively the radius of a single
nucleon which is approximately
10-15 m.
Question
Show that radius of a nucleus is proportional to the
cube root of nucleon number for nuclear matter of a
fixed density.
…………………………………………………………...
…………………………………………………………...
…………………………………………………………...
…………………………………………………………...
The relationship between nuclear volume and nucleon number can be explored using
Activity 130S Software Based 'The density of nuclear matter'. This activity uses File 80T
Spreadsheet Data Table 'Density of nuclear matter data'
The final part of this section explores the forces that hold the nucleons in the nucleus
together. These forces provide the ‘nuclear glue’.
Question
Why must the forces holding the nucleons together be very strong?
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
The alpha particle scattering experiments revealed the existence of the nucleus and the
electron scattering experiments enabled the nuclear size to be established more
accurately. The nucleus was undisturbed by the scattering so this could be described as
‘elastic’ scattering. No energy is taken from the incoming particle.
If much higher scattering energies are used the scattering may be inelastic. Energy is
taken from the incoming particle to create new particles or the target nucleus is excited to
a higher state if energy. Given that the rest energy of a nucleon is approximately 1 GeV,
scattering energies in excess of several GeV can create new particles including nucleons
(in the form of proton-antiproton pairs for example).
Page | 23
Question
Show that the rest energy of a nucleon is about 1 GeV. Use the data sheet to find the
values needed.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
As accelerator energies rose many new particles were created. These together with the
more familiar neutrons and protons all interact through the strong nuclear force are
collectively referred to as hadrons. There were patterns found in the properties of these
particles and it was thought that all of these new particles must be combinations of more
fundamental components. Gell-Mann suggested that these new particles were
combinations of a yet-to-be-discovered new particle that he named the quark.
Gell-Mann’s model would only work if the proposed quarks had charges of either 1/3 or 2/3
the charge of the electron. The concept of a fractional unit of charge was at odds with all
conventional wisdom.
We now think that there are six different quarks (described as different ‘flavours’), each
with an associated anti-quark. Three of them carry a charge of + 2/3 e. The other three
carry a charge of – 1/3 e:
1.
2.
3.
The up quark (+ 2/3 e) and the down quark (– 1/3 e) make up protons and neutrons.
The charmed quark (+ 2/3 e) and the strange quark (– 1/3 e) are more massive than
the up and down quarks.
The top quark (+ 2/3 e) and the bottom quark (– 1/3 e) are the most massive known
quarks.
You are not required to know about ‘flavours’ other than ‘up’ and ‘down’.
The quark model explains all known baryons, antibaryons and mesons.
1.
A baryon consists of three quarks. An antibaryon consists of three antiquarks. For
example, a proton consists of two up quarks and a down quark (i.e. uud) and a
neutron consists of one up quark and two down quarks (i.e. udd).
2.
A meson consists of a quark and an antiquark. For example, a pion or  meson
consists of an up or a down quark and an up or down antiquark.
Family of particles
hadrons
Type of hadron
baryons
mesons
Number of quarks
Three
Two
Page | 24
See page 204 of the student’s book.
The search for quarks was carried out with very high energy electron scattering – up to 20
GeV. Electrons were used because their only interaction with the charged quarks would be
via the electromagnetic force. The intention was to map the charged particles inside the
nucleons (protons or neutrons).
With deep inelastic scattering such as this, particle creation occurs; quark-antiquark pairs
are produced from the energy of the interaction. As the quark is given a huge kick by the
incoming electron a ‘jet’ of new particles emerges in the direction you would expect to see
the emerging quark. This ‘jet’ is mainly mesons which are quark-antiquark pairs.
Page | 25
At this level of energy the effects of relativity tends to simplify the situation. To the
incoming electron, the target proton appears to be approaching it at a speed close to the
speed of light. Time dilation slows down the motion of the quarks within the proton so that
they seem to be almost at rest – like ‘sitting’ targets. Also, length contraction shrinks the
nucleon in the direction of approach making it appear like a flat disc with three target
pancakes inside.
Display Material 150O
OHT 'Deep inelastic scattering'
Deep inelastic scattering
M ed ium en ergy: elastic scatter ing
elec tron
proton
Q uark s mov e rapidly inside proton.
T he interaction tim e is long enough
for the proton to behav e like a blur of
c harge.
elec tron
sc attered
at large angle
H igh ener gy: deep inelastic scatterin g
u
electron
d
An electron can hit one quark and be sc attered.
Ex change of high-energy photons leads to the creation
of a jet of partic les and antipartic les .
u
jet of partic les, mainly m es ons
At high energies individual quarks scatter electrons
Quarks as relativistic stationary pancakes
proton as seen by observer not moving relative to it:
rapidly moving spherical quarks fill a sphere
coming towards
the electron
proton as seen by electron moving rapidly towards it:
almost stationary pancake quarks filling a flat disk
Page | 26
What holds the quarks together inside a particle? We know that charged particles exert
forces on each other by exchanging photons. The same principle applies with quarks
except that quarks attract each other by exchanging gluons. These provide the ‘glue’ that
keeps the hadrons intact. They also lead to the forces that make nucleons attract each
other thus holding the nucleus together. This gluon attraction is known as the strong
interaction.
Electric charge comes in two forms: positive and negative. The charge carried by quarks is
known as ‘colour charge’ which comes in three forms: red (R), green (G) and blue (B).
You should be aware that, when mixing colours of light, which has nothing to do with quark
colours, white is made by adding red, green and blue. That means R + G + B = W.
Free particles are colourless (white, W, in this context) combinations of charge.
Baryons have three quarks, one red, one green and one blue. This makes the combination
white, so baryons are ‘colourless’.
Mesons have only two quarks; they cannot be made ‘colourless’ by adding just two colour
charges. However, a ‘colourless’ meson can be made by adding a colour quark with its
corresponding anticolour quark e.g. red and antired.
Gluons differ from photons in that photons are not electrically charged but gluons carry
colour charge. Gluons emit and absorb gluons. This leads to a situation where the quark
colour interaction gets stronger with distance. There is an analogy here with the stretching
of a rubber band. This is why quarks cannot be knocked out of particles to exist on their
own. Pulling two quarks apart simply stores more energy in the gluon field until a quarkantiquark pair is released. A quark-antiquark pair is a meson. That is exactly what is
emitted with deep inelastic scattering.
Quarks and electrons are ‘matter-like’ particles and
are fermions.
Gluons and photons are ‘force-like’ particles and
are bosons.
Activity 60P Presentation ‘Quark models’ is optional but gives some flavour of the way that
the rules apply to the combining of quarks.
Page | 27
Display Material 160O
OHT 'Quarks and gluons'
Quark–gluon interaction
red quark
blue quark
red–blue gluon
red quark
blue quark
Quarks interact by exchanging gluons, which change the quark colours. Here a red quark and a blue quark
exchange a red–blue gluon. The red quark becomes blue and the blue quark becomes red. The quarks exchange
energy and momentum.
Quarks pulled apart make more quarks
quark
quark
two quarks held together by
the gluon field...
gluon field
...pull the quarks apart. The
gluon field increases in
energy...
quark
quark
quark
quark
antiquark
quark
...a quark–antiquark pair
materialises from the gluon
field
Page | 28
Section 17.3 The music of the atoms
Learning outcomes
●
Electrons confined in a region of space can be modelled as standing waves, with
wavelengths determined by the size and shape of the confining region.
●
The de Broglie wavelength is given by
h

p
●
Discrete atomic energy levels correspond to discrete electron standing waves in an
atom.
●
Electrons can make quantum jumps between allowed energy levels, emitting or
absorbing a photon whose energy is given by E = hf = Einitial – Efinal.
●
The energy level spacings in hydrogen are given by
En = -13.6eV
n2
where n is the principal quantum number.
Standing waves in atoms
This section builds on the understanding of electrons acquired in Chapter 7 and shows
how a wave-like description can account for the energy levels in atoms.
We start with some observations of the effects of the de Broglie waves associated with
electrons. These images were obtained with a scanning electron microscope by Crommie,
Lutz and Eigler of IBM. Here you see an image of individual iron atoms being arranged in a
circle on the surface of copper. An electron standing wave appears inside the circular trap.
To see these images and many more, visit
http://www.almaden.ibm.com/vis/stm/corral.html
The construction of this ring is shown on page 209.
The standing wave seen gives an indication of the electron density as determined by a
scanning tunnelling electron microscope. The electron density indicates the probability of
finding an electron. In practice, a high electron density produces a large tunnelling current
and vice versa. The current readings are used to construct the image shown.
Page | 29
Activity 140E Experiment 'Standing waves – for electrons?'
This activity is used to revise the wave nature of the electron and to show how electron
standing waves can be formed when electrons are trapped in an atom.
The following diagram shows standing waves in more everyday situations. In all cases, the
only standing waves that can exist are those that ‘fit’ into the ‘box’.
Display Material 170O
OHT 'Standing waves in boxes'
Standing waves in boxes
Waves on a string
fixed end
fixed end
Waves on a circular diaphragm
rigid edge
Only certain field patterns are possible because the waves must
fit inside the box
Page | 30
The guitar string standing wave shown above is in one dimension. A one-dimensional
electron standing wave on the carotene molecule as shown in the next diagram.
Display Material 180O
OHT 'Colours from electron guitar strings'
Molecular guitar strings
Carotene molecule C40 H56
electrons spread
along the molecule
Analogy with guitar string
electrons make standing waves
along the molecule
length L
electron wavelength proportional to L
Long m olecules absorb visible wavelengths of light
The ‘string’ is the carotene molecule and delocalised electrons ‘spread’ along this string.
The molecule is quite long which leads to standing waves with a wavelength that is
relatively large. The energy of the standing waves is such that the carotene molecule
absorbs blue light. This makes a group of such molecules appear yellow or orange.
The concept of electron standing waves can be used to explain why atoms have discrete
electron energy levels (see pp 210-215).
The diagrams below show how spectral evidence leads to our understanding of electron
energy levels.
Page | 31
Display Material 195O
OHT 'Atomic line emission spectra'
These atomic line emission spectra are taken from the Advancing Physics A2 student’s
book, chapter 17. This shows the line emission spectra of a number of elements. Note that
white light from a tungsten filament lamp produces a continuous spectrum.
Display Material 190O
OHT 'Energy levels'
Spectral lines and energy levels
low energy
long wavelength
energy levels of an atom
energy
n=4
4–3
n=3
3–2
4–2
n= 2
2–1
3–1
4–1
n=1
high energy
short wavelength
E = hf
photon emitted as
electron falls from
one level to a
lower level
Spectral lines map energy levels. E = hf is the energy difference between two levels.
Page 212 describes the spectrum aptly as a scrambled ‘map’ of the electron energy levels
inside the atom.
Page | 32
We can make a first attempt at understanding how these discrete electron energy levels
come about by imagining an electron trapped in a simple box with vertical walls. The
diagram below shows that, because an electron standing wave will be formed, only certain
values of wavelength would be allowed.
Waves and energy levels
Standing waves allow only
discrete values of wavelength 
If motion is non-relativistic
momentum:
L=

2
p = mv
L=
kinetic
energy:
E K = 12 mv2
...etc
For each wavelength there is a
discrete value of momentum p
Kinetic energy and
momentum are related by
p2
EK =
2m
h
p=

The wavelength determines the
momentum of the electron since
p = h/ λ.
Once the momentum is known,
the kinetic energy (which
represents the total energy in
this case) can be found. The
diagram shows how the kinetic
energy can be found directly
using
Ek = h2/(2m λ2).
For each wavelength there is a discrete
value of kinetic energy
p2 =
h2
2
so
EK =
h2
2m 2
small , high energy
large , low energy
Discrete wavelengths imply discrete levels of energy
Question
Show that Ek = n2 (h2/8mL2) where n is the number of loops in the standing wave and L is
the size of the box. You are advised to calculate Ek for 1 loop, 2 loops and 3 loops in turn.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
Page | 33
The energy levels predicted by this model have discrete values with no ‘in between’ states
being allowed. These would represent the electron energy levels within this simple atom.
Unfortunately, this simple model predicts a result that does not agree with experimental
measurement! It predicts that E α n2 where n is the number of loops in the standing wave.
Another weakness is that the model is only one-dimensional.
The next diagram shows how this comes about.
Display Material 200O
OHT 'Standing waves in atoms'
A guitar–string atom
Sim plify:
energy
C hange the 1/r
potential well
of the nucleus
into a pair of
fixed high walls
energy
d
trapped
electron
nucleus +
Energy
trapped
electron
1/ r
potential
well
 4 = 1 /4
n=4
levels increase
energy  n2
3 = 1 /3
n=3
2 =  1 /2
n=2
1 = 2d
n=1
E4 =
h2
= 4 2E 1
2m 2 4
E3 =
h2
= 32 E 1
2m 2 3
E2 =
h
= 22 E 1
2m  2 2
2
E1 =
d
h2
2m2 1
no levels at all
below n = 1
In general:
E n = n 2E 1
For the nth level
2
En = n
E n = n2
h2
2m 2
1
h2
2m(2d) 2
Each level has a quantum number n. The energy depends on the quantum
number.
In reality, the electron does not move inside a potential well with vertical walls. The
potential actually varies as 1/r. The next diagram shows how discrete energy levels can be
obtained with such a potential well. In this model it is necessary to consider the total
energy of the trapped electron as the sum of the potential and kinetic energies. The kinetic
energy is positive and the potential energy is negative. The total energy must be negative
for the electron to be ‘trapped’ within the potential well of the nucleus.
Page | 34
Display Material 210O
OHT 'Size of the hydrogen atom'
How small c ould a hydrogen atom be?
im aginary
box
Replace 1/r potential by a box of
width d = 2r
Calculate kinetic energy for waves
 = 2d = 4r
Calculate potential energy at r
d = 2r
 = 4r
1/r potential +
nucleus
potential energy E p = –
e2
4 0r
standing wave /2 = d
momentum p = h/
2
kinetic energy = p /2m
kinetic energy E k =
h2
2m 2
Find the m in im um rad iu s of an atom , fo r total energy < 0
s hort wavelength
kinetic
unstable
energy
120
kinetic energy =
100
+
small radius
Ek + Ep > 0
potential
energy
medium wavelength
just
stable
kinetic
ener gy
h2
2m 
minimum
radius of
bound atom
2
Note that the single ‘loop’ of the
standing wave must fit inside the
‘box’ that is defined by the walls of
the potential well. The wavelength
varies with the position of the
standing wave within the potential
well and this in turn defines the
kinetic energy of the electron.
The rules are:
Ek + Ep = 0 electron is just free
Ek + Ep > 0 electron is not bound
Ek + Ep < 0 electron is bound
80
total energy > 0
unstable
60
40
total energy = 0
just stable
20
+
potential
energy
medium radius
Ek + Ep = 0
0
0.02
long wavelength
kinetic
energy
stab le
–20
0.04
0.06
radius r/nm
total energy < 0
bound
–40
potential energy = –
potential
energy
large radius
Ek + Ep < 0
–60
0.08
e2
40 r
+
If size is too small, th e kinetic en ergy is too larg e for electrical po ten tial en ergy to
bin d th e electron
Question
Use the formulae given above to calculate the electrical potential energy, kinetic energy
and total energy for hydrogen atoms having the ‘radii’ given. Remember that λ = 4r.
Atom ‘radius’ / m
Electrical potential
energy / J
Kinetic energy / J
Total energy / J
1 × 10-11
1 × 10-10
1 × 10-9
Page | 35
You should now check your calculations and explore other atomic sizes using Activity
150S Software Based 'Sizing up a hydrogen atom' which uses File 90L Launchable File 'A
crude hydrogen atom calculator' [Warning: the graph rescales every time you change the
data so you should not be fooled by the actual heights shown in each graph.]
An atom cannot be smaller than a certain size. This is because a small atom will have
short de Broglie wavelengths for its electrons, which will as a result have too much
momentum and kinetic energy to be held in the atom by the electrical attraction of the
nucleus.
Further analysis, originally carried out by Schroedinger in 1926 (see p 214) shows that this
model predicts that E α 1/n2 which matches experimental results exactly for the case of a
simple hydrogen atom. The number n now takes on the meaning of principle quantum
number.
Some of the standing waves in hydrogen are shown below (see p 215). Chemists will
recognise these as s-orbitals.
An excellent programme for viewing models of orbitals in different ways can be found on
http://winter.group.shef.ac.uk/orbitron/. Click on the orbital labels on the left-hand side to
explore the orbital shapes.
Page | 36
The next diagram shows how a single set of energy levels in the hydrogen atom can
explain how many different series of spectral lines occur with hydrogen (see page 214).
For example, the familiar Balmer series (mainly in the visible part of the spectrum) can be
understood using the formula:
1 - 1
2 2 n2
where n has integer values greater than 2.
The corresponding formula for the Lyman series would be:
1 - 1
1 2 n2
Further details are given on pages 214-215.
Other elements in the Periodic Table have more electrons than hydrogen. When adding
electrons to build up atoms above hydrogen we must remember that electrons are
fermions which means that no two electrons can exist in the same quantum state. This is
known as the Pauli exclusion principle.
To make helium, the next electron can go in the lowest energy level (n = 1) because it has
opposite ‘spin’. However, it is now impossible to add more electrons to that level. The third
electron needed to make lithium must go into the second energy level, n = 2. There are
three dumbbell-shaped (p) orbitals as well as the spherical s-orbital at this level and each
of these can accommodate two electrons with opposite ‘spin’. That means that eight
electrons can be added to the n = 2 level in total before this level becomes filled. You
should see that this explains the ‘shells’ of electrons that were described in GCSE
Chemistry.
Remember that each orbital is actually an electron standing wave. The properties of the
elements are governed by the nature of the possible standing waves and the fact that
electrons are fermions.
Page | 37
Protons and neutrons are also fermions so energy levels are also to be found when
building up the nucleus. The proton and neutron are two different states of nucleon, each
of which has two possible ‘spin’ directions. That makes it possible to put four nucleons into
the lowest energy state. In fact, this cluster of four particles takes on much significance
when building up the nucleus, and we must not lose sight of the fact that this particular
combination makes up the very stable alpha particle.
The nucleus has nuclear energy levels in the same way that atoms have electron energy
levels. However, the energies involved are much larger so the photon emitted when a
nucleus changes from a higher to a lower energy level is a gamma ray and not visible light
for example.
Page | 38
Section 17.4 Known and unknown
Learning outcomes
●
Matter is ultimately made from just two different classes of particles: quarks and
leptons, both fermions.
●
These interact by exchanging bosons (e.g. photons in electromagnetic interactions).
●
Particles are created and annihilated in particle–antiparticle pairs.
This final section is the briefest possible overview of the state of particle physics and some
questions which remain tantalisingly unanswered.
Display Material 220O
OHT 'What the world is made of'
These displays show all the particles which go to make up the whole world, so far as we
know. The masses of the quarks are not well-defined, because they are never seen as
free particles. The values here are rough estimates of the so-called 'current mass'.
The particles of the everyday world
Everything you touch around you is made of just these particles:
The world
around you
Leptons
charge
e– electron
–1
e neutrino
0
rest energy / MeV
Quarks
charge
0.511
u up
+
0?
d down
2
3
1
–3
rest energy / MeV
6?
10?
A complete picture of your world should include their antiparticles too:
The world
around you
charge
rest energy / MeV
e– electron
–1
0.511
u up
e neutrino
0
0?
d down
antiparticles e+ positron
+1
0.511
u anti-up
0?
d anti-down
particles
Leptons
e antineutrino
0
Quarks
charge
+2
3
–1
3
–2
3
+1
3
rest energy / MeV
6?
10?
6?
10?
To account for all known matter, the pattern of a pair of leptons and a pair of quarks
repeats three times:
Generation
Leptons
charge
1 The world e– electron
around you
e neutrino
–1
– muon
 muonneutrino
–1
 – tau
–1
2
3
 tau-neutrino
0
0
0
rest energy / MeV
Quarks
0.511
u up
0?
d down
106
0?
1780
0?
s strange
c charmed
b bottom
t top
charge
+2
3
–1
3
–1
3
+2
3
–1
3
+2
3
rest energy / MeV
6?
10?
200?
1500?
5000?
90 000?
Page | 39
The other particles that make up the world are the bosons, the carriers of interactions:
interaction
force carrier
electric charge
rest energy / GeV
explains
electromagnetism
photon
0
0
Everyday interactions
including all chemistry
weak interaction
Z0
W+
W–
0
+1
–1
93
81
81
Radioactive
decays; changing
particle nature
0
0
What holds nucleons
and mesons together
0
0
Conjectured, but not
detected
8 different ‘colour
strong interaction combinations’ of
gluons
gravity
‘graviton’
The hunt is on for another, the Higgs boson, which is thought to be responsible for
particles having mass.
Finally, take a look at the unfinished business; at the way solving one problem throws up
another, often deeper problem:
●
●
●
●
●
●
What decides the masses of the various particles? At present, masses of particles
have to be found experimentally. No theory predicts them from more basic
principles.
Where does mass come from, anyway? There is a theory (the Higgs field) of how
particles acquire mass. By the time you read these words, the boson which carries
the Higgs interaction (Higgs boson) may – or may not – have been observed.
Why do fundamental particles come in pairs of two leptons and two quarks? Is there
any relationship between the leptons and the quarks? The energies required to test
ideas about this could be so large that the theories might be effectively untestable.
Why are there three and only three generations of fermions? Nobody knows.
Can the strong interaction be unified successfully with the weak and
electromagnetic interactions? Can there be a unified field theory?
Can gravity be related to the other interactions? Can its exchange particle, the
graviton, be detected?
These remain unsolved mysteries at present.
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Questions and activities
Section
Essential
17.1
17.2
17.3
Optional
Read A2 text pp 189-196
Qu 1-6 A2 text p 197
Question 50C
annihilation'
Comprehension 'Creation from
Question 10S Short Answer 'Things that
don’t change'
Question 20S Short Answer 'Beta decay
and conservation’
Question 30S Short Answer 'Creation
and annihilation'
Activity 20P
when?'
Presentation 'Who, what and
Question 40M Multiple choice 'Particles
and interactions'
Question 60M Multiple choice 'Keeping
momentum and energy unchanged'
Reading 20T Text to Read 'Three poems about
particles'
Reading 30T Text to Read 'Early ideas about
elementary particles'
Reading 60T Text to Read 'Symmetry and
conservation laws'
Reading 10T
Text to Read 'The
discovery of beta decay'
Read A2 text pp 198-207
Qu 1-6 A2 text p 208
Question 70S Short Answer 'Rutherford
scattering: energy and closest approach'
Question 80S Short Answer 'Rutherford
scattering: directions of forces'
Question 90S Short Answer 'Electrons
'measure the size of nuclei'
Question 100S Short Answer 'Scattering
and scale'
Question 110S Short Answer 'Putting
quarks together' [If time]
Read A2 text pp 209-215
Qu 1-6 A2 text p 216
17.4
Question 150S Short Answer 'Spectra
and energy levels'
Question 160S Short Answer 'How small
could a hydrogen atom be?'
Question 170S Short Answer 'Carrots
and guitar strings'
Question 180S Short Answer 'The
hydrogen spectrum'
Read A2 text pp 217-218
Summary
Qu 1-9 A2 text p 220
Activity 40S
Software Based 'Time and the
Muon' using File 30S
Spreadsheet Models
'Models comparing trip measurements'
Question 120C Comprehension 'Finding parts of
protons'
Activity 100S Software Based 'Where scattered
probes go'
using File 60L Launchable File 'Destinations of
scattered probes'
Reading 70T
atom'
Reading 80T
Text to Read 'Towards a nuclear
Text to Read 'Tracking particles'
Question 140C Comprehension 'How Niels Bohr
began quantum theory'
Question 180D Data Handling 'Products of the
Big Bang'
Reading 40T Text to Read 'Quantum theory in
the twentieth century'
Reading 50T Text to Read 'Where did all the
antimatter in the universe go?'
These notes draw almost exclusively on the resources to be found in Advancing Physics A2 Student’s Book and CD-ROM published by
Institute of Physics Publishing in 2000 and 2008. They are intended to be used in conjunction with these resources and others not
specified.
John Mascall
The King’s School, Ely, Cambs
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