14.3 Compton Effect

Ok, we're going to continue with what we started before, with the photoelectric effect and with Planck and Einstein's theories. Basically, what has happened so far is that a lot of evidence has been coming out saying that light and EMR are particles. So, if light has particle nature and acts as a particle, as Compton explained, then we can say that if a particle has energy, which it does, we see it as equals HF. Therefore, that thing must have momentum as well. Compton then asked, "Well, how can we calculate the momentum of these particles?" So, if it's a bundle of light, a photon, it should have some momentum. The tough part is proving it because we don't feel the momentum of light. It is so small that we wouldn't be able to feel it, like you would if you were hit by a ball. So, because light and energy have no... Mass, we can't use our old formula peak with em B. So what Compton did was basically modify Planck's formula and came up with P equals H over wavelength. Or, if we were using frequency, Vp equals H times frequency over C. So, how do we prove this? He started shooting high-frequency x-rays, both x-rays on graphite. The idea, or common sense, would be that if this was a wave and it diffracted or refracted, it would slow down as it goes through the graphite. So, if a wave slows down, it should get compressed and have a smaller wavelength, similar to what the picture at the top shows. That's what was expected. However, when Compton actually did it, he found out that the x-rays scattered at a different angle. He found that they actually... Got faster, so the wavelength increased. So, it's kind of weird that what's happening is our... it's kind of the opposite of what we'd expect. So when Compton predicted was, we've got some conservation. What he said is we have some sort of momentum of the energy coming in. I'll just leave it as momentum. So, we have this x-ray coming in that has momentum. He says it's gonna lose some of it as it goes through the graphite, and the momentum that it loses is gonna get transferred to the electrons that get ejected from the metal. So, if we look at this, we can calculate the momentum coming in as P equals H over lambda. We can calculate the scattered waves as H over lambda, and they're going to be different because it loses some of the energy. And then the momentum of the Electron. Though we still have to use our old formula, we use F to use P equals MB because the electron has mass. We would use our old momentum formula. So, Cough did this, and he actually calculated two dimensions and all these different things. He found out that it was basically a perfect hundred percent conservation of momentum. If you remember back when we started in the first unit, momentum quite often wasn't conserved because of the extra losses of energy, heat, and things like that. But, in this case, at this microscopic level, Compton found out that it worked almost perfectly. This is sort of the main turning point to say that this whole particle theory actually has to be true. If our particles are acting as, or if our life is acting as particles in that momentum, and we're getting conservation of momentum, then that has to be true. So, from this point on, this whole wave-like [Music] theory was starting to be abandoned and particle theory was taking over. Particle duality theory sort of took over one other quick calculation or one other formula. Compton figured out that because we're always getting this light coming in, it's gonna hit the metal and scatter off at a different angle. Our electron is gonna get shot off at a different angle. He said we can basically always figure out what this angle is gonna be, kind of like a refraction type of topic. So, we'd have this particle coming in, hitting the metal and refracting at an angle. He said we can calculate, or we could calculate, the difference in the wavelength using H / MC times 1 minus the cosine of the angle. Do you think to remember that? The angle is always going to be the angle to the normal, not the angle between the total. So, because of that, we can now solve all kinds of questions using what we learned in unit 1: the conservation of momentum, but now using these other formulas. The first thing we want to figure out is, we have... Our particle is coming in, so we want to figure out its momentum. It's H over lambda. So, it says we have an X-ray with a wavelength of 1.2. So, we'd have 6.63 x 10^(-34) divided by 1.28 x 10^(-12). If we do that calculation, we get 5.18 x 10^(-22) kilograms meters per second. Okay, so now what we want to do is figure out the scattered momentum. We don't know anything other than we know the angle is going to be 15 degrees. So, we're going to have to go back and use our new formula. Our new wavelength will be H over MC times 1 cosine of our angle. To figure out the new wavelength, you input the values: 6.63 (mass of the electron), speed of light, and 1 minus the cosine of 15 degrees. It's important to be careful because this calculation determines the change in wavelength. When plugging in these values, we obtain 4.27 x 10^-12. This means that our incident wavelength, which was 1.28, has gained 4.27. Therefore, when we add that to the initial wavelength, we would get 5.55 x 10. To the negative twelve meters. So that's our scattered wavelength of our outgoing ray. So we have our five point one eight going in. Remember, this one's going to scatter at fifteen degrees. Now we can calculate that momentum the same way we go each divided by our answer. So that gives us our 6.63. We're going to divide this by five points by five. So to do that on our calculus, we get a momentum of 1.19 to the negative 22.\n\nOkay, so then we also know that the electron is going to go off at some sort of angle. That's what the question is asking, what is the velocity of an electron? So the first part, what is the wavelength of the scattered x-ray? We've got that here, so now the last part is we need to figure out the velocity of that electron. So let's kind of redraw this: we have our incoming momentum, which was 5.18x10^-22, and we've got our angle, which is 15 degrees, and it was 1.19x10^22. And we know our electron is going to go off something like that. So, just like we did in the first unit, now what we want to do is we have to break this into vectors. So we need to figure out the horizontal and vertical components of the scattered photon. And we know them. That's going to be the horizontal and vertical of our electron. And those momentums have to combine to what we started with, so let's do that first. We'd figure out our 1.19 and multiply it by the sine of 15. I think I've got my calculator in radians, which actually screws up this calculation. So I'm not going to continue because when I did the cosine of 15 in this step, I was in radians. This whole answer part here is going to be wrong. So you get the idea. I'm just going to leave it at this so we get the idea of s. Then, you kind of go back to this diagram. If we assume these numbers were right, the first thing we'd want to do is figure out our X and Y vectors of that one. Then we know the X of that one plus the X of the electron would have to be the same as what we started with. And then our two Y's would have to be equal to each other to cancel out. All we'd have to do is figure out the hypotenuse and angle. That would give us the momentum of the electron. Once we have the momentum, we can calculate the velocity by simply moving my mouse.