link to lecture transcript - UT-H GSBS Medical Physics Class Site

Today, we continue our talk about Compton scatter.
1

Let us first review what we’ve covered so far about Compton scatter cross sections.
We’ve said that the Compton scatter cross section is equal to the classical scatter
cross section in the limit of zero energy. It’s also equal to the classical scatter cross
section for forward scatter, that is, θ = 0. The Compton scatter cross section
becomes more peaked in the forward direction as the photon energy increases. The
Compton scatter cross section for backscattered photons, that is, θ = π, is lower than
that for θ = 0 by about one order of magnitude for photons of energy 1 MeV, and
about two orders of magnitude for photons of energy 10 MeV. This is what we
know so far about Compton scatter.
What we are going to do now is try to get an equation, first of all for the total scatter
cross section, and then we will look at the energy gy dependence p of the cross section.
The total scatter cross section is the attenuation coefficient per unit incident photon
fluence. Thus, from the scatter cross section we will be able to obtain the
attenuation coefficient.
2

We previously derived an equation for the differential scatter cross section per unit
solid angle. The first step toward obtaining the total scatter cross section is to
convert the differential scatter cross section per unit solid angle to the differential
scatter cross section per unit plane angle. We are going to do the exact same thing
that we did with coherent scatter and use the relationship dΩ = 2π sin θ dθ.
So dσ/dθ is equal to dσ/dΩ times dΩ/dθ, which is dσ/dΩ times 2π sin θ. This is
exactly what we did before with the cross section for coherent scatter.
3

Note that the sin θ term suppresses the cross section at θ = 0 and θ = π as it had done
for the expression for the cross section for coherent scatter.
So the final thing that we need to do is integrate over θ to obtain a total cross section
which is equal to this quantity. We see here σ 0, which is the cross section for
coherent scatter, and then we see a lot of factors involving α, where α is related to
the energy of the photon beam.
4

So σ 0 is the classical cross section, which is 8/3π times the classical radius of the
electron squared, and the remaining term is the integral of the Klein-Nishina
coefficient.
The total Compton scatter cross section is directly related to the electronic
attenuation coefficient. Remember, we’re looking at interactions with the electrons.
Therefore, the attenuation coefficient we’re getting is an electronic attenuation
coefficient, which is the total Compton coefficient. From this attenuation
coefficient, we can derive the energy transfer coefficient, the energy absorption
coefficient, etc.
If we want these numbers, we can look them up in J&C Table A2A second column.
5

Take a look at that table, and observe the energy dependence of the Compton scatter
coefficient for free electrons. Remember we said that the classical cross section was
independent of energy. Now let’s look at the correction due to multiplying the
classical attenuation coefficient by the integral of the Klein-Nishina coefficient.
At 1 keV the Compton cross section is roughly equal to the classical value – it’s
roughly 10 -28 m 2 /electron.
At 7 MeV, we are down 1 order of magnitude; at 100 MeV we are down two orders
of magnitude.
Now notice this. From 1 keV to 100 MeV, the scatter coefficient drops down two
orders d of f magnitude. it d As A a function f ti of fenergy we go down d five fi orders d of f magnitude it d
of energy from 1 keV to 100 MeV, but we only drop down two orders of magnitude
in the Compton scatter coefficient. Because of this we say the Compton scatter
coefficient is roughly independent of energy.
6

Let’s look at this graph. Notice that this is a log-log plot. The total Compton scatter
coefficient is on the vertical axis, and the photon energy is on the horizontal axis.
7

At least for a while it’s very flat and then it starts trailing off. So we drop two
orders of magnitude for five orders of magnitude of energy and we say the Compton
scatter coefficient is essentially independent of energy.
8

What about the dependence of the Compton coefficient on atomic number Z? The model we are using is a free
electron gas. We are looking at the probability of an interaction per electron. We are not talking about the
nucleus. Nor are we talking about the size of the nucleus or the number of electrons in the atom. Consequently,
the electronic coefficient for Compton scatter is essentially independent of Z, and, as we have seen previously, it
is also roughly independent of photon energy.
So what is it dependent on? About the only quantity the linear attenuation coefficient is going to be dependent
on is density. The mass attenuation coefficient is going to be essentially independent of everything.
Consequently, when Compton scatter is the predominant interaction, the linear attenuation is going to be linearly
dependent on the density.
AAnother th way of f saying i this thi is i “Equal “E l masses attenuate tt t equal l amounts.” t ” That’s Th t’ actually t ll a very good d thing; thi and d
the reason why that’s good results from the fact that radiation dose is defined to be energy absorbed per unit
mass. If we look at the absorption of energy in one gram of bone versus one gram of soft tissue with the same
incident photon fluence, we find that one gram of bone absorbs approximately the same amount of energy as one
gram of soft tissue. Equal masses absorb equal amounts.
How is that different from a situation in which the photoelectric effect is the predominant interaction?
Remember that for the photoelectric effect, absorption is going to be very strongly dependent upon atomic
number. So high-Z materials absorb a greater fraction of photons than do low-Z materials. This difference in
absorption gives rise to contrast in diagnostic imaging and an enhancement of dose to bone in low-energy
radiation therapy therapy. This was a problem in the pre-megavoltage pre megavoltage era when we were treating patients with 100 kVp
or 250 kVp x-rays. Because there was so much photoelectric interaction, for a given amount of radiation, bone
would receive a much higher dose than soft tissue.
When we’re in the Compton regime, bone receives the same dose as soft tissue, and that’s good.
Now, from the point of view of imaging, Compton interaction is not good because we don’t have as much
contrast. Recall that equal masses absorb equal amounts, so that because bone is a little denser than soft tissue,
we get a little bit of contrast between bone and soft tissue.
From the point of view of shielding, the Compton interaction is very good because equal masses absorb equal
amounts. So an equal mass of lead is as good a shield as an equal mass of concrete. Now if you were a builder
what would you rather work with, lead or concrete? Concrete is a standard building material; builders know
how to pour concrete. They need to make sure there aren’t any air bubbles or anything like that. It is easy to
pour four feet of concrete. Consequently when we shield a radiation therapy facility we are going to be looking
at conventional building materials to do our shielding; we just use thick walls. Several inches of lead shielding
may be just as effective a shield as a few feet of concrete, but the lead is much harder to work with and builders
don’t like to use it.
9

Next we want to calculate the energy transfer coefficient. We have calculated the
total attenuation coefficient, which tells us the fraction of photons attenuated per
unit path length. We want to calculate the energy transfer coefficient to determine
how much energy is transferred to the electrons as a result of the interactions. In
order to do so, we take the linear attenuation coefficient and weight it by the
fraction of the incident photon energy that is transferred to the electrons. This
fraction is equation to the amount of energy transferred to the electrons divided by
the incident photon energy. energy
10

The energy transferred to the electron is given by this equation; we derived that in
the previous lecture. The fraction of energy transferred to the electrons is this
quantity divided by hν. So dσ tr/dΩ is ½ r 0 2 times 1+ cos 2 θ , that is, the classical
quantity, times the Klein-Nishina coefficient times the ratio α(1 – cos θ) divided by
1 + α(1 – cos θ).
11

If we want to calculate the total cross section, hence the attenuation coefficient, we
have to integrate over angles. The integration is rather messy.
I am not going to integrate this for you as we can spend a couple of days doing the
integration; I’d rather not do it.
Th The bottom b tt line li is i that th t we still till have h this thi ¾ σ0, which hihi is what htwe had hdf from the thlinear li
attenuation coefficient times a horrendous quantity that has a very complicated
energy dependence.
12

If we plot the energy transfer coefficient as a function of energy, this is what we get.
Notice that, unlike the linear attenuation coefficient, the energy transfer coefficient
is suppressed at low energies.
13

For high energy photons, that is, large values of α, the energy transfer coefficient
approaches the classical coefficient.
14

The energy transfer coefficient has a maximum in the vicinity of ½ to 1 MeV. This
means we have an enhancement of energy transferred to kinetic energy of charged
particles at energies fairly close to what we like to use for radiation therapy. That’s
really neat, because we want to transfer as much of this photon energy as possible to
charged particles, that is, the electrons, to produce secondary ionizations. So that’s
good. We have a maximum energy transfer where we would like such a maximum
to be.
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At low energies, however, the energy transfer coefficient is strongly suppressed.
Compton scatter is not a good energy transfer mechanism at low energies. To recap,
energy transfer via Compton scatter is very efficient at high energies, but at low
energies Compton scatter is not an efficient mechanism for energy transfer.
16

What about the scatter coefficient? Remember when we have attenuation, two
things can happen. Some of the energy can be transferred to charged particles,
while some of the energy can go to a scattered photon. So the scatter coefficient is
going to be equal to the total coefficient minus the energy transfer coefficient.
At low energies, the energy transfer is small, so the scatter coefficient is
approximately equal to the total attenuation coefficient. Compton scatter is not a
good process for transferring energy to charged particles at low energies, but this is
also consistent with classical scatter, so we do wind-up getting the coherent scatter
in the limit of low energies.
At high energies, almost all of the energy gets transferred to charged particles.
Consequently, q y the scatter coefficient approaches pp zero. So the scatter coefficient
goes between the total attenuation coefficient at low energies and zero at high
energies.
Now, let’s use this information about coefficients to deduce some properties of
energy transfer and energy absorption.
17

For example, how do we determine the mean energy that’s transferred to charged
particles?
We now know how to determine mean energy transfer based on the definition of
coefficients. The mean energy transfer is the initial photon energy multiplied by the
ratio of energy transfer coefficient to attenuation coefficient.
18

Using this equation we can now plot mean energy transferred via Compton
interactions vs photon energy.
19

More important is the energy distribution. We have incident photons interacting
with absorbing material via Compton interactions. Let’s look at a monoenergetic
beam of photons and determine the distribution of energies of the Compton
electrons that are going to be produced.
Let’s determine the probability distribution of electrons generated from a
monoenergetic photon.
20

This probability is represented by dσ/dE. We can write this as the quantity dσ/dθ
times dθ/dE. We have already calculated dσ/dθ, so all we need to calculate is
dθ/dE. We have an equation of energy as a function of scatter angle, and we can use
this equation to get dE/dθ.
21

We find that dE/dθ is the incident photon energy, hν, times α sin θ, divided by the
quantity [1 + α(1 – cos θ)] 2 . We combine this quantity with dσ/dθ to obtain an
expression for dσ/dE, the distribution of energies of Compton electrons.
22

This figure is a plot taken from Johns & Cunningham. For each value of incident
photon energy, we have a different curve for dσ/dE.
Note first, that for each photon energy, there is a maximum energy of the Compton
electron. What is this maximum electron energy? Recall that the maximum energy
imparted to the Compton electron occurs when the photon is scattered 180°, or
backscattered. We determined this quantity in the previous lecture to be equal to the
energy of the incident photon multiplied by the ratio 2α divided by 1 + 2α. For
example, for a 1 MeV photon, the energy of the scattered photon is 204 keV, so the
maximum energy of the Compton electron is 796 keV. For a 500 keV photon, we
can calculate the maximum electron energy to be a bit less than 350 keV.
23

We also observe there are peaks in dσ/dE at both zero energy and the maximum
energy. Low energy and high energy electrons are more likely than medium energy
electrons.
Notice, for example, for a high energy incident photon the differential energy cross
section is relatively flat but as you get to the maximum energy you have a peak. For
very low energy photons we find there is a significant decrease in differential cross
section as we increase the energy. We observe a distinct minimum in the energy
distribution of the scattered electrons and then again a peak.
24

We can always calculate what the maximum electron energy is from the Compton
equation; this is the quantity that we call a Compton edge, an observed increase in
the energy distribution of the scattered electrons.
So if we are able to look at the scattered electron energy distribution using some
kind of multi-channel analyzers or something like that we will find there’s going to
be a peak right at the maximum energy. This maximum is called the Compton edge.
I think you are going to be seeing Compton edges in subsequent courses when we
talk about scintillation detectors.
25

How do we determine the total cross section?
We can integrate the differential cross section over energies. It turns out we can do
this integration numerically rather simply.
26

Remember dσ/dE is relatively constant with electron energy. If you want to
estimate the electronic attenuation coefficient from a differential coefficient, it is
rather easy to do so. This coefficient the electronic attenuation coefficient then is
going to be directly proportional to the mass attenuation coefficient.
27

Here’s an example of how we can estimate the electronic attenuation coefficient.
We want to determine the electronic attenuation coefficient for a 0.8 MeV photon in
a free electron gas. How do we determine this? First of all, the maximum energy of
the electrons can be estimated from the graph or calculated from a formula to be
0.606 MeV.
28

Now we also look at the graph and see that dσ/dE is approximately constant with a
value of 40 10 -30 m 2 /e MeV.
To a first approximation, let’s forget about the Compton edge; we will just
approximate the integration by assuming that dσ/dE is constant. To do the
integration, we multiply dσ/dE by the energy interval, 0.606 MeV, and find that σ is
the value of 24 10 -30 m 2 /e. Now if we look at the number in Johns & Cunningham
we find it’s 23.5 10 -30 m 2 /e. So simply by assuming dσ/dE to be constant, we get
a very good estimate of the total cross section.
29

Here’s another example.
Let’s determine the number of Compton interactions and the number of electrons
that are set into motion with energies between 0.15 MeV and 0.25 MeV when we
have a slab of bone 6 mm thick, and bombard the bone with 10 4 photons of energy
0.5 MeV.
We are going to do some accounting now. We are going to pretend we are
accountants and we want to follow the energy and figure out where these Compton
electrons go and what kind of energy these Compton electrons have.
30

First, let’s recall that the attenuation coefficient is the fraction attenuated per unit
absorber b b thickness. hi k How many photons h are attenuated? d? We could ld use the h
exponential approximation but when we’re estimating let’s use the thin absorber
approximation. The number of photons attenuated is the number of incident
photons times the attenuation coefficient times the absorber thickness.
How many incident photons do we have? We have been told that the number of
incident photons is 104 cdetp otoss 0. . WWhat at is s the t e Compton Co pto attenuation atte uat o coefficient coe c e t for o ½ MeV eV
photons. We look that up in Table A2a in Johns & Cunningham and find that to be
0.2892 10-28 m2 per electron.
What’s the thickness of the absorber? The thickness of the absorber is the thickness
of the bone times the density of bone. We multiply the thickness of bone, 0.6 cm,
by the density of bone, 1.650 kg/m3 , which we get from the Table 5.3. Next we
multiply by 10-2 multiply by 10 to convert the bone thickness from meters to cm
2 to convert the bone thickness from meters to cm.
Let’s convert that into centimeters times the electron density. Remember we have
to be able to calculate the electron density to be Z over A times Avogadro’s number.
Z over A is about a half. So electron density is always going to be roughly half of
Avogadro’s number. That’s a very good approximation for low to moderate Z
material, such as bone. Electron density comes out to be 3.19 10 23 electrons per
kilogram. So the electron areal density is 3.16 10 27 electrons per meter squared.
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So 3.16 10 27 electrons per meter squared times the Compton coefficient times the
number of incident photons gives us the number of photons that are scattered.
The result is 9.14 10 2 photons.
We find the assumption of a thin absorber is reasonable in this case. So now we
kknow hhow many electrons l t are scattered. tt d
32

What we now want to find out is how many electrons have energies between 0.15
and 0.25 MeV.
What is the differential cross section for photons of incident energy ½ MeV
producing electrons of energy around 0.2 MeV? We find the differential scatter
cross section is roughly 70 10 -30 meters squared per electron per MeV. That’s a
ballpark figure. Again, we’re estimating here. We will not do an exact calculation.
To do the exact calculation you would have to plug and chug into the equation for
differential cross section, which I am not going to have you to do.
33

We now have the differential scatter cross section. The energy interval we’re
looking at from 0.15 to 0.25 MeV is 0.1 MeV. To obtain the cross section we
integrate the differential cross section over the energy interval. Assuming the
differential cross section is constant over that energy interval, a reasonable
assumption, we find the cross section is 7 10-30 meters squared per electron. To
calculate the number of electrons generated in this energy range we multiply the
number of incident photons, 104 , times the cross section, 7 10-30 meters squared
per electron times the absorber thickness 3 16 1027 per electron, times the absorber thickness, 3.16 10 electrons per meter squared squared,
giving us an answer of 220 electrons.
In general when we want to calculate the number of electrons that are liberated, we
take the differential energy cross section, multiply it by the energy interval, multiply
it by the number of incident photons, multiply that by the electron density. The cross
section section, then then, is the probability of an interaction per unit incident photon fluence. fluence
Is everyone comfortable with this kind of a calculation? The problem sets will give
you more practice in doing calculations of this nature. Our basic goal in doing these
problems is to be able to know where the energy is going. Remember, we need to
follow the energy.
34

So far, our approximations have been based on the model of a free electron gas. For
a free electron gas we have no binding energy. There is no force attracting the
electrons to the atom, but that’s not completely correct.
So now we are going to make a minor refinement and account for binding energy.
To do this, we have to use quantum mechanics. We need the wave functions for
atoms. You can only do quantum mechanics relatively accurately on fairly simple
atoms.
The effects of binding energy are important only for low energy incident photons
for which the binding energy is comparable or at least not very much less than the
energy of the incident photon.
35

To account for binding energy, we multiply the differential cross section, dσ/dθ,
based on the Klein-Nishina theory, times a quantity that amounts to an atomic form
factor. It represents the probability that an electron actually leaves the atom. It’s a
function of x, which is related to the scatter angle and the photon energy, and it’s a
function of atomic number Z.
36

These values were tabulated by Hubbell in the Journal of Physical Chemistry
Reference Data.
So if we take this quantity, take everything together, and integrate over angles, we
can obtain the Compton coefficient in the real world. We are going to have some
slight deviations from the Compton coefficient for a free electron gas that account
for binding energy.
37

We can’t evaluate this integral in closed form, but if we look at Johns &
Cunningham Table A4, we’ll see different values for Compton coefficients for each
material. These values differ slightly from those for a free electron gas because of
the fact there is binding energy.
38

So what is the effect of binding energy? Here is a plot of Compton coefficients
versus photon energy for a free electron gas, oxygen, and lead. We see the curve
for the free electron gas on top; below that is the curve for oxygen; the lowest
values are for lead.
First of all, notice the binding energy only has an effect on the Compton coefficient
at low energies; once you get above 100 keV, the binding energy essentially has no
effect.
Binding energy is more important for higher Z materials. This also makes sense.
So the impact of binding energy on Compton coefficient is most noticeable for high-
Z materials.
In general, if we look at energies where the Compton effect is the predominant
interaction, the effect of binding energy is not significant.
39

Let us now summarize what we know about Compton scatter.
First of all, if we base our determination of the attenuation coefficient on the free
electron model, the attenuation coefficient is nearly constant to 100 keV. It falls less
than 2 orders of magnitude out to 100 MeV. So the coefficient is proportional to hν
raised to a power something smaller than -1, maybe -2/3, from 100 keV to 100 MeV.
To a ballpark approximation we can say it’s roughly independent of energy.
At low energies, the binding energy of the electrons affects the Compton scatter
coefficient suppressing attenuation in high-Z materials such as lead. But this
suppression only occurs at lower energies.
40

At higher energies there is no difference between any of the Compton attenuation
coefficients so then σ/ρ is essentially independent of Z.
The dependence of the Compton coefficient on atomic number is really an effect of
the binding energies. At higher photon energies you can forget about the binding
energies; we are basically dealing with free electrons. So in summary, σ/ρ, the total
attenuation coefficient, is roughly independent of Z, and depends on the hν to
something a little bit less than the magnitude of -1, say -2/3 or something like that.
41

In this graph we see the differential angular cross section for Compton scatter as a
function of scatter angle. Thompson scatter is the curve that’s symmetric in angle,
Rayleigh scatter peaks at low angle scatter.
42

Note that this graph is for very low energy photons incident on a low-Z target, 10
keV photons incident on carbon.
Coherent scatter at these low energies is significant and stronger than Compton
scatter. Also the binding energy has a significant effect on Compton scatter at these
low energies and that Compton scatter approaches the classical limit.
43

Compton scatter involves the interaction between a photon and a single electron.
The probability of Compton scatter is almost independent of Z; it decreases slightly
with increasing energy.
When Compton scatter occurs, some of the energy is scattered and some of the
energy is transferred into the Compton electron. The amount, the distribution of
energy, depends on the angle of emission of the scattered photon and the energy of
the incident photon.
44

The fraction of energy transferred to kinetic energy of charged particles per
collision increases with increasing photon energy. For low energy photons, energy
transfer is low, σ tr is much smaller than σ; Compton scatter is not an efficient means
of energy transfer. For high energy photons, σ tr is approximately equal to σ, and
Compton scatter is an efficient method of energy transfer. And finally, in soft tissue
Compton scatter is the most important process in the energy range of 30 KeV to 30
MeV.
We’ve spent a lot of time on Compton scatter. The reason we’re spending a lot of
time on it is because Compton scatter is a very important process. It’s the most
important process in the therapy energy range and matches the photoelectric effect
for importance in the diagnostic energy range.
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