For the next several lectures, we will be looking at specific photon interactions with matter. In today s lecture, we begin with the photoelectric

For the next several lectures, we will be looking at specific photon interactions with matter. In today s lecture, we begin with the photoelectric effect. 1

The objectives of today s lecture are to identify and describe the photoelectric effect and look at the dependence of the attenuation coefficient for the photoelectric effect on atomic number and energy and determine some consequences of this dependence. 2

Before we look at the photoelectric effect, specifically, let us look at this classification of photon interactions with matter. This classification is based on what the photon interacts with, and the effects of the interaction. Photons can interact with atomic electrons, with nucleons, with the electric field surrounding the nuclei or electrons, or with the meson field surrounding the nucleons. Interactions can result in either complete absorption of the photon, ejection of a photon with the same energy as that of the incident photon (That s called elastic, or coherent scatter), or ejection of a photon with a lower energy as that of the incident photon (That s called inelastic, or incoherent scatter). 3

That gives us a total of 12 possible interactions processes. It turns out, however, that only 5 of these processes are of any significance in radiological physics. 4

The five processes are as follows: First, there is classical scatter. This process can be described as a 1b process in which the photon interacts with atomic electrons and is scattered elastically. The second is the photoelectric effect, a 1a process in which the photon interacts with atomic electrons and is completely absorbed. The third process is Compton scatter. This is a 1c process in which the photon interacts with atomic electrons and is scattered inelastically. The fourth process is pair production. In this process, the photon interacts with the electric field surrounding the nucleus and is completely absorbed. The fifth process is photonuclear disintegration, in which the photon interacts with nucleons and is completely absorbed. I have listed these interactions in roughly the order of increasing energy. Classical scatter tends to occur at the lowest energies and photonuclear disintegration occurs at the highest energies. It is important to know that some of these interactions can occur simultaneously. For example at very low energy we will have primarily classical scatter. As we go higher in energy, the probability of photoelectric effect will take place and for this range of energies we will have both photoelectric effect and classical scatter. As we go to higher energies, Compton scatter starts playing an important role. In fact, when we are in the range of energies that we encounter in radiation therapy we have primarily Compton scatter, but some amount of photoelectric effect, and essentially no classical scatter. As we increase the energy, we start getting pair production playing a role. So there are situations where we will have photoelectric effect, Compton scatter and pair production, all of which have some significance in the interaction process. Then at even higher energies, photoelectric effect becomes negligible, pair production increases in probability, and eventually we also get photonuclear disintegration. So a lot of these interactions are simultaneous interactions, but we are going to be looking at the various interactions and how the probability depends on energy. In the next several lectures, we will get a better understanding of these interactions. 5

Because some of these processes occur simultaneously, when we look at the total mass attenuation coefficient, we find that mass attenuation coefficients for each process are going to add. We are not really going to be looking at photodisintegration from a quantitative point of view, so we are not going to look at attenuation coefficients for photodisintegration. Those do have a role but a fairly small role. We will be looking primarily at coherent scatter, photoelectric effect, Compton scatter, and pair production, and for these interactions, we find that the total mass attenuation coefficient is the sum of the mass attenuation coefficients for each one of the processes. 6

We will start with the photoelectric effect. Here is a very qualitative description of the photoelectric effect: a photon comes in to an atom with incident energy of hν. The photon interacts with the atom, transferring all of its energy to the atom and ejecting an orbital electron from the atom. 7

We must make sure that energy is conserved. Consequently, the energy of the ejected photoelectron must be equal to the energy of the incident photon minus the binding energy of the electron. The binding energy is the energy required to overcome the attractive force of the nucleus. For soft tissue, the binding energy of an inner shell electron is about a half kiloelectron volt, a fairly small value. So, when a photon comes in and interacts with soft tissue, most of the photon energy is going to be transferred to the photoelectron. 8

We also need to know what happens to the target atom. A small amount of kinetic energy goes to the target atom and we can probably account for this using regular kinematics. But recognize that the target atom is very, very massive, at least compared to the photoelectron. If we recall one of our situations where we have some kinetic energy that was divided up between an ejected particle and a much more massive nucleus, we calculated that almost all of the kinetic energy went into the ejected particle. In this case, the target atom has a mass of at least 2,000 times that of the ejected electron and certain very likely a lot more than that. So the recoil energy of the target atom is essentially zero. We could make a very good approximation by saying that all of the kinetic energy goes into the photoelectron. 9

We have now ejected an inner-shell electron. What happens to the atom once we ejected an inner shell electron? We are left with a vacancy, and whenever we have a vacancy, we emit a characteristic x-ray and we also emit Auger electrons. The characteristic x-ray can go downstream and ionize additional atoms. In tissue, that s not really very much of an issue, because the energy of the characteristic x-ray is going to be fairly low, so most of the energy is going to be deposited locally. The Auger electron is very low in energy as well, so its energy is also going to be deposited locally. So, when the photoelectric effect occurs, we eject an electron with on the order of many kilo-electron volts of energy. The electron can travel some distance away from the target, producing more ionizations, depositing energy along its path, and we have to be concerned with that additional energy deposition. The characteristic x-ray and the Auger electron deposit their energy in the immediate vicinity of the interaction. This is a qualitative picture of the photoelectric effect. One quantitative piece of information that we can extract from all of this is the kinetic energy of the ejected electron. We can also determine energies of characteristic x-rays and energies of Auger electrons. So those are numbers we can extract from this. 10

If we wanted to go further with a theoretical approach to the dynamics of the photoelectric effect, we would need tools of relativistic quantum mechanics. This is beyond the scope of this course, so, from here on in, our quantitative description of the photoelectric effect will be largely empirical. 11

Let us look at a plot of the energy dependence of the mass attenuation coefficient for the photoelectric effect, τ/ρ, which is a measure of the probability of the interaction. If we plot the mass attenuation coefficient versus energy on a log-log plot, we would find that for the most part it is a straight line. Look in particular at the line for water which is the blue line. It is roughly a straight line with a negative slope of some value, n. This straight line on a log-log plot implies that the mass attenuation coefficient for the photoelectric effect is proportional to the energy of the photon raised to some power -n. 12

Now it turns out that this slope is about -3. The quantity τ/ρ will fall by approximately three orders of magnitude for every order of magnitude increase in energy. So the probability of a photoelectric interaction occurring goes as 1 over the cube of the energy. That s a dependence that you need to remember. Notice it is a very, very strong dependence on energy. The probability of photoelectric effect occurring is highly dependent on energy and as you increase the energy this probability decreases as the cube of the energy. What is the consequence of this energy dependence on penetrating ability of the radiation. The penetrating ability of the radiation increases significantly with energy. We see this energy dependence, in particular, in diagnostic x-ray, where a rule of thumb says that if you increase the kilovoltage by 10 kv, you double the number of photons passing through the patient and reaching the detector. 13

We have seen that the first approximation to the dependence of photoelectric mass attenuation coefficient on energy goes as energy to the -3 power. Notice, however, that for lead we observe discontinuities in the behavior of the mass attenuation coefficient versus energy. We see discontinuities near 16 kev and near 90 kev, where we observe steep increases in the attenuation coefficient. What s going on here? Why are we seeing these discontinuities? 14

It turns out that the discontinuities in the curve of mass attenuation coefficient versus energy occur at energies equal to the binding energies of the electrons in the various shells of the target atoms. The K-shell binding energy of lead is about 88 kev, and that s where we see a discontinuity. The L-shell binding energy is at about 16 kev, and that s another location for a discontinuity. So, we observe that discontinuities in the mass attenuation coefficient occur at energies corresponding to binding energies. Why do these discontinuities occur? If we are looking at an incident photon with energy less than the binding energy, say, of a K-shell electron, can the photon ionize the K-shell electron? No, we don t have enough energy in that incident photon to overcome the K-shell binding energy and ionize a K-shell electron. However, if the photon energy is just greater than the binding of the K-shell electron, we do have enough energy to ionize the K-shell electron. At energies below the binding energy of the K-shell electron, the incident photon can ionize L, M, etc., shell electrons, but just above the binding energy of the K-shell electrons, we now can add K-shell electrons to the electrons that can be ionized. Because we can ionize more electrons, the attenuation coefficient, which is a measure of the probability of an interaction, undergoes an increase. That spike in the attenuation coefficient is roughly a factor of about 5. If you look at the tables at the back of the Johns and Cunningham text, you will see mass attenuation coefficients for different materials. In fact, if you look at the table for lead, you will see that spike in mass attenuation coefficients. The discontinuity corresponding to the K-shell binding energy is called the K-edge for photoelectric absorption, and the discontinuity corresponding to the L-shell binding energy is called the L-edge for photoelectric absorption. We need to keep in mind that we will observe these increases in the absorption coefficient at energies that correspond to the binding energies of inner-shell electrons. So now we understand, I hope, the energy dependence of the attenuation coefficients for the photoelectric effect. You should be able to draw the curve of energy dependence of mass attenuation coefficient, at least from a qualitative point of view. 15

Observe also how the attenuation coefficient depends on the atomic number of the absorber. The attenuation coefficient for lead is much greater than that for water. Typically, we will see about a 3 order of magnitude change in the attenuation coefficient per 1 order of magnitude change in atomic number. For water the atomic number is 7.5; for lead the atomic number is 82. That s roughly 1 order of magnitude a factor of 10. We see that the attenuation coefficient for the photoelectric effect in lead is roughly 1000 times that in water; it is roughly proportional to the cube of the atomic number. 16

This Z-dependence is experimentally obtained. We find that τ/ρ proportional to Z n. The exponent n is approximately equal to 3 for high-z materials; closer to 3.8 for lower Z materials. So for a good ballpark figure we can say that the attenuation coefficient is roughly proportional to Z 3 ; that rule of thumb will do you pretty well. 17

So, if we combine the proportionalities, we get that τ/ρ, the mass attenuation coefficient for the photoelectric effect, is proportional to Z 3 divided by (hν) 3. That s a very important relationship to keep in mind. Because of the nature of the dependence of the probability of the interaction on atomic number and energy, we observe that photoelectric absorption is most probable at low energies and high atomic numbers. Where do we take advantage of this? Well, for example, let s do some diagnostic imaging. We are able to interpret images because of differences in contrast. This contrast that we get in a radiograph is related to differences in the attenuation of the photon beam passing through different materials in the patient. Let s compare the attenuation coefficient of a photon beam passing through soft tissue to that passing through bone. What s the atomic number of soft tissue? 7.5. What s the atomic number of bone? There s a lot of calcium in bone, so that will increase the atomic number of bone. The atomic number of bone turns out to be around 11. So 11 divided by 7.5 is around 1.5. That s not a hugely different amount. But, we recall that the photoelectric interaction coefficient is proportional to the cube of the atomic number, so we now cube that difference. So now, 1.5 cubed turns out to be somewhat greater than 3. We see there s a factor of 3 difference in the probability of interaction of a photon passing through bone versus the probability of interaction of the photons passing through soft tissue. Consequently, far fewer photons pass through bone, and we see a great deal of contrast between the beam that passed through bone versus the beam that passed through soft tissue and hence we are able to detect these differences in a radiograph. Why do we use contrast media in radiography? Contrast media are materials that we incorporate into the patient to try to increase attenuation of the photon beam. One kind of contrast medium that we use, for example, is barium. First of all, barium has a higher atomic number than soft tissue, so we get enhanced absorption of the beam as a result of the higher atomic number. The other neat thing about barium is that its K-edge is at an energy just slightly below the average energy that we use for the diagnostic x-rays. So, in addition to the increased absorption due to atomic number, there is again an increase in absorption due to the fact that we are slightly above a K-edge. So we get some additional attenuation of the beam because of the barium. This is all you are going to get on the use of contrast media here. I think you will get that in much more painful detail in Med Phys II, but these are the fundamentals that lead up to that. 18

Now we have some idea of how the photoelectric effect depends on the atomic number of the absorbing materials and the energy of the incident radiation. The other aspect of photoelectric effect is the angular dependence of the ejected photoelectrons. At low energies, it turns out that the photoelectron is ejected at roughly 90 relative to the direction of the incident photon. At higher energies the photoelectron is ejected in a more forward direction. Photoelectrons are really ejected over a whole distribution of angles but the angular distribution is more peaked toward 90 at low energies and more peaked in a forward direction as we increase the energy. How do we explain this angular behavior? Recall that the incident photon can be looked at as an electromagnetic wave with an electric field perpendicular to the direction of propagation. This electric field induces motion of the atomic electron in the same direction of the field, that is, at 90 to the motion of the incident photon. So, if nothing else affects the motion of the electron, it will most likely be ejected at a 90 angle to the motion of the incident photon. At higher photon energies, we also need to consider the momentum of the incident photon, since we also need to conserve momentum. This momentum is in a forward direction, so the ejected electron must carry of some momentum in the forward direction to conserve momentum. The greater the momentum of the incident photon, the more momentum in the forward direction carried off by the photoelectron. 19

This graph illustrates the angular distribution of the photoelectrons as a function of energy. Note that at an energy of 0.020 MeV, that is, 20 kev, we see the most probable angle to be about 60, whereas at 2.76 MeV, the most probable angle is around 5. 20

One final note about momentum conservation. Consider an incident photon with energy much greater than the binding energy of the electron. In that case, the kinetic energy of the photoelectron will be approximately equal to the energy of the incident photon. But, since the electron carries a finite rest mass, the momentum of the ejected electron would wind up being greater than the momentum of the incident photon. We cannot have this, as momentum must be conserved. Consequently, some momentum gets transferred to the recoil atom to ensure conservation of momentum. 21

Finally, there is the additional radiation that I had mentioned earlier in the lecture. The ejected photoelectron leaves a vacancy behind. An Auger electron can be emitted, but that energy is deposited in the immediate vicinity of the interaction. A characteristic x-ray can be emitted, and its energy is deposited near, but not adjacent to, the interaction. Keep in mind then that in tissue, characteristic x-rays have very low energy, so the energy is deposited locally. I think we now have pretty good idea of how energy is deposited and where it is deposited as a result of a photoelectric interaction. Finally, of course, we need to follow the photoelectron. It travels some way downstream and deposits energy along its track until it runs out of energy. 22

Let s summarize the photoelectric effect. First of all the photoelectric effect involves bound electrons. Next, the probability of ejection of a bound electron is at its maximum if the photon has just enough energy to eject the electron from its shell, that is, enough energy to overcome the binding energy. The mass attenuation coefficient for the photoelectric effect varies inversely as the cube of the photon energy. The mass attenuation coefficient for the photoelectric effect varies directly as the cube of the atomic number. 23

And finally, in tissue, the energy transferred is approximately equal to the energy absorbed. That is, there s very little energy that s radiated by means of ejected photons. The characteristic x-ray is absorbed locally and we have no scattered photons. So with all these things in mind, we should have some idea of what goes on in the photoelectric effect. This is basically the only interaction process that we talk about on a fairy qualitative basis. From here on in, we will look at some of the other processes. We are going to start doing some derivations to actually get some numbers out. But for photoelectric effect, this is really as far as we are going to go, at least for the time being. 24