This is the first of two lectures in which we will cover the topic of Compton Scatter.
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The objectives of the lecture are fairly straightforward.
First, we will identify and describe the Compton Scatter process. Next, we will
calculate the attenuation coefficients for Compton scatter. Finally, we will
determine the dependence of the attenuation coefficient on the nature of the
absorber and on the photon energy. This is pretty much what we do for the
attenuation coefficients for each of the interaction processes.
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Compton scatter is the most important interaction that we deal with in radiation oncology. In soft
tissue, it is the dominant interaction in the range of 30 kiloelectron volts to 30 million electron volts.
In this lecture and the next one, we are going to see that certain consequences resulting from some of
the properties of Compton scatter are very useful for radiation oncology, which is one of the reasons
why we treat patients with photons in this energy range.
As we recall for the photoelectric effect, one of the biggest issues was the dependence of the
attenuation coefficient on atomic number. Consequently, when the photoelectric effect is the
dominant interaction, we get high contrast from higher Z materials. It turns out that the greater
attenuation with higher Z is a problem when we are dealing with radiation therapy. Because of the
enhanced attenuation in high-Z materials, if we deliver the same amount of radiation to bone and soft
tissue, bone will absorb more radiation than soft tissue, resulting in an enhancement of dose. This is
not desirable. We would like bone and soft tissue to get the same dose or, at least something close to
it, and we are going to see later on in this lecture and the next lecture that that is indeed the case.
On the other hand, we don’t get much image contrast when Compton scatter is the dominant
interaction, because we can’t take advantage of differences in Z, and that’s going to degrade image
quality when we try to image using photons at therapy energies. We also get scattered radiation,
which degrades image quality even more. We will see more of that as we go along in the lecture.
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In analyzing the process of Compton scatter, we will look first at the kinematics of
the process. The kinematics will allow us to track the particles involved in the
process, their energies, and the directions they are ejected. Next we will determine
cross sections for the process, which will enable us to predict the probability that a
Compton interaction will take place.
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With that in mind, we will first look at the process of Compton scatter from a qualitative point of view. Just as we have done
with every other process, we are going to follow the energy.
The Compton scatter interaction involves a photon interacting with a single electron. It is inelastic scatter in which some
energy is transferred. Compare this to coherent scatter, in which the energy transferred is negligible. With Compton scatter
some of the energy of the incident photon is transferred to the kinetic energy of the electron. Thus, we now have both a
scattered electron as well as a scattered photon with significantly lower energy than that of the incident photon. Some kinetic
energy is given to the electron and some energy is given to the scattered photon.
The energetic electron is ejected from the atom. What can now happen to it? The electron can deposit more energy into the
target material by causing further ionizations downstream. When we talk about electron interactions we are going to look at
that process in a lot more detail. The other thing the electron can do is interact with nuclei and cause Bremsstrahlung.
So we have a fraction of the incident photon energy converted into kinetic energy of charged particles. Hence we can calculate
an energy transfer coefficient as well as kerma. Recall that kerma is the amount of kinetic energy transferred to secondary
charged particles.
Some of that electron energy gets re-radiated in the form of Bremsstrahlung and some of it causes further ionization and so we
have an energy absorption coefficient. It is that energy absorption coefficient that ultimately is going to be the handle that
connects the amount of radiation that comes in, the fluence, and the amount of energy that is absorbed, which is the dose.
Further ionizations will take place—we will worry about those later—and as a result of the ionizations we get biological
changes, which is what we are looking for in radiation therapy.
Now the scattered photons could also cause further ionization downstream. We are going to figure out what the energy of that
scattered photon is, and it is non-negligible. Unlike the photoelectric effect, in which almost all the energy of the incident
photon is transferred to the secondary electron, in the Compton effect, we see a division of energy between the secondary
electron and the scattered photon.
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We have just described the process qualitatively. Let us now try to obtain a more
quantitative description of Compton scatter.
Here is a diagram of the process. We start with an incident photon with an energy
hν, a corresponding wavelength λ, and momentum p equal to the energy hν divided
by c. Knowing the energy and momentum of the incident photon, we can do some
kinematics on the interaction process.
What comes out of the interaction? We have a Compton electron with an energy Ee,
momentum q, and velocity v, ejected at an angle φ to the direction of the incident
photon. We also have a scattered photon with an energy hν’, a wavelength λ’, and a
momentum p’ equal to hν’/c. The scattered photon is ejected at an angle θ to the
direction of the incident photon.
Does everyone follow so far? We see where things are going and we are going to
keep track of what’s going on.
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We are going to try to follow both the energy and the momentum. Energy is conserved; momentum is conserved. The
application of conservation laws, in fact, will tell us how the two particles, the scattered photon and the Compton electron are
ejected with respect to each other.
The first approximation we shall use, although we may refine this later, is that the electron is a free electron. So we are dealing
with a free electron gas. Why can we make this approximation?
Calling the electron a free electron means we neglect the binding energy of the electron. What’s a typical electron binding
energy? We may guess around 90 keV, but that is the binding energy for a K-shell electron in lead.
But the Compton effect is more pronounced for outer-shell electrons, and we really want to look at interactions in soft tissue.
A typical outer-shell binding energy for an electron in soft tissue may be only a few electron volts. Now compare that value to
the energy of the incident photon, which may be a few million electron volts. Comparing the binding energy of the electron to
the energy of the incident photon, we can neglect a few electron volts binding energy when compared to a few million electron
volts photon energy.
So what’s left? We are going to conserve energy and momentum. The first equation that we see here is for momentum in the
x-direction. The momentum in the x-direction before the interaction is the momentum of the incident photon, which is p =
hν/c. The momentum in the x-direction after the interaction is p’ cos θ + q cos φ. This is all freshman physics; you all learned
how to conserve momentum.
Now let’s look at the momentum in the y-direction. The momentum before the interaction is zero, because the incident photon
is moving in the x-direction. Consequently, the momentum of the Compton electron in the y-direction is equal to the
momentum of the scattered photon in the y-direction, or p’ sin θ is equal to q sin φ.
Finally, the energy of the incident photon is equal to the energy of the scattered photon plus the kinetic energy of the electron.
That’s the third equation that we need.
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Things now get a bit more complicated because we need relativistic mechanics to
do the energy and momentum relationships for the electron. These electrons are
ejected with quite a bit of kinetic energy. However, it does not take a huge amount
of energy for an electron to be relativistic. If we’re looking at electrons with ½
MeV of energy, we need to take relativistic effects into account.
How do we write down the relativistic momentum and the relativistic energy? The
relativistic momentum is equal to the total mass times the velocity where, if we
recall our kindergarten relativity class, the total mass is the rest mass, m0, divided
by the square root of 1 – β2. Recall that β is v/c. The kinetic energy from a
relativistic point of view is the total energy minus the rest energy. The total energy
is the relativistic mass times c2, which is equal to the rest mass times c2 divided by
the square root of 1 – β2. From this quantity we subtract off the rest energy m0c2.
We now have the relativistic momentum and the kinetic energy of the electron, and
we can go ahead and solve the kinematic equations.
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Just like the scientist in the cartoon, we will not be explicit in step 2, and just show
the results.
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When we solve the equations, we find the following:
The energy of the scattered photon is equal to the energy of the incident photon
divided by 1 plus α times 1 – cos θ, where α is the energy of the incident photon
divided by the rest energy of the electron, or 0.511 MeV. We will see this quantity α
in many of the equations to follow.
The energy of the Compton electron is equal to the energy of the incident photon
multiplied by α times 1 – cos θ and divided by 1 plus α times 1 – cos θ.
Finally, if you add the energy of the scattered photon to the energy of the Compton
electron, you get the energy of the incident photon, which is the way it should be,
since energy is conserved. You can also show that momentum is conserved as well.
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Remember, what we want to do is follow the energy. We start off with an incident
photon of energy hν and we want to know how much energy goes into the scattered
photon and how much kinetic energy is transferred to the Compton electron. We
have now found that the way that energy is distributed is going to depend on two
quantities. It’s going to depend on α, the energy of the incident photon and it’s
going to depend on θ, angle of the scattered photon. So we would like to determine
how the deposition of energy depends on the energy of the incident photon and
where the energy is scattered.
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Now, one thing we note is that if we solve these equations to look at the change in
wavelength of the photon, which we denote as λ’ – λ, we see that change in
wavelength of the photon is independent of the energy of the incident photon and
it’s only a function of the angle of scatter. In fact, its dependence on angle of scatter
is very simple. A good problem set question would be for you to derive this
equation for the change in wavelength of the photon.
So regardless of what the energy of the incident photon is, we can always determine
what the change in wavelength of the photon is, the quantity Δλ. We have a rather
simple formula to get that quantity, and it’s always dependent solely on the angle of
scatter.
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Let us now talk a little bit about energy transfer in some limiting cases. We will
start with considering a direct hit; the photon comes in, hits the electron head-on,
and ejects it in a forward direction. We have a forward-scattered electron, and, to
ensure that momentum is conserved, the photon is scattered in a backwards
direction, that is the angle θ is equal to π.
Remember our equations for energy transfer. The cosine of θ is minus 1, so that we
find when we have a direct hit, the largest amount of energy gets transferred to
kinetic energy of the Compton electron. The amount of energy transferred to the
Compton electron is the energy of the incident photon multiplied by 2α over 1+2α.
Recall that α is the photon energy divided by the rest mass of the electron, or 511
keV.
If the maximum amount of energy is transferred to the Compton electron, then the
minimum amount of energy is transferred to the scattered photon. The amount of
energy transferred to the scattered photon is the energy of the incident photon
divided by 1+2α.
So the maximum amount of energy we can impart to a Compton electron is equal to
the energy of the incident photon, hν, times 2α, divided by 1+2α.
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Here we are looking at the maximum energy that can be transferred to a Compton
electron, and consequently, the minimum energy that can be transferred to the
scattered photon.
Let us look at how this energy transfer depends on the energy of the incident
photon. Remember, we are looking at a direct hit, in which the electron is scattered
in a forward direction.
For high photon energies, that is, large values of α, the quantity 2α swamps the 1 in
the denominator. We see, then, that the fraction of energy transferred to the electron
approaches 1, that is, all the energy of the incident photon gets transferred to kinetic
energy of the Compton electron, and, of course, almost no energy gets transferred to
the backscattered photon. So when we have a head-on collision at high energies,
almost all the energy goes to the Compton electron.
What happens for low incident photon energies? For low energies the energy of the
Compton electron approaches zero. Almost no energy goes to the Compton electron
and almost all the energy gets back-scattered by the photon.
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Note that this relationship is true irrespective of the angle of scatter – let’s go back
to the previous slide.
To summarize, for large alpha, as long as the cosine of the photon scatter angle is
not equal to 1, the energy of the scattered electron will be approximately equal to
the energy of the incident photon and the energy of the scattered photon is zero. For
small alpha, all the energy goes to the scattered photon.
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In particular, we are looking here at energy transferred to the Compton electrons.
Remember, it’s the energy transferred to the electrons that’s going to be depositing
the radiation dose. For low energies, Compton scatter doesn’t transfer energy well,
but for high energies, it is an efficient method of energy transfer. This is an
important consequence for you to keep in mind and we’ve now actually derived
that.
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The next issue is right-angle scatter. The photon is scattered at right angles. The
quantity 1 – cos θ is equal to 1. The fraction of initial photon energy that is
transferred to kinetic energy of the Compton electron is α divided by 1 + α, and the
fraction of energy transferred to the scattered photon is 1 over 1 + α.
These are our limits of right-angle scatter.
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And finally, we want to look at a grazing hit. For a grazing hit, θ is approximately
0. If a photon is not scattered very much, then the photon energy is essentially
unchanged. The electron takes minimal energy, but keep in mind, the electron may
not be ejected at an angle greater than π/2.
We cannot backscatter electrons; and the reason is we cannot conserve both
momentum and energy with a backscattered electron. The electron may not have
any momentum in the negative direction; all the momentum must be in the positive
direction. So that’s a grazing hit.
We have now looked at three cases of angles. We have looked at a direct hit –a lot
of energy transfer. We have looked at 900 photon scatter, a kind of intermediate
case. We have looked at 00 photon scatter where essentially all the energy goes to
the scattered photon.
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Let’s look at some numbers to help us understand the energy dependence of
Compton scatter.
First, we’ll look at high energies, say a photon with incident energy 5.11 MeV.
Let’s plug in some numbers. Here α is equal to 10.
At 1800 photon scatter, that’s a direct hit on the electron, 2α = 20. When 2α = 20,
the energy of the scattered electron is 20/21 times the energy of the incident photon.
That is, about 95% of the energy of the incident photon beam gets transferred to
Compton electrons. We have a high-energy electron, energy about 4.7 MeV, which
can cause a lot of ionizations before it finally comes to rest.
If the photon is scattered at right angles, that is, θ = 90°, the fraction of energy
transferred to Compton electrons is α/(1+α) or 10/11. A bit over 90% of the incident
photon energy goes into the electron.
For a photon energy of about 5 MeV we see that Compton scatter is a very efficient
way of transferring energy from the incident photon to the Compton electron.
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Let’s now look at a photon with less energy. We’ll drop down a factor of 10 to a
photon energy of 0.511 MeV, that is, α=1. For photon backscatter, that is θ=1800,
2α divided by 1 + 2α is equal to 2/3, so 67% of the photon energy goes into the
scattered electron. For a sidescattered photon, for which θ=900, only 50% of the
incident photon energy goes into the scattered electron. So the electron receives
50% to 67% of incident photon energy.
From this information, once you have the Compton scatter attenuation coefficient,
you should be able to calculate the Compton scatter energy transfer coefficient.
Remember how we get that. We take the attenuation coefficient and multiply it by
the fraction of energy transferred. So here the fraction of energy transferred is
something between .5 and .67 so the energy transfer coefficient is roughly ½ to 2/3
the linear attenuation coefficient.
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Now, what happens at low energy? Here we have a 5 keV photon, a very lowenergy incident photon. For this photon, α is .01, 2α is .02, so for θ=180°, a
backscattered photon, the fraction of energy transferred to the electron is .02 over
1.02 or 0.0196. So only 2% of the photon energy is transferred to the Compton
electron and 98% of the energy goes to the backscattered photon.
For theta equal to 90°, the fraction of energy transferred to the electron is equal to α
over 1+α, or less than 1%. So at an incident photon energy of 5 keV, the scattered
photon holds on to 98% to 99% of the energy of the incident photon energy.
Compton scatter is a pretty lousy method of transferring energy at the very low
energies.
At high energies, such as energies encountered in radiation oncology, over 90% of
the energy gets transferred to kinetic energy of electrons. This is good because it
means we are causing more ionizations. However, at very low energies, even below
energies used in diagnostic imaging, 98% to 99% of the energy remains with the
scattered photon.
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Finally, if we look at the limit of zero energy, no energy is transferred to the
electrons and that’s the classical limit.
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Now, I want to look at the high energy limit. We are going to look at the
consequence of high energy photons that are scattered. In particular, we want to
look at photons that are scattered 90° and 180°. For example, these could be
photons that are scattered from a primary barrier in a treatment room. So we are
interested in what happens to these side- or back-scattered photons.
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Let’s do some numbers. What is the energy of a photon that’s scattered 180°? The
energy in the scattered photon is hν over 1+2α, which we can approximate by hν
over 2α. What is α? α is hν over m0c2, so the energy of the backscattered photon,
the photon that is scattered 180°, is approximately ½ m0c2, or one-half the rest
energy of the electron.
Notice something very important here. The energy of the backscattered photon is
essentially independent of the energy of the incident photon. So, for the highenergy limit there is a maximum energy that a photon scattered 1800 can have and
that maximum energy is about 256 keV. You can’t have a backscattered photon of
energy greater than 256 keV irrespective of the energy of the incident photon.
Now, let’s look at 90° scatter. For a photon that’s scattered 90°, the energy of the
scattered photon is hν divided by 1+α, roughly hν over α, and what is that equal to?
m0c2, or 511 keV. So for photons scattered 90°, the maximum energy of those
photons is again independent of the energy of the incident photons, and is equal to
511 keV. What is the consequence of this?
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It’s nice to know there are maximum energies; but are there any consequences to
this information?
The answer is Yes.
When we are doing shielding calculations and trying to calculate the amount of
shielding required to protect against scattered radiation, we are especially worried
about 90° scatter. Why are we interested in 90° scatter? Photons coming from a
linac incident on the patient are all scattered off to the sides. What’s the maximum
energy of 90° scatter? A number you should have memorized by now, 511 keV. So
the radiation scattered from a patient will not be higher in energy than 511 keV, and
it’s going to be on the order of magnitude of 1/10 of 1% of the incident radiation. If
we shield a therapy room for primary photons and leakage photons—that’s photons
coming out of the head of the linac, which are high-energy photons, that shielding
will generally be adequate and we will already have shielded for scattered photons.
So, in general, when you are designing shielding for a therapy room, you don’t have
to shield for the scattered photons, because you are already shielding for much
higher energy photons when you are shielding for leakage and primary. That’s a
very important consequence which all of you are going to encounter later when you
take your radiation protection class.
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So far so good. We understand energy transfer. Are we comfortable with it? Once
you look at the notes a couple more times, I hope you’ll be comfortable with it.
Now, let’s look at the probability of scatter as a function of angle. How much
radiation is going off in what direction? What we need to do is look at the quantity
called the differential scatter cross section, dσ/dΩ or dσ/dθ.
When we wanted to determine the energy transfer, we were able to use just simple
energy and momentum conservation with a little bit of relativistic mechanics thrown
in. If we want to determine what the probability of scatter is as a function of angle,
we need to use quantum mechanics. In particular, we need to use relativistic
quantum mechanics.
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If you want to see the derivation, I can refer you to the original paper by Klein and
Nishina in the Zeitschrift für Physik, volume 52, page 853, from 1929. On the other
hand, for the purposes of this course, we want to know the consequences of the
derivation.
Let’s look at a first approximation, that the electrons are free electrons. With this
approximation, the cross section that we obtain is an electron differential scatter
cross section. This cross section is equal to the classical differential scatter cross
section multiplied by a fudge factor, which we call the Klein-Nishina coefficient.
We find that dσ/dΩ is simply the Thomson scatter cross section, r02 over 2 times 1 +
cos2θ times the Klein-Nishina coefficient.
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This is what the Klein-Nishina coefficient looks like. I’m sure it is easy to derive,
but we won’t have to do it in this course. I’ll leave the proof up to the reader and
you don’t have to memorize this formula. I may ask you on a problem set to
calculate some Klein-Nishina coefficients, but you will never have to memorize this
formula. Perhaps the only time you would have to use this expression is if you are
developing some Monte Carlo code and explicitly calculating scattering cross
sections.
Let’s look at some properties of this coefficient. By looking at some properties of
the Klein-Nishina coefficient, we can appreciate what happens at low energies, what
happens at high energies, what happens at different angles.
The first thing we note is that the Klein-Nishina coefficient is less than 1.
Therefore, the scatter is going to be reduced from that we would calculate
classically.
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Now, let’s look at some limits. First of all, what happens at low energy? For α very,
very much less than 1, the terms in α in the Klein-Nishina coefficient can be
neglected. What’s left is that the Klein-Nishina coefficient for very, very low
energies is equal to 1. What has happened here is that we have derived the
Thompson scatter cross-section as a special case of Compton scatter for very, very
low energies.
Why is the low-energy limit the Thomson scatter cross-section rather than the
Rayleigh scatter cross-section. Recall that here we have assumed a free-electron
model and have not accounted for electron recoil.
What happens for small angle scatter? For small angle scatter, θ is 0. This is a
glancing blow in which the photon just comes right through and imparts just a little
bit of energy to the electron. Most of the energy, however, stays with the photon.
So θ is 0, cos θ equals 1, so the factors of 1-cos θ all go to 0. The Klein-Nishina
coefficient then is equal to 1, so we are back to the classical limit again. So for both
low energy scatter as well as small angle scatter, we reduce things down to classical
scatter.
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Let’s look at high-energy scatter. For high-energy scatter, α is much, much greater
than 1, so we can drop some terms that are small compared to α.
Let’s first look at backscatter, that is, 180° scatter, a head-on collision of the photon
with an electron. The cosine of 180° is -1. In both the first and second terms, we
drop the 1 compared to the 2α. We also drop the 1 compared to the 4α. A lot of α’s
cancel out, giving us a Klein-Nishina coefficient of 1/4α.
So the differential scatter-cross section, the correction to classical scatter, is going to
be dependent on the energy, and in fact, inversely proportional to the energy of the
incident photon. What we are starting to look at is the idea that as we increase the
energy, the probability of backscatter is going to go down.
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What about sidescatter. We’re looking a high energy photon, for which α is much
greater than 1. We have θ equal to π/2, thus cos θ is equal to 0. The first factor is
1/(1+α)2, which is approximately 1/α2. In the second factor, the numerator of the
second term is α2, and the denominator is α, so the second factor goes to α, and the
Klein-Nishina coefficient is equal to 1/α.
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Finally, for zero-angle scatter, for which cos θ = 1, the Klein-Nishina coefficient is
equal to 1 and is independent of energy.
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Let’s look at some differential cross sections for Compton scatter versus angle for
various energies. Notice that when θ=0 we get classical scatter, recalling that the
Klein-Nishina coefficient, which is the correction to classical scatter, is equal to 1.
When the photon energy is equal to 0 we also have classical scatter.
Notice that when the photon energy is about 1 MeV the differential scattering cross
section goes down by a factor of 10 at π/2 and is essentially independent of angle of
scatter for angles greater than π/2. For incident energies of 10 MeV we go down 2
orders of magnitude at an angle of π, that is for backscattered photons.
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Let us then summarize Compton scatter cross sections.
One, Compton scatter cross sections equal classical scatter cross sections in the
limit of zero energy.
Two, Compton scatter cross sections equal classical scatter cross sections in the
limit of 0 angle, that is, forward scatter.
Three, Compton scatter cross sections are peaked in the forward direction as the
photon energy increases, that is, most of the photons are scattered in a forward
direction. You will see some very important consequences of this in radiation
therapy.
Finally, Compton scatter cross sections at θ = π, that is backscatter, are lower than
cross sections for forward scatter by one order of magnitude for photons of energy 1
MeV and about 2 orders of magnitude for photons of energy 10 MeV.
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