1
PH2 – Waves and Particles
General
PH2 takes the student into new and, we hope, exciting territory. The main ‘blocks’ in PH2 are
Waves and Photons. Having these re-united in the same module means that the nature of light
becomes an important theme of PH2.
The old PH2 section on energy levels and line spectra is now seen to have interesting
applications: it has been extended to include a treatment of lasers, while line absorption
spectra are shown to give us information about the atmosphere of stars. In fact, there is a
small section devoted to the physics of stars. It mainly concerns what we can learn about
them from the radiation they emit.
In an extension to the old PH2 section on nuclear structure, the basic constituents of matter
are acknowledged to be quarks and leptons. The four fundamental forces are also introduced
in this section. One teaching strategy would be to generate interest by starting PH2 here, as
the ideas are quite modern, new to most students, and raise tantalising issues.
PH2.1 – WAVES
Most of the content of the Waves and Light sections of the old PH1 is included here. To make
way for new material elsewhere, some gentle pruning has been done. Beats have disappeared.
So has the relationship between intensity and amplitude. Students are no longer required to
describe experiments showing diffraction or interference of water waves, sound and
microwaves, though they are required to be familiar with such experiments; an examination
question could probe understanding of a set-up given in the question. Students do still have to
be able to describe a version – any they please – of a two-slits experiment for light.
Waves are well covered in standard textbooks.
2
PH2.2 – REFRACTION OF LIGHT
What is required has been spelt out in the specification in rather more detail than previously.
The Snell’s Law equation is written in such a way as to emphasise symmetry.
The only application of total internal reflection now required to be learnt is the step-index
multimode (‘thick core’) optical fibre. Step-index means simply that the core is glass of one
refractive index and the cladding is glass of a lower index (students need to know why it has
to be lower), with an abrupt change in index at the interface.
While such fibres are fine for conveying light for illumination, students need to know that
they can’t be used for transmitting a rapid sequence of data over a long distance. Multimode
dispersion is the problem. Light travelling at an angle to the axis of the fibre (ray A in the
diagram) will travel further for a given axial length of fibre, than light (such as ray B)
travelling parallel to, or at a smaller angle to, the axis, and so will arrive later. Thus the
arrival time of an element of data encoded in the light is smeared out. The element could start
to arrive (by the shortest route) earlier than the previous element has finsished arriving by its
longest route. Even worse confusion can occur.
There are two ways round the problem of multimode dispersion…
• Make fibres with graded index cores. This means cores which have a progressively lower index as
we go out from the axis towards the interface with the cladding. The lower the index the faster the
light travels so, if the grading is correctly calculated, the longer, more zigzaggy, paths cash in on
the ‘faster medium’ and take no longer than the short, axial, route. Clever stuff, but note that
graded index fibres are not in the WJEC specification.
• Make the core very thin. Its diameter must be no more than a few wavelengths of the light
(or infrared) being carried. Such fibres are monomode. Light travels parallel to the axis.
There are no zigzag modes. Students are required to know this. They are not required to
know why very thin fibres are monomode. This is just as well, because it cannot be shown
by ray optics, nor even by simple application of Huygens Principle. Electromagnetic wave
theory is needed.
The website http://www.techoptics.com/pages/Fiber%20Optics%20-%20Optical%20Fiber.html
gives an excellent summary of fibres for data transmission, with some facts and figures.
PH2.3 – PHOTONS
This is a heavily modified, and, we hope, more satisfying, version of the ‘Quantum Physics’
section of the previous specification. Particles behaving as waves, and the electron diffraction
demonstration have disappeared. The electron-volt will not be used in PH2. The origin of the
continuous and line X-ray spectra will not be tested.
3
The photo-electric effect is still arguably the most direct and easily understood evidence for
photons. Although the wording of our requirements [new PH2.3(a)-(d)] has been re-thought,
the same ground is covered, except that Millikan’s role need not be mentioned. This should
discourage students from spending valuable time in the examination drawing elaborate
diagrams of Millikan’s ‘laboratory in a vacuum’.
Students find the photoelectric effect difficult, especially the experimental determination of
KEmax. We have therefore written a detailed account – below – of the material to be tested, so
that the level of approach required is as clear as we can make it.
The section on atomic energy levels and line spectra now culminates in a treatment of lasers.
Detailed notes follow the photoelectric notes.
PH2.3 (a)-(d) Detailed notes on Photons and the Photo-electric Effect
Photons
Everything in nature seems to come in lumps or quanta (singular: quantum). For example,
ordinary matter is made of atoms, and electric charge comes in units of e. This lumpiness was
only becoming fully accepted a hundred years ago. But in 1905 Einstein made the bold
suggestion that light, too, was ‘lumpy’. Light quanta are now called photons.
A photon is a discrete packet of electromagnetic radiation energy. The energy of a photon is
given by
Ephoton = hf
In which f is the frequency of the light and h is a constant called Planck’s constant.
[h = 6.6 = 10-34 Js] [This is given in the WJEC list of constants.]
[For interest only… The constant h had first arisen in the earlier (1900) work of Max Planck
on the radiation inside a cavity with hot walls. Planck had shown that the energies of
oscillating particles in the wall seemed to be quantised.]
Einstein suggested some experiments in which the quantisation of light should reveal itself.
The simplest to understand involved the photoelectric effect, a phenomenon known about
since the late 1880s.
The Photo-electric Effect
When electromagnetic radiation of high enough frequency falls on a metal surface, electrons are
emitted from the surface.
For most metals, ultraviolet is needed. For some (including sodium, potassium, caesium), light
towards the violet end of the spectrum) will release electrons.
4
Demonstrating the Photo-electric Effect
EITHER Using a zinc plate with a gold leaf electroscope (or a coulombmeter)…
•
•
Clean a zinc plate with fine emery paper or steel wool.
Attach the plate to the top disc on a gold leaf electroscope, so there is good electrical
contact.
• Charge the zinc plate and inner assembly of the electroscope negatively1, e.g. by rubbing the
zinc plate with a polythene rod which has been rubbed with wool or fur. [Charging by
induction using a perspex rod is more reliable, but might be considered too confusing!]
The leaf should now be raised, because the leaf and the back plate are both charged negatively
and repel each other. The leaf should temporarily rise further if the charged polythene rod is
brought near the zinc plate.
•
Place an ultraviolet lamp near the zinc plate. Switch it on. The leaf should be seen to fall.
[Safety note: Don’t look at the ultraviolet lamp (when it’s turned on!)] Clearly the plate (and
inner assembly of electroscope) is losing charge.
•
Repeat the procedure, but charging the zinc plate and inner assembly of the electroscope
positively, e.g. by rubbing the plate with a charged perspex rod.
This time the ultraviolet does not affect the leaf. Charge is not lost.
The simplest explanation is the correct one… The ultraviolet causes electrons to be emitted from
the zinc plate. If the plate is charged positively, the electrons are attracted back again. If the plate
is charged negatively the emitted electrons are repelled and lost from the plate for ever.
1
Note that it is also possible to charge up the electroscope using an EHT power supply. It is
imperative that the output with the current-limiting resistance [usually several M] is
employed if doing this.
5
OR Using a vacuum photocell…
•
The apparatus can be supplied by Philip
Harris (Unilab division).
•
Note the polarity of the power supply.
Any electrons emitted from the caesium
surface will be collected by the
‘collecting electrode’.
•
If the photocell is covered the current is
zero; if light falls on the caesium
electrode there is current.
Photo-electric Puzzles
Before 1905, the energy of a beam of light was thought of as distributed uniformly across
broad wavefronts. Calculations showed that it should take some time before an electron in a
metal surface could absorb enough energy from the light to escape from the surface. Yet
emission is observed as soon as the light falls on the surface.
Another puzzle was why, for a given surface, we find that light of frequency below a certain
value (the threshold frequency) causes no electron emission at all.
Einstein’s theory of the photo-electric effect solves both these problems…
Einstein’s Photo-electric Equation
Although the free electrons in a metal have no allegiance to particular atoms, there are forces
‘bonding’ them to the lattice of ions as a whole. In order to escape from the metal an electron
has to do work against these forces. Some have to do more work than others, but there is a
certain minimum quantity of work to be done, so no electron can escape unless it is given a
certain minimum energy.
The work function, , of a metal is the minimum energy needed by an electron in order to
escape from the surface.
Einstein’s key idea was that any electron which leaves the surface is ejected by the action of a
single photon. Photons don’t co-operate in the process.
Recall that a photon of light of frequency f has energy hf.
Suppose that a photon gives its energy hf to an electron, and that the electron is able to
escape. The minimum energy used in escaping is , so the maximum kinetic energy the
escaped electron can have is what’s left over of the photon’s energy. So we have the simple
equation…
KEmax = hf – 
This assumes that the photon energy is greater than (or equal to) the work function; in other
words that hf   , so f   / h.
If f <  / h, the photon energy will be less than the work function so no electrons at all can
escape – a simple explanation of the phenomenon of threshold frequency.
6
The threshold frequency, f0, for a metal is the minimum frequency of electromagnetic
radiation needed to produce electron emission from the surface.
From the argument just given,
f0 =  / h.
This relationship can also be deduced from Einstein’s equation (without using ‘’ or ‘<’
signs!)… At the threshold frequency even the most energetic electron will only just manage
to escape, so KEmax = 0, and therefore hf 0    0  f0 =  / h.
Provided that the light is above the threshold frequency, as soon as it falls on the metal
surface electrons will start to be emitted, as emission results from individual photon ‘hits’,
and is not a cumulative process as supposed before Einstein.
Experimental test of Einstein’s Equation
We use the arrangement with the vacuum photocell given earlier for demonstrating the
photoelectric effect, but with the power supply polarity reversed.
•
Use white light with a coloured filter, or a light emitting diode, to illuminate the caesium
surface with approximately monochromatic light. [Its wavelength can be found using a
diffraction grating, hence its frequency, using f = c / .]
•
Increase the p.d. between the
collecting electrode and the caesium
surface until the current drops to zero.
At this point the p.d. is called the
stopping voltage, Vstop, because it stops
all emitted electrons, even those with
the most K.E., from reaching the
collector electrode.
•
The maximum K.E. of the emitted
electrons is simply given by
KEmax = e Vstop
[How do we justify this? Because of the applied voltage, emitted electrons are subject to
repulsion by the positive collector electrode and attraction by the emitting surface, hence to a
resultant force towards the emitting surface. The electrons therefore get slower and slower as
they cross the gap. When the stopping voltage is applied, even the most energetic of emitted
electrons have no K.E. left when they have made it across the gap. The K.E. lost is equal to
the P.E. gained for these electrons. That’s what the equation states.
It is just like finding the K.E. of a ball thrown upwards in the Earth’s gravitational field by
measuring the its maximum height, and using the energy conservation equation K.E. lost =
mgh]
•
Repeat the process using two or three more frequencies of light.
•
Plot a graph of KEmax against frequency, f. If Einstein’s equation is correct it should have
a positive slope equal to h and a negative intercept, equal to 
7
We can see this by comparing Einstein’s equation with y = mx + c.
A sample graph is presented below.
Graph of KEmax against frequency of light for a caesium surface
It is useful practice to find from the graph…
• a value of Planck’s constant,
• the threshold frequency for caesium,
• the threshold wavelength for caesium
• the work function for caesium
[In practice it is difficult to obtain a good value for h or  using a commercially available
vacuum photocell. Slight impurities on the caesium surface (e.g. a thin oxide film), and
unwanted electron emission from the collector electrode, both affect the stopping voltage.]
The first convincing verification of Einstein’s photo-electric equation, leading to an accurate
value of the Planck constant was completed in 1916 by R.A. Millikan, working in the United
States. The secret of his success was a remotely operated knife working in the vacuum to
skim off surface layers from the caesium surface as they became contaminated.]
[For interest only: http://focus.aps.org/story/v3/st23]
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Effect of changing the Light Intensity
If we bring a monochromatic light source towards a surface we increase the light energy
falling on the surface, per m2, per s. We are said to be increasing the intensity of the light.
Clearly we can apply the same idea to ultraviolet or any other electromagnetic radiation. We
find that…
(1) For light or ultraviolet of a given frequency, changing the intensity has no effect on the
maximum K.E. of the emitted electrons.
This is exactly what Einstein’s theory predicts. The energy given to individual electrons
comes from individual photons, and a photon’s energy, hf, depends only on the frequency
(or, equivalently, the wavelength) of the radiation. It doesn’t, then, depend on its
intensity (provided we don’t change the frequency).
(2) For light or ultraviolet of a given frequency, increasing the intensity increases the
number of electrons emitted per second.
Again, this is just what we’d expect from Einstein’s theory. Increasing the intensity
means increasing the number of photons arriving at the surface, per m2, per s. Naturally
this means that more electrons will be emitted. [Each identical photon has the same
probability of emitting an electron.]
We can show the effect with the same
vacuum photocell arrangement used
for demonstrating the photoelectric
effect. Note that the polarity of the
power supply is arranged to
encourage electrons to cross the gap.
•
Use a monochromatic light source to
illuminate the caesium surface.
•
Check that increasing the p.d. does not
affect the current, I. This means that
all the electrons emitted per second are
being collected.
•
Bring the light source closer and observe the effect on I.
•
I is the charge flowing per second, so the number of electrons emitted per second is I/e in
which e is the charge on each electron.
9
PH2.3 ( l)-(t) Detailed notes on Laser Physics
Raison d’être
Lasers, by now, are in nearly every household as they were once in every self-respecting
science fiction film. Although, the death ray mystique will appeal to most A-level students
there is real value in studying the physics which forms the foundation of laser construction.
Unlike other subjects such as relativity or nuclear physics which can only be touched upon
within an A-level course, the essential physics underpinning lasers can be taught reasonably
well in a few lessons. Also, these fundamentals of lasers can be taught at the right level while
not oversimplifying the subject.
Summary
(l)-(q) in the syllabus are related to the theory and construction of a simple laser. The new
concepts here are:
1. Stimulated emission of radiation.
2. Lifetime of energy levels.
3. Population inversion.
4. 3 and 4-level laser systems.
5. Basic construction of a laser.
(r)-(t) are related to semiconductor lasers and their uses.
Note: some of the content has been put in for general interest but will not be examined. Those
details in square brackets or small print will not need to be learned and will not be examined.
LASER is an acronym and stands for Light Amplification by Stimulated Emission of
Radiation. Which leads us nicely onto what is stimulated emission?
The Three Important Atomic Processes
These three processes [two of which are already known from 2.3(j)] are:
1. Absorption of light
2. Spontaneous emission of light
3. Stimulated emission of light
Absorption of light by an atom is shown in the diagram below – a photon of the correct
energy is absorbed by the atom and an electron gains enough energy to move from the
ground state to the excited state (Note: for the moment we are only considering the ground
state and the first excited state only).
Excited state
Electron excited to
higher energy
Photon absorbed
Ground state
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Spontaneous emission is the reverse process – an electron drops spontaneously (and
randomly) from the excited state to the ground state and emits a photon of the same energy.
These photons have random phase and random direction.
Excited state
Electron drops to
Lower energy
Photon emitted
Ground state
However, there is also a third process [which was originally proposed by Einstein in 1917].
This process is known as stimulated emission – an electron is ‘stimulated’ to drop from its
excited state by an incoming photon.
Excited state
Electron stimulated
to drop by photon
Incoming photon and emitted photon exit
in phase and in the same direction
Ground state
The reason that the electron is stimulated to drop is that the incoming photon is an
electromagnetic wave and its e-m field will exert an oscillating force on the excited electron.
If the incoming photon is of the correct frequency, this oscillating force will cause the excited
electron to drop and both photons will exit with the same frequency, phase and direction.
Note: again, the incoming photon needs to be of the correct energy.
Inverting The Population
In order to get as much light out of a system as is possible we need to get as many atoms
excited as is possible. Obviously, the more electrons we have in an excited state the more will
drop and emit photons (either spontaneously or through stimulation). However, there is one
serious problem that arises when we produce a lot of light – the very photons that we produce
are the actual photons that can be absorbed (they have the correct energy to produce both
effects). If we have photons being absorbed all the time then our laser beam isn’t getting any
stronger.
Forget, for the moment about spontaneous emission (we are allowed to but we’ll explain why
later). When a photon arrives at an atom one of three things can happen:
1. It can pass by and do nothing.
2. It can be absorbed (if the atom is in the ground state).
3. It can cause stimulated emission (if the atom is in the excited state).
11
When it comes to producing a laser beam with a high intensity the three options above will
have the following effect on the beam.
1. No change in the beam.
2. Net loss of one photon from the beam.
3. Net gain of one photon in the beam.
We need to arrive at a situation where stimulated emission is more likely than absorption so
that the laser beam increases in intensity. Since stimulated emission occurs if the electrons
are in the upper level and absorption when electrons are in the lower level we need to get
more electrons into the upper, excited level. This is called population inversion (or N2 > N1
as stated in the syllabus, where N2 and N1 are the number of electrons in the excited state and
the ground state respectively).
Unfortunately, this goes against what happens in nature – lower energy levels are always
more heavily populated than higher energy levels when we have thermal equilibrium (as we
go up 1eV to higher energy levels the probability of occupation of the level drops by a factor
of 1017). There’s only one thing for it – get rid of this thermal equilibrium. How do we do
this? We continue to pump energy into exciting electrons to higher energy levels to maintain
a population inversion and to break the conditions of thermal equilibrium.
Population inversion is not usually possible if we only have two energy levels (if pumping is
carried out by light). As we start to pump our system we have the following situation:
Excited state (N2 = 0)
Pumping light
Ground state (N1 = 8)
Many electrons will be promoted to the higher energy and all seems fine. Unfortunately, if
we succeed in exciting half the electrons we are now in the following situation:
Excited state (N2 = 4)
Pumping light
Ground state (N1 = 4)
In this situation the incoming flood of photons is just as likely to cause an electron to drop
(stimulated emission) as it is to cause an electron to rise (absorption). The best we can
achieve here is N2 = N1 which is not quite good enough.
12
The 3 Energy Level Laser System
E3
2
1
2
3
Pumping. Electrons are promoted from the
ground state (E1) to E3 usually by using an
E2
external light source or by electron collisions.
Electrons drop quickly (because E3 is chosen to 1
have a short lifetime of the order of anoseconds)
3
to the metastable (E2). Calling E2 metastable
means that it has a long lifetime and electrons
stay there for a long time (not that long really
E1
around a millisecond but that’s a very long time
for an electron).
This is the transition that produces the laser photons so we must have N2 > N1. Note that,
although stimulated emission still reduces our population inversion, the pumping is at a
different wavelength. We have to make sure that the pumping [1] exceeds the stimulated
emission [3] to maintain a population inversion.
Other things to note:
 E3 (to E2) has to have a short lifetime because E3 cannot start to fill up – pumping
won’t then be possible. Also, we don’t want the electrons to stay in E3 and have them
stimulated to drop back to E1 by the pumping light – that’s back to the 2-level system
again which wasn’t quite good enough.
 More than half the electrons from E1 must be pumped to E2 (via E3) in order to obtain
a population inversion – that’s a lot of electrons!
The 4 Energy Level Laser System
E4
1
2
3
4
Pumping again.
Quick drop to the metastable state E3.
This is the laser light producing transition so this
time N3 > N2. However, because E1 is the ground
state, E2 is practically empty initially so
obtaining population inversion is far, far easier
(definitely no need to pump half the electrons!).
Another quick transition so E2 has a short
lifetime. This is because we want E2 to be empty
so that we have a population inversion (if N2 is
small it’s easier for N3 to be larger than N2).
2
E3
3
1
E2
4
E1
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Laser Construction
Laser beam
Amplifying
medium
100% reflecting
mirror
Mirror: approximately 99% reflecting
(approx. 1% gets transmitted)
In order to ensure that the laser produces light of a high enough intensity, the above set up is
used. The amplifying medium is the region where the population inversion exists. This means
that the conditions are right in the amplifying medium for stimulated emission. Under these
conditions one photon has the potential to produce two photons and these can produce 4
photons, then 8 photons etc. Like a chain reaction, this process will lead to an exponential
increase in output energy. Laser physicists aren’t happy with this, they go even further – they
use mirrors to ensure that this exponential increase happens many times. Because only 1% of
the light exits each time it reflects back and forth between the mirrors, on average, the beam
will pass through the amplifying medium a hundred times before it exits. Now, considering
that each time the beam passes through the amplifying medium it is increasing exponentially,
this factor of 100 makes an enormous difference. [Try calculating e0.1 and then e10 on your
calculator and see the difference!]
This all leads to very high light intensities inside the amplifying medium and this is why (as
was said earlier) we can forget about spontaneous emission. Imagine that you’re an excited
electron sitting happily in your higher energy level. Normally, you’ll just drop down
spontaneously when your time is up. But, inside a laser, there’s so much light that you never
drop spontaneously because before your time’s up you’ve been disturbed by another photon,
stimulated to join in with all the other light and join in coherently as well!
Efficiency
Usually, lasers are very inefficient beasts. Because of the large energies required to maintain
a population inversion, their efficiencies are generally far below 1%. Some reasons for this:
 The pumping energy (see [1] in the 3 and 4 level systems) is considerably larger than
the output photon energy.
 High intensity pumping combined with the high intensity of the laser beam means that
the amplifying medium will get very hot. So, there will be large heat losses. To make
this matter worse, we need to cool the amplifying medium usually so that it, or its
container, doesn’t melt. By cooling the system we just transfer more heat and increase
our losses but better this than destroy a £50 000 laser!
14
Semiconductor Lasers
The basic structure of a standard ‘edge emitting’ semiconductor laser is shown below. The
whole block shown below is a semiconductor chip with dimensions approximately 0.5mm x
0.5 mm  1 mm.
metal
contact
‘sandwich’ of area with
population inversion i.e.
amplifying medium
light emission
not silicon, usually gallium arsenide.
This surface and its opposite are mirrors due
to the air-solid boundary
metal contact below also
[Although the chip itself has dimensions
approximately 0.5mm x 0.5mm x 1mm,
the beam has an approximate cross
sectional area of 30µm x 5µm and passes
along the ~1 mm length of the chip.]
The above laser fits the basic shape of a normal laser (shown below).
mirror
Amplifying
medium
Mirror
The mirrors, however, are far from the 100% and 99% reflecting ideals discussed earlier. The
mirrors are simply due to the semiconductor-air boundary at the edges of the chip. [This in
fact gives 40% reflection only (at both sides).] This would be disastrous for highly inefficient
gas lasers but not for our semiconductor laser. The reason why:
 The population inversion inside the semiconductor sandwich area is millions of times
higher than in gas lasers [~1025 electrons/m3].
 The exponential increase in light intensity (i.e. 1 photon becoming two, becoming
four etc.) occurs far more quickly because of the higher population inversion.
 So the fact that we lose 60% of the light at each reflection is compensated for by
having huge gains between the mirrors.
15
How can a population inversion be set up just by applying ~3 V d.c. voltage?
[You don’t need to know this but you might find it interesting]
Some of this next part used to be in the A-level syllabus many years ago. It shows a couple of band
diagrams of a p-n junction in a very highly doped semiconductor.
Electrons
p-type
semiconductor
n-type
semiconductor
Holes
Note: This is a diagram for a horizontal p-n junction but the junction in the laser diode diagram is
vertical.
After a small (~3 V) p.d. has been applied, we get:
Electrons
ve
+ve
Holes
Can you see the area of population inversion? If you look carefully, there’s a small region in the
middle between the p-type and the n-type where we have a high concentration of electrons above
holes. These are the electrons that can be stimulated to drop and provide laser light.
Advantages and Uses of Laser Diodes
These are straightforward and can be summarised as follows:
Advantages:
Some Uses:




Cheaper
Smaller
More efficient
Easy to mass produce





Inside DVD and CD players
Barcode readers
Telecommunications (via optical
fibres)
Image scanning
Laser surgery
The usefulness of laser diodes is ‘reflected’ in the number of them produced annually –
around 1 billion (109) laser diodes are produced worldwide per year!
16
For more than enough further reading see:
http://members.aol.com/WSRNet/tut/ut1.htm
Physics PH6 written paper 2007, comprehension question.
Wikipedia http://en.wikipedia.org then type ‘laser’ or ‘semiconductor laser’
Google http://www.google.co.uk/ then search ‘laser theory’
PH2.4 – MATTER, FORCES AND THE UNIVERSE
We start in familiar territory. PH2.4(a) deals with the simple model of the atom as nucleus
with orbiting electrons, in which the nucleus is composed of nucleons (protons and neutrons).
A
Z
X notation is required.
The rest of PH2.4 is devoted to the modern quark and lepton picture, with a short review of
the fundamental forces.
PH2.4 ( b)-(g) Detailed notes on Quarks, Leptons and Fundamental Forces
Nucleons have structure
This section, until a third of the way down the next page, is non-testable background
material which could provide a lead into PH2.4 (b).
1.
Although electrons are point-like at scales of 10-18 m, nucleons have diameters of the
order of 10-15 m.
1
Nuclear radius varies approximately as A 3 – electron diffraction experiments – so
nuclear volume is proportional to A, showing that the volume of each nucleon is, at least
approximately, the same.
Exercise: Radii of nuclei in fm: 6 C 3.16,
28
Si 3.93, 56 Fe 4.85,
120
Sn 5.99,
208
Pb 7.16
1
3
Show that r  A , and determine the density of nuclear material. [An A2 exercise would be to use
a log-log plot - and then go on to determine g at the surface of a 10 km diameter neutron star!]
2. Free neutrons decay into protons with a half-life of about 10 minutes.
The equation is 01 n  11 p  01 e  ν e and is clearly energetically favourable, as the rest mass
of the neutron is greater than the combined rest masses of the proton and the electron.
This is suggestive of a structure for at least the neutron and therefore the proton. [The
very similar masses of the neutron and proton suggests that the particles are closely
related – hence the use of the generic term nucleon.]
3.




The Stanford/MIT Experiment of 1968
20 GeV Stanford Linear Accelerator (SLAC) in 1968
Electrons fired down 2 km long accelerator tube
Target was a 7 cm container of liquid hydrogen.
Looked at interactions between the incident electrons and the protons in the hydrogen,
in which there was a large momentum transfer, shown by a large scattering angle for the
electrons (looked at 10 and 6 which are quite large in these experiments.)
17

Results
A large number of interactions was observed at these angles (just like the GeigerMarsden-Rutherford -particle scattering experiment). More technically, the
distribution of scattered electrons was similar to that in the -particle scattering,
suggesting a point charge distribution. In other words – just as in the Geiger-Marsden
experiment – the apparently solid object was in fact a collection of tiny ones.
These large momentum-transfer interactions resulted in the production of new particles
– typically the 0 meson.
The model of nucleon structure which best fits these and other results is the quark model,
proposed by Gell-Mann and Zweig in 1964, in which nucleons are composed of three quarks
bound together by the strong (nuclear) force. The work of Richard Feynman was instrumental
in this interpretation.
Quarks, antiquarks and quark combinations
There are two types of quark in everyday matter – the so-called up quark (u) and down quark
(d). The properties of these quarks are:
quark
charge / e
u
 23
d
 13
Note that there is a potential confusion between the down quark (d) and the deuteron (d) i.e
2
1 H – their symbols are the same. The distinction is usually clear from the context, but care
will be taken over the framing of examination questions.
Every particle has an antiparticle, which is identical to its corresponding particle apart from
having opposite charge. Particles and corresponding antiparticles annihilate. In general the
antiparticle has the symbol of the particle, but with a bar over it. Thus for quarks we have:
antiquark
charge / e
u
 23
d
 13
Quarks are bound together into composite particles called hadrons. These are further
classified…
Combinations of 3 quarks are called baryons [meaning ‘heavy ones’]. Neutrons and protons
are baryons. A proton is composed of uud, and a neutron udd. The order of the letters doesn’t
matter – it carries no information. [Check charges!]
proton
u
u
d
u d
d
neutron
What would be the composition of (i) an antiproton, p (ii) an antineutron, n ?
18
There does exist a family of four very short-lived baryons also composed of up
and down quarks: the  particles (delta particles). They are designated ++, +, 0, and .
The + and 0 have familiar sets of quarks – these two particles are essentially higher energy
(excited) states of the proton and neutron respectively. What are the quark constituents of the
Δ++ and Δ ?
A combination of a quark and an antiquark is called a meson. A family of mesons that
consists of only 1st generation quarks is the pions: +, 0 and . These contain only up or
down quarks and their antiparticles. So the charged pions must, have these quark
structures…

u
d

u
d
The composition of 0 is ambiguous. It could be either uu or dd .
It is actually – not for examination – a mixture of the two! Why don’t the uu and the dd
self-annihilate? They do! The lifetime of the 0 is much smaller than that of the charged
pions.
No such thing as a free quark
Our evidence for quarks is indirect, in the sense that it appears impossible to obtain a free
quark, i.e. a quark by itself. Within a hadron, indeed within the nucleus, the quarks can move
around, but they are bound together by the so-called strong force which does not decrease if
the separation between the quarks increases. [Contrast the electromagnetic force between
charged particles.] Hence the impossibility of obtaining free quarks.
Instead of freeing the quarks, when the bond between quarks is sufficiently stretched, a
quark-antiquark pair is created using the potential energy in the stretched bond, and a new
particle, a meson is formed. This is what happens in a so-called inelastic collision when a
high [kinetic] energy particle is fired at a nucleon. For example…
[π0]
Key
up quark
down quark
19
Students will not be expected to recall a host of different hadrons, but could be asked to work
out the quark constituents of specified hadrons using the table below, which will be available
in the examination. For example, a student could be asked to determine the composition at
quark/antiquark level of the +, given that is a meson, or of the Δ , given that it is a baryon.
[Students are expected to know that individually observable particles (i.e. excluding quarks)
have charges in units of e.]
Leptons
The up and down quark are the so-called first generation of quarks. The first generation of
leptons consists of the electron, e- and the electron neutrino, e. ‘Lepton’ means ‘light one’
[The literal meaning of ‘meson’ is now easily guessed!] The table contains the first
generation of quarks and leptons.
particle
(symbol)
Leptons
electron neutrino
electron (e)
(e)
up
(u)
Quarks
down
(d)
charge / e
1
0
 23
 13
Lepton
number
+1
+1
0
0
There is also a second and third generation, each with two quarks, one charged lepton and
one uncharged lepton. No questions will be set involving generations higher than the
first.
The electron’s antiparticle is the positron, e+. [Note that, for historical reasons, we don’t use
the bar notation in this case.]
The electron was the first fundamental particle to be discovered [in 1896-7 by J J Thomson].
A-Level students should need no introduction to the electron. It is probably a different matter
for the electron neutrino – or neutrino for short...
Historically, the evidence for the existence of neutrinos was indirect and arose from the fact
that, unlike -particles, -particles have a range of energies – suggesting that another particle
was present which shared the energy in a random way. This evidence is not available in AS
physics as it relies on the application of both mass/energy and momentum which are not
available until the A2 course.
Nonetheless this cloud chamber photograph of a 6He disintegration (beta decay) – see photo
below – is highly suggestive.
The photograph [from a cloud chamber],
shows paths of charged particles
produced from the decay of a 6He
nucleus at X. The short fat track is that of
the daughter 6Li nucleus. The thin curved
track is the electron (- particle). Clearly
something must have been ejected
upwards and to the left……
Further evidence comes from a
consideration of the - energies [more in
A2!]
Source:
Sang 1995
20
Lepton number
The electron and the (electron) neutrino are each assigned a lepton number of 1. Their
antiparticles, the positron and the antineutrino, νe , are each assigned a lepton number of –1.
The point of doing this is that we find lepton number to be conserved. Thus in any interaction
the sum of the lepton numbers before is equal to the sum of lepton numbers afterwards.
For example, the beta decay of the 6He nucleus is
6
2
He  63 Li  e  νe
The lepton numbers add to zero on the right. The lepton number is clearly zero on the left.
[Note also the conservation of charge, another infallible conservation law.]
In this interaction, a neutron in the nucleus is replaced by a proton. So we could write the beta
decay as
n p + e + νe
Or, at an even more fundamental level, recalling that a proton is uud and a neutron is udd, we
could write the beta decay as
d u + e + νe
Again, check charge and lepton number conservation.
Although the n, p version of the beta decay equation is fine for demonstrating the
conservation laws, it should not be taken as meaning that beta decay involves one neutron
decaying in isolation. We know that the neutron’s nuclear environment is involved, because
different unstable nuclei have different half lives for beta decay. A similar remark applies
even more strongly to the quark version: an isolated neutron has a half life of about 10
minutes, whereas a proton is to all intents and purposes completely stable – no tendency for
its d to undergo the change shown. So the intra-nucleon environment matters. These
subtleties will not be tested.
Exercise on applying Charge and Lepton number conservation
Over the past 30 years, scientists have detected neutrinos from the Sun by their interactions
with dry-cleaning fluid: specifically their interactions with a neutron in 37Cl nuclei.
→
[X is unidentified for the moment.]
This equation could be written at the nucleon level:
n + e → p + X.
It could also be written at the quark level:
d + e → u + X.
Candidates could be asked to identify particle X. Solution: In order to balance with the other
particles, it has to have a charge of 1 and a lepton number of +1, which means that it is an
electron. Note that we could also say that it cannot be a baryon, because baryon number is
also conserved, but that is not a requirement of the specification so it will not be asked.
21
The four forces or interactions experienced by particles
Interaction
Experienced
by
Range
Gravitational
all particles
infinite
Weak
all particles
very short
range
Electromagnetic
all charged
particles
infinite
Strong
quarks
short range
Comments
Extremely weak – negligible except in the context of
large objects such as planets and stars
Not considered here
A very weak interaction – only significant in cases
where the electromagnetic and strong interactions do
not operate. Interaction governed by this are of low
probability [in the case of collision interactions] or of
long life-time [in the case of decays]. Governs any
interactions including both hadrons and leptons, e.g. 
decay and the p-p part of the p-p chain (see PH2.5).
Much stronger [and therefore more likely / shorter
lifetime] than the Weak. Governs interactions
composed entirely of charged particles and photons
[N.B. Also experienced by neutral hadrons because
they are composed of quarks].
Responsible for electric attraction and repulsion.
The strongest interaction – only experienced by quarks
and particles composed of quarks (i.e. hadrons).
Responsible for the production of new particles from
nucleon-nucleon interactions.
Examples of interactions – which interaction is responsible for each?
 Neutron decay: n  p  e  e
Governed by the weak interaction because, (i) includes neutrino, which is not affected
by the strong or electromagnetic interactions, and (ii) it has a long lifetime [~ 10
minutes, which is long in particle terms, so the probability of the interaction is low.]
 Proton – proton collision: p  p  p  p   :
Governed by the strong interaction because all the particles present are hadrons
[consist of quarks]. Given enough energy to create a 0 this interaction is almost
certain to occur, although p  p  p  n   is also a good possibility.
 Electron-electron repulsion: e  e  e  e
Governed by the electromagnetic interaction because (i) the strong doesn’t affect
leptons and (ii) although it would be possible to occur via the weak interaction, this is
so much less likely.
 Proton-proton fusion: p  p  d  e   e
Governed by the weak interaction. Given the presence of the neutrino it is the only
candidate. Also the fact that the average lifetime of a proton in the centre of the Sun is
~ 109 years, means that the process is very unlikely in any individual collision.
 0 decay:      [i.e. decays into two photons]
Governed by the electromagnetic interaction because, (i) the pion is composed of a
charged quark and a charged antiquark, which both feel the electromagnetic
interaction, and (ii) neither the strong nor the weak interactions affect both quarks and
electromagnetic photons. The lifetime of a 0, at 10-12 s, is very short compared to the
charged pions which decay by a slower [less likely] weak process, e.g.
  e  νe [lifetime ~10-8 s], because there is no electromagnetic process available.
22
Useful websites
These are written at an appropriate level for AS/A students:
Standard Model: http://en.wikipedia.org/wiki/Standard_Model
from the bbc. : http://www.bbc.co.uk/dna/h2g2/A666173
Particle adventure: http://particleadventure.org/frameless/standard_model.html
The following site is worth it for one memorable sentence in brackets:
http://engr-sci.org/pnu/1990/physnews.004.htm
These sites contain a plethora of information – for self-motivated students only:
http://hypernews.slac.stanford.edu/slacsite/aux/HiPPP/scattering/
http://www.physics.ox.ac.uk/documents/PUS/dis/index.htm
http://physics.nmt.edu/~raymond/classes/ph13xbook/node194.html
23
PH2.5 – USING RADIATION TO INVESTIGATE STARS
This section of the specification is traditional 19th century physics applied to a current
research area. The emphasis is on the use of spectra to investigate stars – the idea that some,
in principle, very simple observations are enough to pin down important characteristics of
stars including their surface temperature, power output and composition. The line spectra
aspects of this unit are designed to fit in with section PH2.3 and so could be taught as a
continuation of that section. The proton-proton chain section introduces the idea of remote
sensing via subatomic particles and ties in well with section PH2.4. The black-body section is
free standing. Detailed notes follow.
Detailed notes on PH2.5: Using radiation to investigate stars
A star’s spectrum consists of a continuous spectrum, from the dense gas of the surface of a
star, and a line absorption spectrum from the passage of the emitted electromagnetic radiation
through the tenuous atmosphere of the star.
We start with some background material to assist in the interpretation of the continuous
spectrum.
PH2.5 (b), (c) The Black Body Spectrum
Good absorbers of radiation (of a given wavelength) are also good emitters (of that
wavelength). A black body is the name physicists have given to an ideal surface which
absorbs all the radiation that falls on it. Matt black surfaces approach this ideal. No surface at
a given temperature can, purely because it is hot, emit more radiation, at any wavelength,
than a black body.
The power emitted, per unit interval of wavelength, per m2 from a black body depends on the
wavelength as shown by the curves for various temperatures of the body.
24
Two of the most important laws that we find to apply to black body radiation are…
Stefan’s Law
The total power P of electromagnetic radiation emitted from area A of a black body at
(kelvin) temperature T is given by
P =  A T4
In which  is a constant called the Stefan constant. Its value is 5.67  10-8 W m-2 K-4.
P is the total of the power emitted at all wavelengths. It is proportional to the area under the
curve for the particular temperature.
Wien’s Displacement Law
The wavelength, p, at which the maximum power is emitted by a black body is inversely
proportional to the temperature of the black body.
That is
p 
W
T
In which W is a constant called Wien’s constant. We find W = 2.90  10-3 K m.
Three straightforward exercises
1. Use data from the curves to check Wien’s displacement law.
[The easiest way is to multiply p for each curve by the temperature of the black body.
Similar results, close to the accepted value of Wien’s constant, emerges each time.]
2.
Draw vertical lines on the graph grid to show the extremes of the visible region of the
electromagnetic spectrum. Hence explain why the black body at 3000 K appears
yellowish compared with the body at 5000 K (even though the black bodies at all three
temperatures shown are loosely described as ‘white hot’).
What would be the appearance of a black body at 1000 K?
What would be the appearance of a black body of a body at 8000 K?
3.
A white hot lamp filament approximates to a black body. On this basis, estimate its
temperature if the wavelength for peak emission is 1.16  10-6 m.
[Re-arranging Wien’s displacement law…
T = W / p = 2.90  10-3 K m  1.16  10-6 m = 2500 K.]
Estimate the surface area of the filament if the bulb emits radiation at a power of 70 W.
[To estimate the surface area of the filament we will assume that the filament is a black
body at 2500 K. So, re-arranging Stefan’s Law…
A = P /  T4 = 70 / (5.67  10-8  25004) = 3.2  10-5 m2.]
[Note that most of this radiation will be in the infrared (see curves) and is therefore, from
the point of view of illumination, wasted. What is more, the electrical power input to the
lamp would have to be more than 70W, as some energy is lost as heat conducted along
the wires from the filament and as heat convected from the filament by the inert gas
surrounding it. Hence the phasing out of filament lamps as energy wasters.]
25
PH2.5 (d) A Star’s Continuous Spectrum
Using a telescope, a special diffraction grating, and a detector sensitive across a wide range
of ultraviolet, visible and infrared wavelengths we can study the way in which a star’s
radiation is distributed among different wavelengths. [Allowances have to be made for
selective absorption by the Earth’s atmosphere.] None of the experimental method is
required, but students need to know the result: a star has a continuous spectrum almost
identical to that of a black body. We can usually assume that a star radiates as a black body.
On this assumption we can find out some key information about a star light years away by
studying its continuous spectrum. We need to extract two figures: (i) the total power (across
all wavelengths) reaching the vicinity of the earth, (ii) p, the wavelength of peak power.
How we could proceed is best shown by an example. This also illustrates the level at which
we feel it would be reasonable to ask examination questions…
Example
Consider the following observational data, which relates to Arcturus, a bright reddish star in
the constellation of Boötes:
Wavelength of peak emission in continuous spectrum = 674 nm
Power received per m2 at the Earth = 309  10-8 W m-2
Distance from Earth = 367 l-y [= 347  1017 m]
a.
b.
c.
d.
Sketch the continuous spectrum – a nice easy intro.
Calculate the Kelvin temperature.
Calculate the total power emitted [This called the star’s luminosity.]
Calculate the radius of Arcturus.
(b) To find the Kelvin temperature: use Wien’s law:
2  898 103 m K
 4300 K[3s.f.]
max
674 109 m
(c) To find the total power emitted, consider that, at the distance of the Earth, the power is
spread out over the surface of a sphere of radius 347  1017 m, so that the total power is
given by:
P  4 r 2  3  09 108 W  4  68 1028 W .
T
W

[Note that the total Solar power output is 390  1026 W, so that Arcturus is ~ 120  as
powerful as the Sun on these data.]
(d) To find the radius use Stefan’s Law: P  4 r 2 T 4 , where r is the stellar radius this time.
P
4  68 1028
r

 1 38 1010 m
4
8
4
4 T
4  5  67 10  4300
[ i.e. 138 million km, approx 20  that of the Sun.]
Additional remarks on PH2.5 (a) – (d)
“How science works” and the 19th century ‘ultra-violet catastrophe’ suggest themselves as discussion
topics. Students will not be tested on the ‘ultra-violet catastrophe’!
The cosmic microwave background radiation is an almost perfect black body spectrum. We should
really have mentioned it in the specification. We didn’t – but that doesn’t mean it couldn’t be included
in questions, given sufficient - background - information. It fits well in PH5 in the comprehension
passage…..
26
PH2.5 (e) (f) A Star’s Line Absorption Spectrum
Dark lines cross a star’s continuous spectrum. These co-incide in wavelength with the known
emission lines of particular elements. [Allowances may need to be made for Doppler shifts,
but this is not to be tested in PH2 examinations.] The implication is that these elements are
present in the tenuous outer ‘atmosphere’ of the star, and absorb specific wavelengths from
the continuous spectrum of light passing ‘outwards’ through this atmosphere.
This ties in with PH2.3 (h), (j). It would now be fair to test the work on absorption spectra in
this stellar context. For example, students could be given wavelengths of a dark lines, and
simplified atomic energy level diagrams, and asked to make reasoned deductions about
element(s) present in the star’s atmosphere.
http://casswww.ucsd.edu/public/tutorial/Stars.html (up to, but not including, ‘spectral
classification’) presents some useful pictures and diagrams.
http://jersey.uoregon.edu/vlab/elements/Elements.html lets you click on lines to find their
wavelengths.
Almost two hundred years ago Joseph Fraunhofer made a study of dark lines crossing the
Sun’s spectrum, and in particular noted dark lines at exactly the same places in the spectrum
as the yellow [sodium] lines in a candle flame. Although he couldn’t explain what was going
on, he had made an exceedingly important discovery: up to that time no-one had the slightest
idea that the Sun’s composition shared anything in common with that of the Earth. For
interest only, go to
http://astronomy.neatherd.org/Fingerprints%20of%20light.htm
Analysis of stellar spectra reveals that 75% of the universe by mass is Hydrogen, and 24%,
helium, with very small quantities of the other elements.
The question naturally arises: where have the other elements come from? This is not required
to be learnt, but for those interested, a very readable account is given in
http://fire.biol.wwu.edu/trent/alles/Origin_of_Elements.pdf .
PH2.5(g) Source of the Sun’s Energy
In a star like the Sun, energy is transferred from the Sun’s core to its surface layers by a
combination of radiation and convection. But what happens in the core? Energy is ‘produced’
through fusion reactions. The released energy is in the form of gamma ray photons, neutrinos
and kinetic energy of the product nuclei, quickly randomised by collisions to random internal
energy,
We require recall of the main branch (p-pI) of the proton-proton chain, which is the main
energy production mechanism in stars like the Sun. There are 3 steps…
p  p  d  e+  νe
(where d  deuteron 21 H)
p  d  23 He  γ
(where γ  photon)
3
2
He  He  He  p  p
3
2
4
2
It helps to remember that at each step the nucleus acquires one more nucleon.
Neutrinos from the first step occurring in the Sun’s core are detected on Earth. Their release
indicates that the weak force is involved. This means that the probability of the step occurring
is very low for a given proton. That’s why the Sun doesn’t emit radiation at a greater rate
than its miserly 390  1026 W, and is expected to survive for thousands of millions of years.
27
Information on various sub-branches of the p-p chain [ppII and ppIII] might be used as
examples in questions, as might the CNO cycle (important in stars heavier than the Sun).
The ppII chain, which accounts for ~14% of the helium produced takes over at the last step of
the ppI chain. The ppII chain is:
3
4
7
2 He  2 He  4 Be  
7
4
Be  e  73 Li   e
Li  11 He  24 He  24 He
Candidates would not be expected to recall this chain, nor even its existence, but could be
expected to, say, infer that step 1 is governed by the electromagnetic interaction, and step 2
by the weak process, or to write step 2 at the level of quarks  u  e  d   e 
7
3
Useful Websites
Note: Googling “proton-proton chain” gives a plethora of sites, many of which give far too
much detail, but the following are interesting:
P-p chain : http://csep10.phys.utk.edu/astr162/lect/energy/ppchain.html
From Wikipedia http://en.wikipedia.org/wiki/Proton-proton_chain_reaction
Nice – with animations: http://burro.cwru.edu/Academics/Astr221/StarPhys/ppchain.html
Another animation http://www.physics.mun.ca/~jjerrett/protonproton/pp.html
Part of a whole course on stellar physics:
http://www.shef.ac.uk/physics/people/vdhillon/teaching/phy213/phy213_fusion3.html
Dry, but links the production of solar neutrinos to their detection using Chlorine [they put the
dry into dry-cleaning fluid].
http://www.sns.ias.edu/~jnb/Papers/Popular/Scientificamerican69/scientificamerican69.html [nice
diagram of neutrino energies from the different branches of the p-p chain – this then related to the
possibility of detecting using Cl-37]. If it’s not on this site, you didn’t want to know it – that goes for
much of the material that is on the site too!
GCE Physics – Teacher Guidance
4 December 2007