The next three lectures will address interactions of charged particles with matter. In today s lecture, we will talk about energy transfer through

Transcription

1 The next three lectures will address interactions of charged particles with matter. In today s lecture, we will talk about energy transfer through the property known as stopping power. In the second lecture, we will talk about energy, range, and depth dose, while in the third lecture we will talk about how we take various averages of stopping powers when we are trying to determine mean energy transfer. 1

2 Our objectives for today s lecture are first to define stopping power. We will see there are two different kinds of stopping power: Collisional stopping power and Radiative stopping power. We will learn how to differentiate between the two. Finally, we will recognize the dependence of the stopping power on the usual parameters; that is, incident particle energy and properties of the target. 2

3 Why are we interested in stopping powers? First of all, as I have mentioned in many previous lectures, it is really the electrons that are set into motion as a result of photon interactions that transfer energy to the target medium. To some extent we almost don t need to look at the photon interaction, but rather at the interaction of the secondary electrons. Remember photons are indirectly ionizing radiations. Later this semester, we will be talking about things like dose in a cavity, which is going to be a very important topic related to the measurement of radiation dose. At that time, we will want to look at what happens to the secondary electrons and how the secondary electrons transfer energy and cause ionizations in the target. In addition we also use electrons, protons, and other charged particles in radiation therapy, and we need to know how these particles transfer energy to the target. The property that describes how energy is transferred from charged particles to a medium is the stopping power. Once we understand stopping power, we will understand how the energy is transferred into a medium from charged particles. 3

4 The order of business is that we will first talk about collisional stopping powers and then radiative stopping powers. Collisional stopping power addresses transfer of energy from a charged particle to an electron in the target. We are going to separate the discussion of collisional stopping power into what happens for heavy charged particles interacting with electrons, and then what happens for electrons interacting with electrons. We observe some differences in the treatment of collisional interactions between those involving heavy charged particles and those involving electrons. One source of difference is due to the differences in mass of the particles, while another source of difference lies in the identity of the particles. When an electron comes out of an interaction with another electron it is not clear which electron is the incident electron and which is the scattered electron. With a heavy charged particle, you know the incident particle is the heavy charged particle; the scattered particle is the electron. But with an electron interaction you don t know which is which so there are going to be some consequences in the interaction. There are also going to be some issues because of the differences in mass. An incident heavy charged particle is massive; it doesn t get deflected when it ionizes an electron. An incident electron has the same mass as a target electron so there could be much more deflection of the incident electron. We will then look at radiative stopping power, the result of a process in which the charged particle is deflected and gives off Bremsstrahlung. 4

5 What is the interaction between an incident charged particle and the target? The target is either an orbital electron or the nucleus of the target atom, so the interaction is the Coulombic interaction between charges. We can look at the interaction as the electric field of the charged particle interacting with the electric field of either an orbital electron or the nucleus. The quantity that we are going to look at is the stopping power, which is the energy lost by the charged particle per unit path length. Energy is transferred from the charged particle to the target medium. Units of stopping power are units of energy divided by units of length, for example, MeV per meter. We will also talk about mass stopping power, which is stopping power divided by density. The stopping power and mass stopping power are analogous to the linear attenuation coefficient and mass attenuation coefficient for photon interactions. The units of mass stopping power might be, for example, MeV m 2 kg -1. If we use mass stopping powers, then the path length is expressed as an areal density with units of, for example, kg m -2. We need to know the mass stopping power if we want to determine the dose because the mass stopping power is going to tell us how much energy is transferred to the target medium per unit mass of medium. Remember when we talked about attenuation coefficients we are looking at number per unit track length; here we are looking at energy transfer per unit track length. So when we talk about attenuation coefficients we had to determine what fraction of the energy was transferred in addition to the mass attenuation coefficient. Here we have the stopping power directly. We will use the stopping powers, and we will see those in our lectures on cavity theory, when we try to determine dose. 5

6 Our goal is to derive an equation for stopping powers. We will start with some definitions. There are two quantities that we need to define. The first is a, the radius of the atom, how far the target electron is from the nucleus of the atom. The second is a quantity, b, the impact parameter. The impact parameter is the distance of closest approach of the charged particle to the atom. Remember this Coulombic force, the 1/r 2 force, is a force that operates at a distance, so the charged particle doesn t have to come directly into contact with the atom. We will see some cases where it does, but it can interact with the atom from quite a distance. 6

7 We are going to look at some different kinds of collisions. First of all, we are going to look at what we call soft collisions. In these soft collisions the impact parameter is very much greater than the atomic radius; these collisions occur at a large distance from the target. When these soft collisions occur, we could have excitation and ionization of the electrons in the target. As a result of a soft collision a small amount of energy is transferred to the target. It turns out soft collisions typically account for around half of the energy transferred. 7

8 Next we have hard collisions. In hard collisions, the impact parameter is comparable to the atomic radius. Where is the differentiation between soft and hard collisions? We are looking at extreme cases. For soft collisions we are actually a long distance away from the nucleus, and for hard collisions we are a short distance away. When we have a hard collision we will have an interaction with an atomic electron and transfer a large amount of energy to this electron and obviously ionize it. This electron that we remove from the atom then has acquired sufficient energy to undergo additional interactions. We call these electrons that are produced from these hard collisions delta rays. If we knock out an inner shell electron using these hard collisions, we can have production of characteristic x-rays and Auger electrons. These hard collisions will account for approximately half of the energy transfer. 8

9 The third type of interaction is interactions with nuclei. In these interactions the impact parameter is much less than the diameter of the atom. The collision is a direct collision. Most of these encounters are elastic encounters where there is no energy transfer to the nucleus. The incident particle, however, is deflected and Bremsstrahlung is emitted. There are also inelastic interactions, in which the target is raised to an excited state. Inelastic interactions are important because they also give rise to Bremsstrahlung as well as other radiations emitted upon de-excitation of the nucleus. We will see these radiative collisions when we have electron interactions. We will find that the probability for radiative collisions with heavy charged particles is very low. Finally, annihilation is another interaction, which obviously is only important for positrons, but we will not be looking very much at annihilation. 9

10 So let s start in discussing interactions of heavy charged particles. The primary interaction of heavy charged particles with a target is an interaction with the orbital electrons. We must keep in mind that the mass of the heavy charged particle is much greater than the mass of the target electron. The fact that the mass of the incident particle is very much greater than the mass of the target is going to govern the kinematics of the energy transfer. 10

11 We are now going to follow the energy and derive an expression for the mass stopping power, looking at the interaction of a heavy charged particle with an electron. Let us start with a heavy charged particle. The incident particle has a mass M and charge Ze. The target electron has a mass of m 0 and a charge of e. Once again, the mass M is very much greater than the mass m 0, a factor of 2000 if the incident charged particles are protons. Our first approximation is going to use classical mechanics. Our initial calculation will be a classical mechanical calculation; we will later refine it with quantum mechanics. Let s consider an incident particle with mass M, charge Ze, traveling with a velocity, v, in the positive x-direction. What is the force that the electron in the target experiences? It s a Coulombic force from the charged particle. This force is equal to a constant, k, equal to nt m 2 per C 2 times the charge on the heavy charged particle, Ze, times the charge on the electron, e, divided by r 2. That s the Columbic force. 11

12 Here s a diagram of what happens. We have a heavy charged particle, mass M, charge Ze, moving in the positive x-direction with velocity v. Here s an atomic electron. This distance of closest approach is the impact parameter, b. We are going to make some assumptions in this model. First, we assume that any movement of the atomic electron during the collision can be neglected. Next, the initial energy of the electron in orbit can also be neglected. Finally, the changes in trajectory and speed of the incident particle during the single collision are small. Very little energy is transferred to the electron. Thus there is very little deflection of the incident particle. 12

13 What are the consequences of these assumptions? First, movement of the target electron during the collision can be neglected. This means that collision times are short and the average position of the electron can be used in our treatment rather than a changing instantaneous position. The second assumption is that changes in the trajectory and speed of the incident particle during a single collision are small. As a consequence, the incident charged particle loses energy in many low-energy-loss collisions rather than in a few catastrophic collisions. We call this the continuous slowing down approximation (CSDA). In this approximation, just a very small amount of energy is lost during the collision and the particle trajectory remains the same straight-line trajectory. The deviation in the trajectory during a single collision is small. This becomes a problem when the incident particle is an electron, but for the time being let s assume it s a heavy charged particle such as a proton or a carbon ion. It s whatever you want it to be, as long as it s heavy. 13

14 Because of the incident charged particle, the electron sees a force in the y-direction. We need to compute the momentum imparted to the electron in the y-direction. The momentum imparted to the electron in the y-direction is the integral of the mass times the change in velocity of the electron. The change in velocity, dv, can be written as dv/dt times dt. The mass times dv/dt is equal to the force in the y- direction. Consequently, the change in momentum is the integral from time equals minus infinity to plus infinity of the force in the y-direction integrated over time. This is simple, straightforward freshman-level classical mechanics. Now, what is the force in the y-direction? The force in the y-direction is the columbic force times the cosine of the angle theta. 14

15 Let s make a substitution. The distance r from the incident particle the target electron is equal to the impact parameter b divided by the cosine of theta. r = b/cos θ. Now we plug this substitution into the equation for the change in momentum. We pull the factors kze 2 out of the integral. That quantity is constant. We have an integral of cos 3 θ divided by b 2 over time. But θ is a function of time. This is the thing that we will have to integrate. Let s write tan θ as x/b. The derivative of tan θ with respect to t is the derivative of tan θ with respect to θ times dθ/dt, which is sec 2 θ times dθ/dt. Taking the derivative of the right side of the equation tan θ = x/b with respect to time, we have 1/b dx/dt, which is v/b, the velocity divided by the impact parameter. The change in tan θ with respect to time is equal to the velocity divided by the impact parameter. 15

16 A bit of manipulation shows us that dt is equal to b dθ divided by v cos 2 θ. We plug this quantity into dt in the expression for the change in momentum, and find that the change in momentum is kze 2 times the integral from -π/2 to +π/2 of cos 3 θ over b 2 times bdθ over v cos 2 θ. Do a bit of elimination and we get that the change in momentum is equal to kze 2 /bv times the integral from -π/2 to +π/2 of cos θ dθ. That integral of cos θ dθ from π/2 to +π/2 is equal to 2, so the change in momentum is equal to 2kZe 2 /bv. 16

17 Let s now plug in the classical radius of the electron into this equation. Recall that r 0, the classical radius of the electron, is equal to ke 2 /m 0 c 2, so the change in momentum of the incident charged particle is given by 2Zr 0 /bv times the rest energy of the electron, m 0 c 2. 17

18 This momentum is transferred to the electron. We now want to determine the kinetic energy imparted to the electron. If the electron is initially at rest, the kinetic energy imparted to the electron is Δp 2 /2m 0, which is 4Z 2 r 02 /2b 2 v 2 times m 0 c 4. 18

19 We see a factor of v 2 in the denominator, where v is the velocity of the incident charged particle. Let us express the change in energy as a function of the energy of the incident particle, so we replace v 2 by 2E/M. If we do this, we have an expression for the change of energy equal to Z 2 r 02 m 0 c 4 over b 2 times the mass of the charged particle divided by the kinetic energy of the charged particle. Let s look at this equation. First of all, the energy transferred is inversely proportional to the kinetic energy of the heavy charged particle. So low-energy heavy charged particles will lose more energy than high-energy heavy charged particles. One way to look at this is the slower the particle, the more times it spends in the vicinity of the target and has more of an opportunity to lose energy. An good analogy is running a gauntlet. You have a bunch of people with clubs and you try to run past them and they are taking swings at you. The faster you run, the less likely you are going to get hit. You run slowly and you will get hit more. This is similar. The slower the proton runs, the more hits, and the more likely it s going to lose energy. We have that energy transferred is inversely proportional to kinetic energy of the incident particle. The second point to observe is that the energy transferred is inversely proportional to the square of the impact parameter. The more direct the hit, the smaller the impact parameter, and the more energy is transferred. 19

20 We have calculated the energy transferred to a single electron that is a given distance from the path of the incident charged particle. Now we want to calculate the total energy transfer to all the electrons in the target. What is the energy loss of this incident charged particle? In order to do that we need to integrate this change in energy, a function of the impact parameter, over all values of the impact parameter for a given path length Δx. 20

21 So how do we do this? Let us assume that electrons are uniformly distributed over all space. The density of electrons is N e times ρ, where N e is the number of electrons per unit mass, say number per gram, and ρ is the mass per unit volume, say number of grams per cm 3. Thus the electron density is expressed in number of electrons per unit volume. Let s calculate the number of electrons in a cylindrical shell of a given impact parameter. We will look at impact parameter of radius b to radius b+db, and length Δx. How many electrons are in that shell? The number of electrons is the density of electrons, that is, number per unit volume multiplied by the volume of the shell. The volume of the shell is the circumference of the shell, 2πb, multiplied by the thickness of the shell, db, multiplied by the length of the shell, Δx. The density of electrons is N e times ρ. The quantity N e is Avogadro s number, N 0, multiplied by Z over A. 21

22 Now we can calculate the energy loss per unit path length. We just plug this all in. We take the energy loss per electron, ΔE(b), multiply it by the number of electrons per unit path length, Δn/Δx, which is also a function of b, and integrate over all impact parameters from 0 to. Now we plug everything in and we get this equation de/dx is equal to 2π times N e ρ times the integral from 0 to of Z 2 r 02 m 0 c 4 /b 2 times M over E times b db. Finally, we write the energy of the incident particle as ½ Mv 2, and we find that the energy lost per unit path length is now equal to 4π N e ρ times Z 2 r 02 m 0 c 4 /v 2 times the integral from 0 to of db over b. At this point, all of you should be very uncomfortable, because what in the world is the integral from 0 to of db over b. 22

23 I remember first seeing this derivation in a nuclear physics class, and I was very uncomfortable with it, so if you re uncomfortable you re not the only one. The problem with this expression is that the indefinite integral of db over b is equal to log b. What is the problem with log b? It blows up at both ends, 0 and. What s wrong with this derivation? Here s what to keep in mind. Let s first look at large impact parameters. We have a charged particle coming in some large distance from the atom. Recognize that the target electron is not a free electron, but is actually bound to an atom. Consequently, for some large value of the impact parameter, the amount of energy transferred to the electron will not be sufficient to overcome the binding energy of that electron. It s not completely a free electron and there will be some maximum impact parameter, beyond which we will not have energy transfer. This whole model breaks down for large impact parameters. Let s call that impact parameter b max. We don t yet know what it is, but at some value of impact parameter that s really big, the whole approximation is going to break down and we can t use this formula for energy transfer. So we are not going to have energy transfer for really large impact parameters. Now, let s see what happens for small impact parameters. A charged particle with a small impact parameter is going to collide with the nucleus, and not transfer energy to electrons. So there will be some minimum impact parameter, let s call it b min, less than which there will be no energy transfer. So for very large and very small impact parameters we re not going to have energy transfer. 23

24 At least, formally, we can perform the integration from b min to b max rather than from 0 to and get the quantity log (b max /b min ). We will worry a little bit later about what those values might be. We integrate from b min to b max, and everything outside the integral makes a lot of sense. We ve put in β = v/c, β being the β for the incident charged particles. And then we have this log (b max /b min ). We will worry about this in a little while, as we suspend our natural-born cynicism as to the legitimacy of this derivation. 24

25 The quantity de/dx is the stopping power. It has units of energy per unit path length. We can view it as one of two things. It s either the energy lost by the incident particle beam per unit path length or it s the energy gained by the absorbing material. Viewing the stopping power as the energy imparted to the absorbing material is going to get us to our dose and that s where we want to be eventually. We really want to know the mass stopping power, so we divide by the density. So the energy gained by the absorbing medium per unit mass can be obtained directly from the mass stopping power. Now keep in mind that so far, we have only looked at ionization. This is the ionization component of stopping power. We will call it the collisional stopping power. There is also a component of radiative stopping power that we will have to worry about a little bit later in the hour. 25

26 Let s talk about this quantity b max. We said earlier that for some value of impact parameter b, the amount of energy transferred to an electron will not overcome the binding energy of the electrons. What s going to happen is the electron will be excited; it s not going to be ionized. So in some way we should be able to relate b max to the binding energy of the electrons. Let s introduce a semi-empirical quantity called the mean excitation energy of the electrons we will call it I. If the mean excitation energy is large the maximum value of the impact parameter is going to be small. This means it going to take a fairly large amount of energy transfer to cause ionization to take place. So we have to come pretty close to the atom for ionization to take place, and b max will consequently be small. As the atomic number Z of the target atom increases, the mean excitation energy, I, will also be greater, that is, for larger Z we have greater mean excitation energy. That makes sense, doesn t it? 26

27 So, in some way the mean excitation energy will go into the expression for mass collisional stopping power. Now we may also have to include relativistic effects because the ejected electrons may have relativistic energies. We also may have to include quantum mechanical effects, because this is fundamentally a quantum mechanical problem. 27

28 I am not going to do the derivation of the correction terms, but I am going to present you with the final equation for the mean collisional stopping power, 1 over the density times de/dx. I should point out that Johns & Cunningham call it ionization stopping power. I think more authors tend to call it mass collisional stopping power. For this course, we will use the term mass collisional stopping power. We will recognize that other people call it different things. In this equation, you see some things that, by now, should be familiar to you. Outside the brackets, we see terms that originated in the classical treatment, 4πr 0 2 times electron density times Z 2 m 0 c 2 divided by β 2. Inside the brackets we see three terms. The first is the logarithmic term, in which the mean excitation energy goes into the denominator. Both the first term and the second term include some relativistic corrections. 28

29 This last term is what we call a shell correction. C i is an empirical constant that corrects for how the electrons are bound. The mean excitation energy shows up inversely only in the logarithmic term. Therefore, mass collisional stopping power is not strongly dependent on I, the mean excitation energy. The term in brackets is a slowly increasing function of energy. It s essentially independent of energy; for protons it s going to vary by a factor of 2 over a range of roughly 3 orders of magnitude; it s a very slowly increasing function of energy. We will look at some graphs of this as a function of energy in a few minutes. 29

30 The classical mechanics term is the term outside the bracket that we derived a few minutes ago. It displays the dominant behavior of the mass collisional stopping power. The most important issue to keep in mind when we re looking at mass collisional stopping power is the dramatic increase in stopping power at low energies. When beta is small, the mass collisional stopping power shoots up. This is the origin of what we call the Bragg peak. The Bragg peak plays a very important role in the deposition of energy from heavy charged particles. Think about a charged particle traversing through absorbing material. When are you going to see it have the lowest energy? Remember, the charged particle enters the absorber with some incident energy. It is going to start depositing energy along its track, basically a uniform amount of energy along its track, until it slows down. When does it slow down? At the end of its range. So if we look at the deposition of energy from a charged particle beam, we will see an enhancement in the deposition of energy, that is an increase in dose, at the end of its range. One of the reasons that we consider the use of protons for radiation therapy is because of the existence of a Bragg peak, and enhanced deposition of energy at the very end of the proton range. 30

31 Let us now look at electron interactions. They should be similar to those of heavy charged particles because they are based on Coulombic interactions, but there are some differences. 31

32 The first difference between electrons and heavy charged particles is that relativistic effects are much more important for electrons than for heavy, charged particles. Let s look, for example, at some beams. The maximum energy of a proton beam that we would use in radiation therapy is around 250 MeV. For this energy beam, β is around 0.6. That would make the beam somewhat relativistic, but not extremely relativistic. Consider, however, an electron beam of energy 1 MeV. This is an energy somewhat lower than that typically used in radiation therapy. A radiation therapy beam may go up to 10 to 20 MeV. For the 1 MeV electron beam, β is Thus it is clear that relativistic effects play a much greater role in electron beams than these effects play in a heavy charged particle beam. The second difference is that the incident electrons have the same mass as a target orbital electron. Therefore, an incident electron can be deflected a large amount as a result of a collision. The assumption made in the model of a charged particle interaction that there is no deflection of the incident charged particle is just not valid here. The third major difference is the identity of the electron. How can you tell which electron is the incident electron and which is the target electron. There is no way to tell them apart. We don t have green incident electrons and blue target electrons. That doesn t work. So we make the assumption when we do a calculation is that the incident electron is the one that leaves the interaction with the higher energy, but in fact we really don t know. 32

33 Electrons undergo two types of interactions with the target. Electrons can interact with orbital electrons giving rise to collisional energy loss. But, electrons can also interact with the nucleus giving rise to radiative energy loss in the form of Bremsstrahlung. Now, heavy charged particles such as protons can also interact with the nucleus, but the probability of energy loss for a proton interaction with a nucleus is much, much lower. When an electron interacts with a nucleus, the electron gets deflected, and we produce Bremsstrahlung. The stopping power for Bremsstrahlung goes as 1 over the square of the mass of the incident particle, so for protons, the radiative energy loss is approximately 6 orders of magnitude lower than that for electrons. We don t see radiative energy loss in a proton beam. We don t produce Bremsstrahlung in a proton beam. However, we do in an electron beam. Radiative energy loss is going to be significant. 33

34 So here is the expression for collisional stopping power for electrons. This is an equation that was developed by Bethe based on both relativistic mechanics and quantum mechanics. As you can see, it s a pretty hefty equation; there are a lot of terms in it. First of all, let s identify some aspects of it that may be familiar to you by now. The quantity outside the brackets looks somewhat similar to what we had for heavy particles. So does the logarithmic term and so does this β 2. 34

35 Let s identify some of these terms. The quantity μ 0 is the rest mass of the incident electron, E is the energy of the incident electron, and I is the effective mean excitation energy. μ 0, E, and I have the same units typically expressed in MeV. The electron density is Avogadro s number times Z over A. 35

36 Now let s look at this equation in a little more detail. First of all the term outside the brackets is almost the same as that seen for heavy charged particles. But we no longer have a factor of Z. We have a factor of 2 instead. Remember for electrons Z is equal to 1, but it turns out we need a factor of 2 in there because two electrons are involved in the interaction. Again it s the issue of identity. That is really the only change we see from the classical term for stopping power for heavy charged particles. 36

37 Now the other major difference is that we now have a term, denoted by δ, which is a density, or polarization, correction. The reason for this correction is that we have an electron far away from the target, and its interaction with target electrons is reduced by the electric field of intervening atoms. This polarization effect causes a reduction in the stopping power. 37

38 The effect is greater for more dense materials; the effect is small for low energy electrons, but it increases to around 20% near 100 MeV. But it turns out the polarization effect makes up a small correction in the calibration of clinical electron beams, about a 1% error. It s small, but, nonetheless, it s there. 38

39 Let s look at the collisional mass stopping power of electrons in water. First of all let s look at low energy electrons, a small value of v, hence a small value β. The stopping power is dominated by a 1 over β term; again, it s 1 over velocity. This increase in stopping power at low energy is the same issue that we saw with heavy charged particles; as the electron slows down it stays in the vicinity of orbital electrons for a longer period of time. There is more time for the electrons to interact and lose energy. 39

40 But look what happens as we increase the energy of the electrons. As β approaches 1, the 1 over β term remains constant. So the mass collisional stopping power flattens out as a function of energy and becomes roughly independent of energy. We observe a slight increase as we go to higher energies, but to a good approximation, stopping power is roughly independent of energy. The mass collisional stopping power for electrons is approximately 2 MeV cm 2 per gram. That s one of those numbers you probably should remember. The reason that you will remember it is to estimate depth of penetration of electrons in tissue. We approximate tissue by water, which has unit density. So electrons penetrating tissue lose energy at the rate of 2 MeV per centimeter. If we start off with an electron beam of some given energy, and lose energy at the rate of 2 MeV per centimeter, we can estimate the depth of penetration of the electron beam. Let s say we start off with an electron beam of 10 MeV, how far is that electron beam going to penetrate in tissue? If we start with 10 MeV and lose 2 MeV of energy per centimeter, we will penetrate 5 cm before we run out of energy. So now you have a way to estimate the penetration of an electron beam in tissue because of the fact that the collisional mass stopping power is essentially independent of energy for energies greater than about 1 MeV. 40

41 Next, let s look at the Z-dependence of collisional stopping power. Our first approximation is that the target material is a free electron gas. In that case, the mass collisional stopping power is essentially independent of Z. 41

42 Let s go to a higher order of approximation. Electron density is not really constant with increasing Z. Remember the fact that electron density is proportional to Z/A. Z/A gets smaller for larger Z, thus electron density decreases for larger atoms. Higher Z also means greater binding energies for inner shell electrons, so the mean excitation energies will increase with increasing Z. We recall that the effective mean excitation energy goes into the expression for collisional stopping power as a logarithm of 1 over the mean excitation energy. Increasing Z will make that logarithmic term smaller. The density effect, however, is greater at higher Z, as higher Z materials tend to be denser. 42

43 Let s put everything together, and we see that as Z increases, in general, the mass collisional stopping power decreases. But notice as we go from water, with a Z of 7, to lead, with a Z of 82, factor of 10, the reduction in mass collisional stopping power is less than a factor of 2. So there is a small reduction of mass collisional stopping power with increasing atomic number, but it s a relatively small dependence a factor of 10 increase in Z resulting in a factor of 2 decrease in mass collisional stopping power. 43

44 We need to talk about a phenomenon called straggling. We know that energy loss due to collisions is a stochastic process. For an individual electron incident on a target, we do not know the specific amount of energy transferred to the target electron. All we know is that there is a distribution of energies peaked around the value of the mass collisional stopping power. Consequently, if we start out with a monoenergetic beam of electrons, after a large number of collisions, the energy distribution of the beam will spread out. With a spread-out energy distribution, the range of individual electrons will vary so that there will be a distribution of ranges in the electron beam. This is the phenomenon of straggling. 44

45 Let s talk about radiative stopping power. What is physically going on here? We have an incident charged particle deflected by a nucleus producing Bremsstrahlung. An energetic charged particle has a mass of M and a charge of z times E and passes near a nucleus with a mass of M n and a charge of Z times e. 45

46 Let s look at the kinematics of the interaction between the two charges. The Coulomb force k z Ze 2 /r 2 is equal to the mass of the incident particle times its acceleration. We can then solve for the acceleration. 46

47 We make the approximation here that the mass of the incident charged particle is much less than the mass of the nucleus, so nuclear recoil can be neglected. If the incident particle is an electron, this approximation is certainly valid. The incident charged particle is deflected and the accelerated charged particle will radiate energy at a rate proportional to the square of the acceleration. So we will find that the atomic radiative stopping power is roughly proportional to the square of the atomic number of the incident charged particle, 1 if it s a proton or an electron, 2 if it s an alpha particle, multiplied by the square of the atomic number of the target nucleus, divided by the square of the mass of the incident charged particle. 47

48 We see here that high Z materials radiate more Bremsstrahlung because of the Z 2 term, but we knew that already. Remember, we talked about Bremsstrahlung production in the target of an x-ray tube. What do we use as target material for the x-ray tube? Typically tungsten, a high-z material. We use a high-z material in the target because we said that the efficiency of x-ray production is greater for larger values of atomic number. The electronic stopping power is the atomic stopping power divided by Z, so the electronic stopping power is proportional to Z. Electrons are more likely to radiate than heavy charged particles because of that 1/M 2 term. If we compare the likelihood of Bremsstrahlung for protons with that of electrons, we note that the mass of the proton is on the order of 2,000 times the mass of the electron. Consequently, the radiative stopping power, a measure of the likelihood of production of Bremsstrahlung, for protons is more than 6 orders of magnitude less than that for electrons. 48

49 So here s an equation for mass radiative stopping power, derived by Heitler and Evans. This is for electron energies less than 100 MeV. We won t worry about greater energies. The mass radiative stopping power, which is 1 over ρ de/dx, is equal to 4r 0 2 multiplied by the electron density, times the atomic number of the target, times the electron energy, divided by 137. The mass of the electron is included in the 137. There is a term in the log of the energy as well. The term outside the brackets is classical, the term inside the brackets is quantum mechanical. If you notice this is very similar to the term that we have for pair production. We mentioned this earlier in our treatment of pair production. In Bremsstrahlung production an electron interacts with the Coulomb field of atom making a transition between two energy states, emitting a photon. In pair production a photon is destroyed by the interaction with the atomic Coulomb field, producing a pair. 49

50 So let s put this together and look at mass stopping power for electrons in water and in lead. First of all, as we now know, there are two components. There is collisional mass stopping power and there is radiative mass stopping power. For low-energy electrons, collisional mass stopping power dominates radiative mass stopping power. We know that; we ve seen this before when we talked about the production of x-rays. Again, collisional mass stopping power for energies greater than about 0.5 MeV is essentially independent of energy. In water the collisional mass stopping power is approximately equal to 2 MeV per centimeter. The collisional mass stopping power for lead is approximately a factor of 2 lower. Notice, however, that when we look at radiative stopping power, we see that the radiative stopping power for lead is at least an order of magnitude greater than that for water. Consequently, radiative stopping power is the dominant means of energy transfer at about 10 MeV for lead. For water, we don t worry about radiative stopping power until we are at about 100 MeV. When we are doing calculations, let s keep this in mind. Remember when we went from energy transfer coefficients to energy absorption coefficients we need to subtract off the component that is due to re-radiation. We see here that when we are dealing with electrons in water or tissue, reradiation is very low. The radiative stopping power is very low for interactions in water. Consequently, the energy transfer coefficient is relatively close to the energy absorption coefficient for most energies. It s only when we get to really high energies that we need to make a correction for re-radiation. 50

51 Some points to keep in mind: Collisional processes dominate for energies in the megavoltage range. Ultimately, at high energies, radiative processes become important. For low Z materials, collisional equals radiative at about 100 MeV, while for lead, collisional equals radiative at about 10 MeV. Finally, radiative stopping power is roughly proportional to Z times E. 51

52 The total mass stopping power is the sum of the collisional and the radiative mass stopping powers. The collisional stopping power creates ions that ultimately deposit energy locally. Consequently, dose is going to be related to mass collisional stopping power. The radiative component escapes the immediate vicinity and these Bremsstrahlung photons deposit dose elsewhere, so we do not need to count the radiative component when we are trying to determine dose deposited locally. 52

53 Let s look at some numbers. Johns and Cunningham Table 6-3 gives mass stopping powers in water for monoenergetic electrons as a function of electron energies and Table A-5 has mass collisional stopping powers for various materials. 53

54 Let s use these numbers to solve a problem. Take a monoenergetic beam of 20 MeV electrons with fluence of 10 4 electrons per square centimeter incident on a water phantom. Determine the energy deposited as ionization and the energy radiated as Bremsstrahlung in the first millimeter of water. 54

55 We find from Johns and Cunningham Table 6-3 that the collisional stopping power for 20 MeV electrons is MeV per centimeter while the radiative stopping power is MeV per centimeter. The energy deposited due to ionization is the collisional stopping power times the electron fluence times the path length. Multiply them all together we get 2,063 MeV per square centimeter being deposited due to ionization. The energy radiated as Bremsstrahlung is the radiative stopping power times the electron fluence times the path length. Multiply these quantities together and we get MeV per square centimeter. We see that for 20 MeV electrons, which is at the high end of the energy range used in radiation therapy, 5 times as much energy is deposited locally as is radiated. So even for high-energy electrons from a practical point of view, most of the energy is still deposited locally. 55