Physics 30 - Compton Effect

So today we're finishing off the EMR unit with the Compton effect. I'll explain what this is in just a moment. So the Compton effect of course the main focus today we're going to talk about the momentum of photons. We're going to talk about the energy of photons as well as particles. Then we're going to talk about something called Compton scattering. Now a huge part of today is going to be about the example questions that we're going to go through. There's going to be a lot of work that's involved with them so it's going to be kind of like a back-end kind of heavy day today. But anyway let's get going. So photons and momentum. Because photons are said to be particles they should have momentum right. We've talked about photons being both a wave and as a particle that's how they behave. But if they are indeed a particle they should logically have some momentum to them as the other particle would. In 1923 Arthur Compton proved that photons do indeed possess momentum. X-rays were directed at electrons at rest. So we had stationary electrons in a carbon target and after colliding with the target both x-rays and electrons were scattered. The scattered x-rays had a longer wavelength than the instant x-rays. So in other words they gained some wavelength or in other words they lost some energy. Meaning some of the initial energy and momentum had been transferred to the electron. Okay so in other words he was able to determine that photons, light beams basically, did indeed have the ability to transfer energy from one thing to another. So here's a very blurry diagram showing this. There's the instant photon coming in here and again it behaves as both a particle and a wave. But it's coming towards the target electron at rest. As soon as it strikes the electron your photon, the like the photon that came in, the light particle that came in, gets scattered off at an angle. Just like as if this was like a billiard ball hitting something right. Gets scattered off in another direction and it now has a longer wavelength which implies that it actually lost some energy. Meanwhile the electron starts going off in another direction and of course that would mean that it had gained energy because it had no kinetic energy initially and it gained some kinetic energy afterwards. Now of course these two things go off at different angles. This one goes off at an angle theta, this one goes off at an angle phi. We'll talk about that in a second as well. But the bottom line is there's actually a formula that does exist on your formula sheet. Change in your wavelength, so delta lambda, change in your wavelength, is h divided by mass times the speed of light. Mass of course referring to the mass of the electron. Times by one minus cosine theta. Theta being the angle that the photon was scattered at. Okay that is on your formula sheet just so you know. Next up, photon collision. In collisions of particles both energy and momentum must be conserved. I've talked about that a million times before. Of course energy and momentum don't just come out of nowhere, they have to be conserved. The total energy and momentum of the incident photon to the x-rays must be equal to the total energy and momentum of the scattered photon and the recoil electron. Compton proved this is what happened in the observed collisions. So with Compton scattering, that's what we call this, summarize the changes in the characteristics of the photon and electron after a Compton scattering collision. Assuming it's a photon striking an electron at rest. Well for the photon the wavelength increases, the energy decreases, and the photon deflects at that angle theta. Okay that's important to recognize because that's one that's used in in a formula of course. As for the electron the energy increases, the velocity also would increase of course because the energy we're referring to here is kinetic, and the electron deflects at an angle phi. Okay it's not the same thing as theta, it's just a different angle altogether. It's another letter, another Greek letter that's used to represent an angle. So with photon momentum, because photons have no rest mass, so they don't they don't exactly have a mass on their own, their momentum must be derived using Einstein's famous formula. So E equals mc squared is what we have to use to derive the momentum. Imagine E equals mc squared being broken down into E equals mcc, and if you think of E equals mcc, you can bundle the m and the c together to make E equals pc, because remember p is mass times the velocity, and c of course is just the speed of light. So that's where that's coming from, it's just becoming your energy equals your momentum times the speed of light. So if the energy of a photon is known, we can find the photon's momentum just divided by the speed of light, right? But what if we don't know the energy? Like what if we want to find the the momentum or how much momentum is existing within a photon, but we don't know even how much energy is existing in that photon? Well there's a bunch of different formulas we can use for this, and it all depends on whether or not you have something that has mass versus something that doesn't have mass, like a photon. So this chart of course, it'd be really a good idea to jot this one down into your notes. If you have the printed notebook that's in there, but this is a really useful chart for sure. It breaks it down from momentum and energy. Particles with mass, so things like electrons, protons, and alpha particles, your momentum is the same old, same old. It's p equals m times c. Now if it's not specified to be moving at a speed of light, of course you're just going to use the given speed instead, because momentum is a mass times a speed. It's not a big surprise. As for energy, if not specified to be the speed of light, use the given speed as well. It's e equals mc squared. That's the energy of a particle at mass times the speed of light. Now for relativistic particles, for instance photons, we would actually have to kind of rearrange this a little bit differently. You could say that p equals h over your wavelength, right? So again, this is if we didn't actually know what our energy was. You could say it's h divided by your wavelength. I think actually that is on your, yeah it is, it is on your formula sheet. I just have to take another look. This is indeed on your formula sheet. Just know that of course, because it's dealing with a wavelength, you should be able to tell that it's a relativistic particle, so something like a photon instead. And as for energy, of course if you're looking for energy, that's the formula we just talked about, e equals p times c. Of course if you wanted energy and you had nothing in the first place other than like the wavelength of the photon you had, well then you have to start here and then jump to here, right? Bottom line. So Compton scattering formula, this is the one that I kind of alluded to in that really blurry diagram earlier. Compton also derived a formula which gave the relation between scattered photons change in wavelength and its scattered angle. So the change in wavelength is equal to h divided by mc times by 1 minus cosine theta. And just note that m is referring to the mass of an electron, it's the mass of the scattered electron of course, and that is on your formula sheet. And also notice that delta lambda is your change in wavelength. Because you're gaining wavelength here, it's going to have a higher wavelength afterwards. You can say it's your final wavelength minus your initial wavelength, okay? Hopefully that isn't too big of an issue for you, but just so you know. All right, so as promised we've got an absolute ton of examples we have to fly through. That was only like a seven minute lesson, otherwise there's going to be a lot of work that comes with these. Calculate the momentum of a 6.0 MeV proton. Well MeV, that means mega electron volts, okay? Mega electron volts. So if we wanted to say this another way we'd say it's 6.0 times 10 to the 6, because mega is times 10 to the 6. That is on your formula sheet as well if you're ever wondering about your metric conversions. Electron volts. Now that of course is referring to the amount of energy in that proton. Electron, you know, electron volts aren't exactly the greatest unit. I'm going to change this to joules because I'm a lot more comfortable with it. So I'm going to times that by 1.60 times 10 to the negative 19, and that's going to give me 9.6 times 10 to the negative 13 joules. So that's how much energy is stored in this proton. Now we can't actually assume that this proton is moving at the speed of light, right? It doesn't actually tell us this. So if we're looking for momentum that's going to have to come out of our speed and stuff. So in other words we're going to need to find our kinetic energy from this and find the speed from there. So kinetic energy is going to be the energy that they gave to us, because that's the only thing that makes any sense with a proton, whether energy would be. You could say it's potential, but if it has momentum why would it be potential? It's got to be kinetic energy. Kinetic energy is one half mv squared. So your kinetic energy of course is 9.6 times 10 to the negative 13. This is equal to one half times m. m in this case is the mass of a proton that is on your formula sheet. It's 1.67 times 10 to the negative 27 kilograms times v squared. Just rearrange this to find v squared. So divide by one half, divide by this thing, and then square root it. You're going to find that v is therefore equal to, I'm going to keep all of my decimals here, this is going to seem silly, but just bear with me, 33907235.2 meters per second. In other words, if we were to break that down, that's like 33,907,235 meters per second. That's your given speed, because it told us how much energy was in this. That's going to be our given speed. The reason we're all doing this of course is since we're looking for a momentum, we know momentum is mass times your velocity, right? We can just assume it was the speed of light here, so I can just say it was p equals mc. We have to find that velocity, which we did, and we can plug in the rest. So to find this momentum now, we can say it's p is equal to your mass. That's 1.67 times 10 to the negative 27 times by that velocity, which is 33,907,235.2. Throwing that in your calculator, you're going to get to two sig gigs, because that's all the question had. 5.7 times 10 to the negative 20. Now it's been a long time since we talked about momentum. The units on momentum are just mass times meters per second. So in other words, kilograms times meters per second. Momentum was something that really had really bizarre units. That was something we haven't talked about since like the beginning of Physics 30 here, but still good to know. 5.7 times 10 to the negative 20. Next one, calculate the momentum of 6.0 mega electron volt photon. So kind of similar to the last one here, but a little bit different, of course. Photons, of course, are light packets. We can assume photons move at the speed of light, so we don't have to worry about finding a speed in this case, because since it's a photon, we can make the very valid assumption that a light packet moves at the speed of light. You sure would hope so. So the momentum of the photon can actually come from the formula E equals P C, right? We're looking for momentum, that's P, we know energy, and we know C because it's the speed of light. So if we were to rearrange this, we could say P is equal to E over C. So in other words, P is equal to, let's change, yeah, let's change this into joules. We did that on the last slide. Again, just in case you're wondering, it's 6.0 times 10 to the 6 electron volts, times that by 1.60 times 10 to the negative 19, and just the same as before, it's 9.6 times 10 to the negative 13 joules. It's just better to deal with joules when you're dealing with everything else on this one as well, right? So 9.6 times 10 to the negative 13, divided by the speed of light, so divided by 3.0 times 10 to the 8, and this is going to give you P is equal to, just two sig digs, 3.2 times 10 to the negative 21, and again, the units on that are the same as before, kilograms times meters per second. That, compared to the previous answer, is actually not, I was going to say it was a lot smaller than the other one, but really it isn't actually all that much smaller. It is certainly smaller than the momentum of a proton, but comparatively speaking, it's actually not as bad as I would have imagined. So kind of interesting, right? So same amount of energy, different amount of momentum, but it has to do with, of course, they're moving at different speeds, so that would do it. All right, so the next one, an x-ray photon of wavelength 0.0500 nanometers scatters at an angle of 30 degrees during a Compton, Compton, Compton, geez Louise, scattering experiment. Calculate the wavelength of the scattered photon. Okay, so this one is that classic Compton scattering formula, the really beefy one. It is on your formula sheet, so this is delta lambda, the change in your wavelength, is equal to h over mc times by 1 minus cosine theta. Remember that theta is actually the angle at which your photon is scattering at, and that actually gave it to us in this question. So that's your theta right there. We know our initial wavelength, so that's your lambda i, I guess you could say, is 0.0500 nanometers. You probably want to turn that to regular meters here. You don't have to do it just all in one go. 0.0500 times 10 to the negative 9, because nano means times 10 to the negative 9, so that'll probably be useful to us at some point. Now, the other things we need in order to get this formula to actually work, because again, we're looking for the wavelength of the scattered photon, so in other words, we're looking for lambda f, the final wavelength. We need to make sure we have h, which we do, m is the mass of an electron, we have that, c is the speed of light. Wow, okay, actually we're pretty good here. See, notice how every time even I use that formula, I'm like, oh, there's like a lot to it. Really, like almost all of it's given to you. Like pretty much the whole darn thing is given to you. Matter of fact, like this whole chunk right here, you could literally just express that as some sort of a constant if you wanted to. I'm not going to, but I mean you could, there'd be nothing stopping you. Anyway, delta lambda is equal to h. Now remember, h has two different values. There's one in electron volts, and there's one with joules. Generally speaking, I think with this, it'd be a better idea to use one with joules because we have nothing else that refers to electron volts here. So I'm going to use the one that's in joules, 6.63 times 10 to the negative 34 over mass. Well, that's mass of an electron, so 9.11 times 10 to the negative 31 times c, which is 3.0 times 10 to the 8. And that whole thing, this whole h over mc, I'm going to put it in square brackets here just so we know we've got to do that thing first. We need to times that by 1 minus cosine of our angle, so 1 minus cosine of 30. Just make sure that since that's 30 degrees that you're in degree mode in your calculator, especially those of you who are in not 30 right now. You don't want to be in radian mode or whatever when you're going through this. Throw that in your calculator, although we're going to have delta lambda is equal to 3.25 times 10 to the negative 13 meters. What this represents is this is how much your wavelength is going to be changing by. So in other words, delta lambda is equal to your final lambda, your final wavelength minus your initial wavelength. Well, we're looking for our final wavelength, so logically speaking, we should just say that this is delta lambda plus lambda initial is equal to lambda final. So throw all this together, we're going to say that that's 3.25 times 10 to the negative 13 plus your initial lambda. Boom, it's a good thing I did that in the first place here. So 0.0500 times 10 to the negative 9. That's going to equal your lambda final. And what I get from my final lambda is I'm going to get lambda final is equal to 5.03 times 10 to the negative 11 meters. That right there would be my final answer. Okay, so in other words, it did change the wavelength just, well, not even just a little bit. It actually changed it quite a fair bit, right? It's a much smaller wavelength now. And there you go. So awesome. All right, moving on. What is the maximum change in wavelength that a 0.010 nanometer x-ray photon can experience by quantum scattering? Okay, so I guess the maximum change would happen if the photon reflected straight back. So it was like if it's like completely literally just bounced back from here. So in other words, that's going to mean that our theta is going to equal 180 degrees. And just explaining that again, since we're talking about a maximum change that an x-ray photon can experience, it'd be basically like if a photon was coming to that electron and then literally just like bounce straight back, right? So in other words, that full angle was a full 180, right? So if we're looking for that maximum change in wavelength, really, we are just looking for delta lambda. So it's going to be delta lambda equals h over mc times 1 minus cosine theta. Theta, of course, like I said, being backscattered, so it's got to be 180 degrees. Let's just throw all that in there all in one go. Delta lambda is equal to 6.63 times 10 to the negative 34 over 9.11 times 10 to the negative 31 times by 3.0 times 10 to the 8. Put that in square brackets, so I know I do that first, times by 1 minus cosine 180. This is going to give me delta lambda is equal to just two sig digs, I guess, right here. Although we didn't even use that, did we? I guess it doesn't even have to be two sig digs. Oh, whatever. We'll just do two sig digs. 4.9 times 10 to the negative 12 meters. So that is the maximum change in wavelength that this thing could experience, the most that it possibly can go. For those of you who are not 30 right now, cosine 180. I don't know if the rest of you probably have no idea why I'm doing this here. Cosine 180 degrees, 180 degrees is over here. Cosine is the x value that means x is negative 1. It's like saying 1 minus negative 1, which is 2. Yeah, that's going to be as big as this is possibly going to get, right? So in other words, since we're timesing this whole thing by 2, boom, you got it, right? If you don't know what that means, you don't know what it means, we're all good. All right, last question here. If a 10 nanometer x-ray scattered by an electron becomes an 11 nanometer x-ray, how much energy does the electron gain in the collision? There's a couple different ways you can go about this. I'll show you the way I do this. Remember that really, we have to realize that even in a collision, your total initial energy is still going to equal your total final energy. So I'm going to say E total I is going to equal your E total F, okay? Now, what does it mean to have total initial energy? What's going on there? We could say it's the initial energy of your photon, so E photon I, plus the initial energy of your electron, so E electron I, and that's going to equal the final for both of those as well, right? So E photon F, plus E electron F, okay? Now, let's see if we can break this down even further. I suppose we could say that the energy gained is really just the energy gained by the electron. The energy gained by the electron is just going to be the final energy of the minus the initial energy of the electron. So if I move this initial energy of the electron over the other side, I can say the energy of the photon initially is equal to the energy of the photon final, plus the energy of the electron final, minus this thing that I just minused over, E electron initial. This chunk right here, your final electron's energy minus your initial electron's energy, that's your gain in electron energy. So that right there, we can just call that electron gain. So why don't we say that it's just E photon I, equals E photon F, plus, you know, E electron gain, okay? Now, before I go any further with that though, let's think about what the energy of a photon is. If we actually want to go back, we might as well go back a couple slides here. Notice the energy of a photon is E equals PC, but we don't know the momentum on this one, so we have to actually just say it's E equals H over lambda C, right? For both of our photons, both before and final. So what I'm getting at here is for E photon I, we could say that that's H C over initial lambda, and this E photon F is H C over final lambda, plus E electron gain, right? Because it's just a change in the energy of the electron. I guess really we now have all the information we need to find this E electron gain. H and C are both constants. We're going to use H of course in joules rather than electron volts. Lambda I is 10 nanometers, so in other words 10 times 10 to the negative 9, and lambda F is 11 times 10 to the negative 9. So just throw those into the formula there and then minus them over. I'm not even going to bother writing down the work here because I think hopefully you can figure out just plug those numbers in here and subtract it over. What I'm going to get is an answer though, and again you can try it in your notes if you wish. E electron gain is therefore equal to 1.8, just two sig digs of course, 1.8 times 10 to the negative 16 joules, okay? And I got that just by plugging in all that information. That was probably the hardest question in here. That was kind of an unreal one, but oh well, whatever. Anyway, we're done. That's it. Content effect worksheet, page 39 to 41. That's it, and you're going to notice that's the end of that whole booklet. We're done. EMR is over. Like we're done it. We only have one unit left. Wow. Anyway, another thing to work on, of course, is your EMR assignment that is due Monday, May 11th. Please hand it in on Google Classroom when you're all done to get it in early. Awesome. If you hand it in Monday evening, whatever. It doesn't really matter. Now that week of Monday, May 11th, it'll actually be Thursday, May 14th. We're actually going to have our unit exam, so be prepared for that. But aside from that, we're done. We're done EMR. Then we only have atomic physics left. Only, only atomic physics, right? But anyway, if you ever need anything, you know where to reach me. And again, best of luck, as always.