字幕表 動画を再生する 英語字幕をプリント 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: So since I know series decay is a difficult topic to jump into, I wanted to quickly re-go over the derivation today and then specifically go over the case of nuclear activation analysis, which reminds me, did you guys bring in your skin flakes and food pieces? We have time. So if you didn't remember, start thinking about what you want to bring in, what you got. AUDIENCE: Aluminum foil. MICHAEL SHORT: OK, so you've got aluminum foil. You want to see what in it is not aluminum-- excellent. Well, what else did folks bring in? AUDIENCE: [INAUDIBLE] MICHAEL SHORT: OK, rubber stopper-- sound perfect. Anyone else bring something in? AUDIENCE: [INAUDIBLE] MICHAEL SHORT: OK, so tell you what, when you bring stuff in, bring it in a little plastic baggie. I can supply those if you don't have them with your name on them just so we know whose samples are what because that's going to be the basis for another one of your homeworks where are you going to use the stuff that we're learning today to determine which impurities and how much are in whatever thing that you looked at. And, of course, you're not going to get all the impurities because in order to do that, we'd have to do a long nuclear activation analysis, irradiate for days, and count for a longer time. So you'll just be responsible for the isotopes on the shortlist, which we've posted on the learning module site. So again, bring in your whatever, as long as it's not hair because, apparently, that's a pain to deal with or salty because the sodium activates like crazy or fissionable, which you shouldn't have, anyway. I hope none of you have fissionable material at home. So let's get back into series decay. We very quickly went over the definition of activity which is just the decay constant times the amount of stuff that there is, the decay constants, and units of 1 over second. The amount of stuff-- let's call it a number density-- could be like an atoms per centimeter cubed, for example. So the activity would give you the amount of, let's say decays, per centimeter cubed per second. If you wanted to do this for an absolute amount of a substance, like you knew how much of the substance there was, you just ditched the volume. And you end up with the activity in decays per second. That unit is better known as becquerels or BQ, named after Henri Becquerel, though I don't know if I'm saying that right. But my wife's probably going to yell at me when she sees this video. But so becquerel is simple. It's simply 1 decay per second, and there's another unit called the curie, which is just a whole lot more decays per second. It's a more manageable unit of the case because becquerel-- the activity of many things in becquerels tends to be in the millions or billions or trillions or much, much more for something that's really radioactive. And it gets annoying writing all the zeros or all the scientific notation. And so last time we looked at a simple situation-- let's say you have some isotope N1 which decays with the k constant lambda 1 to isotope N2, which decays with the k constant lambda 2 and N3. And we decided to set up our equations in the form of change. Everyone is just a change equals a production minus a destruction for all cases. So let's forget the activation part. For now, we're just going to assume that we have some amount of isotope, N1. We'll say we have N1 0 at t equals 0. And it decays to N2 and decays N3. So what are the differential equations describing the rate of change of each of these isotopes? So how about N1? Is there any method of production of isotope N1 in this scenario? No, we just started off with some N1, but we do have destruction of N1 via radioactive decay. And so the amount of changes is going to be equal to negative the activity. So for every decay of N1, we lose an N1 atom. So we just put minus lambda 1 N1. For every N1 atom that decays, it produces an N2. So N2 has an equal but different sign production term and has a similar looking destruction term. Meanwhile, since N2 becomes N3, then we just have this simple term right there, and these are the differential equations which we want to solve. We knew from last time that the solution to this equation is pretty simple. I'm not going to re-go through the derivation there since I think that's kind of an easy one. And N3, we know is pretty simple. We used the conservation equation to say that the total amount of all atoms in the system has to be equal to N10 or N10. So we know we have N10 equals and N1 plus N2 plus N3 for all time. So we don't really have to solve for N3 because we can just deal with it later. The last thing that we need to derive is what is the solution to N2. And I want to correct a mistake that I made because I'm going to chalk that up to exhaustion assuming that integrating factor was zero. It's not zero, so I want to show you why it's not now. So how do we go about solving this? What method did we use? We chose the integration factor method because it's a nice clean one. So we rewrite this equation in terms of let's just say N2 prime-- I'm sorry-- plus lambda 2 N2 minus lambda 1 N1. And we don't necessarily want to have an N1 in there because we want to have one variable only. So instead of N1, we can substitute this whole thing in there. So N10 e to the minus lambda 1t equals zero. And let's just draw a little thing around here to help visually separate. We know how to solve this type of differential equation because we can define some integrating factor mu equals e to the minus whatever is in front of the N2. That's not too hard. I'm sorry, just e to the integral, not minus, of lambda 2 dt. We're just equal to just e lambda 2t. And we multiply every term in this equation by mu because we're going to make sure that the stuff here-- after we multiply by mu and mu and nu and mu for completeness-- that stuff in here should be something that looks like the end of a product rule. So if we multiply that through, we get N2 prime e to the lambda 2t plus, let's see, mu times lambda 2 e to the lambda 2t times n2 minus e to the lambda 2t lambda 1 n10 e to the minus lambda 1t equals zero. And indeed, we've got right here what looks like the end result of the product rule where we have something, let's say, one function times the derivative of another plus the derivative of that function times the original other function. So to compact that up, we can call that, let's say, N2 e to the lambda t-- sorry, lambda 2t prime minus-- and I'm going to combine these two exponents right here. So we'll have minus lambda 1 and 10e to the-- let's see, is it lambda 2 minus lambda 1t equals zero. Just going to take this term to the other side of the equals sign, so I'll just do that, integrate both sides. And we get N2 e to the lambda 2t equals-- let's see, that'll be lambda 1 N10 over lambda 2 minus lambda 1 times all that stuff. I'm going to divide each side of the equation by-- I'll use a different color for that intermediate step-- e to the lambda 2t. And that cancels these out.