# 15. Photon Interaction with Matter II — More Details, Shielding Calculations

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### 15. Photon Interaction with Matter II — More Details, Shielding Calculations

The following content is provided under a Creative Commons license Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu MICHAEL SHORT: Today I want us to go more in-depth into the photon interactions with matter, and we’re going to bring the theory back to something that we can actually start to use in doing shielding calculations If you want to find out how much of this material do I need to shield this much gammas, we’re going to answer that question today First I want to start off, again, with Compton scattering because I messed up a couple of the energy things from last time I got excited due to an energetic coincidence between the photo peak and the continent of our banana spectrum and one of the examples in the book So I’m going to correct that now, and we’ll go through in more mathematical detail why that wasn’t the case, and what the actual quirk of physics is because there is a constant energy thing here that I want to highlight to you guys So skipping ahead on the photoelectric effect, which I think was similar, to review Compton scattering It’s the same thing that we saw between two particles, except now one of them is a photon And like I said, on the next homework, after quiz one, you guys will be doing the balance between energy and momentum, because the photons don’t really have mass, in order to figure out what’s the relationship between the incoming energy of the photon, the outgoing energy of the photon, and the recoil energy of the electron And so these are the relationships we were showing last time It is an interesting quirk of physics that the wavelength shift itself does not depend on the energy of the photon coming in As you can see, it just depends on the angle that it scatters at and a bunch of constants, where that m right there stands for mass of the electron Now that wavelength shift– while that wavelength shift does not depend on the energy of the incoming photon– the recoil energy does You can see it depends on both the energy and the angle And the incoming energy of the photon equals h nu And to give that quick primer on photon things, I want to show you guys here why that’s the case So even if you have a constant wavelength shift that might give you a non-constant energy shift So even though in this constant scattering formula, the wavelength shift only matters with the angle, the energy shift actually depends on the angle and the incoming energy of the photon So now let’s look at a couple of limiting cases So as we have e of the photon equals h nu goes to 0, what does t approach? The recoil energy of the electron Let’s just do out the formula This recoil energy equals h nu, which is the energy of the incoming gamma times 1 minus cosine theta over mc squared over h nu plus 1 minus cosine theta As h nu approaches 0, what happens here? AUDIENCE: [INAUDIBLE] MICHAEL SHORT: Yeah H nu goes to 0 This fraction goes to infinity And this goes to 0 Hopefully that’s an intuitive explanation If the incoming photon has 0 energy, it can transfer 0 energy to the electron Now the more interesting case What happens now as e gamma approaches infinity, as the photon gets higher and higher in energy? AUDIENCE: T just approaches h nu MICHAEL SHORT: T approaches almost h nu I actually want to do a quick calculation without doing all the limit math Let’s say we had energy of the gamma was 1 GeV, an extremely high energy So all we’d plug in– and let’s say we wanted to find out what’s the maximum energy of this recoil electron And this is something I want to ask you guys I can’t remember Yesterday did we say that t is a maximum at theta equals pi or pi over 2? What did we say yesterday? AUDIENCE: Pi over 2 MICHAEL SHORT: Interesting It’s pi, actually The case where you have the largest energy transfer– and sorry for not catching that– is just like in a nuclear collision, if the photon were to back scatter, it transfers the maximum energy to the electron So the analogy here is, like, perfect Between two particles hitting and between a photon and a particle hitting, the maximum energy is when theta equals pi And let’s actually plug that in to find out why If we say t max equals h nu– depends on the electron coming

in– 1 minus cosine theta over mc squared over h nu plus 1 minus cosine theta At theta equals pi, cosine theta goes to minus 1, and so then that also goes to minus 1 And so this becomes 2h nu over mc squared over h nu plus 2 And so that without worrying about the numerator, especially in the limit of very high energy photons, you can see that that actually maximizes the recoil energy of the electron The reason we’re harping so much on this recoil energy of the electron is because that’s what we measure So when we look at our banana spectrum, you’re not measuring the energy of the photon You’re measuring the recoil energy of the electron and the ionization cascade that happens as all those electrons smash into each other, creating electron whole pairs, which are counted as current If you guys remember from last class– in fact, I’ll just bring up the blackboard image because we can do that So I took a picture of the board yesterday, photon interactions Part 1 There it is We’ll just use the screen as a bigger blackboard for now So if you guys remember, let’s say a gamma ray comes in and causes a Compton scatter event or a photoelectric emission, or whatever It doesn’t matter which process And it liberates an electron, either by scattering off of it or just getting absorbed and ejecting it Or it doesn’t really matter how, but it creates this electron hole pair That electron right here has this recoil energy, which depends on theta, the angle that it scatters and the incoming Energy and that electron’s going to keep moving in this material, knocking into other electrons very, very efficiently so that most of the energy of that electron recoil actually gets counted as other electrons being freed We’re going to go over on– well, next Friday– electron nuclear interactions, including what’s the probability in energy transfer when electrons slam into each other or when ions slam into electrons or each other? So let me go back to the slides And so this maximum, as this approaches infinity, this actually approaches a value of h nu minus .255 MeV Just to do the quick calculation to give a numerical example, if I plug in theta equals pi and h mu equals 1 gig electron volt, or 1,000 MeV And can anyone remind me what is mass of the electron c squared? What’s the rest mass of the electron? Sorry? AUDIENCE: [INAUDIBLE] MICHAEL SHORT: Right While I’m usually against memorizing anything– because that’s what books and the internet are for– this is one of those quantities that I want on the tip of your tongue as nuclear engineers You should remember what the rest mass of the electron is because a lot of our quantities are calculated based upon it For example, this ratio– what was it, h nu over MeC squared– gives you the energy the photon in terms of the number of electron rest masses, which is a useful quantity in itself So why don’t I plug all this stuff in So we have .511 over 1,000 plus 2 flipped over the x-axis times 2 times h nu And we get– so that t becomes 999.745 MeV Interestingly enough, 1,000 MeV, our ingoing photon, minus that equals that right there So that’s the interesting quirk of physics is as the photon increases in energy, the maximum amount of energy that it can leave with– or sorry, the minimum amount of energy the photon can leave with, or the maximum amount it can impart to an electron approaches 255 MeV or the photon energy minus that What do you guys notice about that number? AUDIENCE: [INAUDIBLE] MICHAEL SHORT: That’s right It’s exactly half the rest mass of the electron So as your photons hit the GeV range and above,

they can all leave with half the rest mass of the electron of energy, which means you have more and more energy able to be transferred in a given Compton scatter as the photon gets higher and higher in energy So for the limit of low energy, the photons basically bounce off without transferring much energy at all And the higher in energy the photon gets, the higher percentage of that maximum transfer can be So that’s what I wanted to clarify from last time It was an interesting coincidence that our photo peak and our Compton edge were pretty close to that number apart But one MeV isn’t quite infinity But it is pretty close That difference right there, what does that come out to? Before I say something stupid, I’ll just calculate it 1.241 It’s like 0.218 MeV, so we’re already most of the way there So once you reach, like, 10 or 100 MeV, you’re pretty much at that limit And so what that tells you is that the distance between a photo peak and its corresponding Compton edge for high energy photons is going to be half the rest mass of the electron Once you get to lower and lower energies, that distance will start to shrink I’m sorry, other way around That distance will start to grow And I have a few examples I want to show you But first, in order to understand these, now I want to get into the part I told you we’d get to yesterday, which is what’s the probability that a Compton scatter happens at a certain angle? And this polar plot explains it pretty well In the limit of really low photons, like .01 MeV or 10 keV photons, you can see that this forward scattering to an angle of 0, or back scattering to an angle of 180– there’s a 180 that’s blocked by an axis right there– they’re almost the same So forward and back scattering, there’s not really that much of a big difference in probability So if we were going to start graphing theta of this Compton scatter as a function of the– going to introduce this new quantity, this angularly dependent cross-section Before we were giving you cross- sections in the form of just sigmas, like sigma Compton Now we’re actually telling you what’s the probability of that interaction happening in this certain angle? So it’s called the differential cross-section, and you can have all sorts of differential cross-sections, like energy differential cross-sections, angle differential cross-sections, whatever have you So if we try and graph what dpes the shape of this look like, this polar plot on a more understandable graph, we can see that the probability is pretty high Let’s just call that a relative probability of 1 And as we trace around this circle, that value gets lower and lower until we hit 90 degrees, or pi over 2, at which point it starts to pop back up almost to its original value So this was for a 10 keV photon Now let’s take a different extreme example We have a 3 MeV photon right here And it’s that long dashed curve, so that one here in the center So we can see that the relative probability of 0 degrees is the same And if we trace around to 180, we’re almost at the origin, which tells us that, for 3 MeV, it starts off the same and quickly drops really far down What that means is that what’s called forward scattering is preferred So up on this board, we were talking about what’s the maximum energy that a photon can transfer, which is always in the back scattering case The other part to note is that that back scattering probability gets lower and lower as the photon gets higher and higher in energy Yeah AUDIENCE: So it’s that basically saying, since we said forward scattering [INAUDIBLE] as the energy gets higher, you just have a harder chance of it interacting MICHAEL SHORT: Exactly Yeah The cross-section value, well, yeah, it goes down as you increase in energy So a total forward scatter, if you had a true forward scatter where theta equals 0, I’ll call that a miss It means that, because we saw on this formula up here when theta equals 0, there is no energy transfer Nothing And so yeah, that would be to me like a miss If that angles ever so slightly above 0, then there is some scattering, but there’s very little energy transfer But that smaller energy transfer becomes more likely when the photon goes higher in energy These are these sorts of cause and effect relationships I want you guys to be able to reason out

If I were to give you a polar plot of this differential cross-section with angle and energy, I’d want you to be able to reproduce this and tell me what’s really going on If we fill in one of the ones in between– let’s go with 0.2 MeV, this sort of single dashed line– you can see that the probability of back scattering is somewhere between the 3 MeV and the 10 keV Yeah, Luke? AUDIENCE: Back scattering refers to the photon going back MICHAEL SHORT: That’s right Back scattering refers to the photon going back, which means this situation, where the difference in angle between the incoming and outgoing photon is pi, 180 degrees, which means it just turns around and moves the other way Right So that dashed line right there would follow something like this at– what did we say? 0.2 MeV The form, the full form of this cross-section right here is referred to as the Klein-Nishina cross-section And it has been derived by quantum electro dynamics, which I will not derive for you now But there are plenty of derivations online if that’s your kind of thing And I had to make a trade-off in this class of how deep do we go into each concept versus how many concepts do we teach? And I’m going for the latter because if there is any course that’s supposed to give you an overview of nuclear, it’s 22.01 There will be plenty of time for quantum in 22.02 and beyond, should you want At any rate, this is the general form of it And what this actually tells you is that as the energy of the photon increases, the effect of that angle will– you guys’ll have to work that out on a homework problem I just remembered I want to stop stealing your thunder and giving away half the homework Yep AUDIENCE: How does [INAUDIBLE]? MICHAEL SHORT: Yes AUDIENCE: –the quantity D sigma D omega MICHAEL SHORT: The quantity D sigma D omega says let’s say you have a photon coming in at our x-axis You’ve got an electron here What’s the probability that I’m going to scatter off into some small area d omega? So in some small d theta d phi, or some small sine theta, d theta d phi, into an element of solid angle I should probably draw that smaller to be a little more differential looking So gammas are going to scatter off in all directions But this d sigma d omega tells you what’s the probability that it goes through that little patch? AUDIENCE: And then that omega is also a function of [INAUDIBLE]?? MICHAEL SHORT: So that omega has some component of theta in it and some component of phi in it Since it’s a solid angle, it depends on both the angle of rotation and the angle of inclination, which we call theta and phi Now to get from this to the regular cross-section you’re used to, you can integrate over all angles omega of the differential cross-section, and you’ll get the regular total cross-section, which is just what is the probability of Compton scattering, full stop If you wanted to know, then, what’s the probability of Compton scattering into this angle, it sounds kind of boring, right? Why do we care about the angle? Anyone bored yet? You can raise your hands Be honest Interesting OK Well, I’m going to tell you why it’s not boring because I don’t think you’re honest You can actually, if you know the angle at which a Compton photon scatters into– actually I want to leave that stuff up– there is a pretty much one to one relation between the energy and the angle of scattering, which means that let’s say you have a cargo container I’m sorry You don’t have a cargo container You have a cargo ship full of tons and tons of these stacked up cargo containers Has anyone actually ever seen one of these before? OK In case not, I’m going to do something dangerous and go to the internet And hopefully the search for cargo container doesn’t come up with something disgusting Oh, look at that How about cargo container ship? Yeah OK You got one of these, right? And your detector goes off, and it just says there is something radioactive that shouldn’t be here How do you find out which container it’s in without taking the ship apart? Interesting problem, huh? Do you just kind of look– yeah? AUDIENCE: Do you kind of like shield certain angles [INAUDIBLE]? MICHAEL SHORT: That’s one way