字幕列表 影片播放 列印英文字幕 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. PROFESSOR: OK, guys. We're actually slightly ahead of where I thought we'd be at this point, so I'm only going to spend about half of today's lecture finishing up some new material on mass parabolas and stability. I also got a comment in through the anonymous box that said please leave a little bit of time after class for questions. So you can get them out right away, because I'm usually running off to teach some other class at, like, the IDC or some other building. So from now on, I'll try and leave about five minutes at the end of class for questions on today's material, and we'll make up with the second half-hour or 25 minutes of this class being for all the questions on the material so far in the first two weeks. But first I wanted to give a quick review of where we were Wednesday and launch back into mass parabolas, which are ways of looking at nuclear stability in relative numbers and even or oddness of nuclei. So you saw last time we intuitively derived the semi-empirical mass formula as a sum of volume, surface, coulomb, asymmetry, and pairing, or whether things are even-even or odd-even terms with the coefficients in MeV gleaned from data, and the forms of the-- and the exponents right here gleaned from intuition. Here we assume that the nucleus can be thought of like a big drop of liquid with some charged particles in it. And so the droplet should become more stable the more nuclei there are-- or the more nucleons there are. But then you have some more outside on the surface that aren't bonded to the others. All of the protons are repelling each other over linear length scales, because the radius of this liquid drop would scale like A to the 1/3. There's an asymmetry term, which means if the neutrons and protons are out of balance, there's going to be some less binding energy. And then there's this extra part that tells you whether the nuclei are even-even or odd-odd. And this works pretty well. If you remember, we looked at theory versus experiment where all the red points here are theoretical predictions, and all the black points are experimental predictions. And for the most part, they look spot-on. It generates the classic binding energy per nucleon curve that you see in the textbook and can predict from the semi-empirical mass formula. Zooming in and correcting for, let's say, just getting absolute values of errors, you can see that, except for the very small nuclei and a few peaks, which we explained by looking even closer, the formula, well, it predicts nuclear stability quite well on average. Again, this line right here, if there's a dot that lies on this blue line, it means that theory and experiment agree. And a deviation by a few MeV here and there, not too bad. But we started also looking at different nuclear stability trends, and we noticed that for odd mass number nuclei, there's usually only one or sometimes none stable isotopes per Z, whereas for even ones, there's quite a few more. And we're going to be now linking up the stability of nuclei versus what mode of decay they will take in order to find a more stable configuration. We looked quickly at the number of stable nuclei with even and odd Z and noted that these places right here where there are no stable nuclei correspond to technetium and promethium. There's no periodic table on the back of this wall, but behind my back on the other wall, there's that periodic table where you can see the two elements that are fairly light with no stable isotopes. That's what those correspond to. And the peaks correspond to what we call magic numbers or numbers of protons or neutrons where all available states at some energy are pretty much filled. And this goes for both protons and neutrons. So something with a magic number for both n, number of neutrons, and Z, number of protons, is going to be exceptionally stable. And we'll see how that's used as a tool to synthesize the super heavy elements that we believe should exist. And finally we got into these mass parabolas. I found this to be a particularly difficult concept to just get mathematically. If you remember, we wrote out the semi-empirical mass formula and said if you take the derivative with respect to Z, as we did it, you would get the most stable z for a given A. And we started graphing for a equals 93 where niobium is stable. That was just the one I had on the brain from some failures in lab earlier this week. We started plotting where those nuclei-- what is it-- the relative masses are for a fixed A. So let's regenerate that one right now, because we were a little fast at the end of last lecture. Then I want to generate one for a equals 40. And you'll see something kind of curious. I'm going to leave this up here for a sec. If you notice, for odd A nuclei, there's only one parabola, whereas for even A, there are two. Why that is, we're going to see when we look at the table of nuclides. But notice this nucleus right here can decay by either positron emission or beta emission to get to a more stable form. And there are many real examples, and I'm going to show you how to find them. So let's start off by going back to the table of nuclides, finding niobium-93. Just go up one more chunk. And there we are. Niobium-93 is a stable isotope. And if you want to see where it came from, you can scroll down a little bit and see its possible parent nuclides right here. So let's say that niobium-- we'll draw it right there-- is stable. We'll put it at the bottom of this parabola. And let's work down in Z. So we'll move to zirconium. Zirconium-93 ostensibly has a very similar atomic mass. But if you remember that 93 AMU is a rather poor approximation for the actual mass of all nuclei with A equals 93. In fact, if you look very closely at the atomic masses, zirconium-93 is 92.90. Niobium-93, well, it looks like we have to go all the way to another digit there. 92.906. You have to go to, like, even more digits 92.906375, 92.906475. So we go down in Z, and we've actually gone up in mass by looks like the sixth or seventh digit in AMU. If we go up in mass, we go down in binding energy. That tells us that there's something that's less stable. And if you notice, we went down by a very small-- or we went up by a very small amount of mass. Notice also that its beta decay energy is really, really small, 91 kiloelectron volts. So why don't we put zirconium just above? And we note that that decay will happen by beta, where the beta, let's say if we have an isotope with mass number A, protons Z, and let's just call it symbol question mark. In beta decay, we have the same A. We'll have a different Z. We're going to have to give these symbols. Let's call this parent, and we'll call that daughter. Plus a beta, plus an electron antineutrino. And what has to happen to that Z in order for everything to be conserved? It's the same reaction that we've got here for-- I'm sorry-- for zirconium. So you'd have to have one fewer proton to release one electron. And so that becomes same A but different Z. And this is the beta decay reaction. Let's go back a little farther. We'll look at the possible parent nuclide for-- did anyone have a question? AUDIENCE: Don't you need one more proton [INAUDIBLE]?? PROFESSOR: Let's see. Which direction are we going in Z here? That goes to yttrium. It actually looks like it's going down. Oh, yeah, for beta decay. I'm thinking-- I have the reaction backwards. Sorry. I need one more proton to account for the extra negative charge. You're right. OK. Yep, I was thinking backwards, because we're now climbing up the decay chain in reverse order. So this could have come from yttrium-93 with a much higher energy of three MeV. So let's put yttrium right here. That gives a beta decay. And we'll just go one more back to strontium-93 Has an even higher beta decay energy. So let's put that up here. And let's take a look at its mass real quick. The mass of strontium-93, 92.914 AMU. If we go back to niobium-93, now it's noticeably different to, like, four significant digits instead of six. 92.914 versus 92.906. And so that shows you that a tiny bit of mass and AMU corresponds to a pretty significant change in binding energy by that same conversion factor that we've been using everywhere. 931.49 AMU per big MeV per C squared. Let's see. Yeah. OK. So let's go now in the other direction, in the positron direction. Niobium can also be made by electron capture from molybdenum. So let's put molybdenum right here. Let's say that around half an-- what did we have here? It was like half an MeV. Like that. And let's see. Molybdenum-93 could have been made by electron capture from technetium-93 with an energy of 3.201 MeV, even more extreme. We'll go back one more, because there's a trend that I want you guys to be able to see. And this could have come from electron capture from ruthenium. I think I may have said rubidium last time, but Ru is ruthenium-93. And that 6.3 MeV, something like that. And this is where we got to yesterday. Now I'd like us to take a closer look at the decay diagrams, which tells us what possible decay reactions can happen in each of these reactions. Since we're right here on the chart, let's take a look at ruthenium turning into technetium by what it says, electron capture. So note that on the table, you can click on electron capture, and if it's highlighted, then the decay diagrams are known. It's not known for every isotope, but for a lot of the ones you'll be dealing with, it is. And you get something I have to zoom out for-- a lot, a lot of different decays. What I want you to look at is this one here on the bottom that I'll zoom in to. That should be a little more visible. So notice that if you want to go down the entire 6.4-something MeV, it usually proceeds by B-plus or positron decay, by either method. And as you go up the chain, as these energy differences get smaller, look what happens to the probability of getting positron decay. It shrinks lower and lower and lower. So there's a trend that the larger the decay energy for this type of reaction, the more likely you're going to get positron decay.