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  • 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.

  • That cancels these out.

  • And I forgot that integrating factor again, didn't I?

  • Yeah, so there's going to be a plus C somewhere here.

  • And we're just going to absorb this e to the lambda 2t

  • into this integrating constant because it's

  • an integrating constant.

  • We haven't defined it yet.

  • Did someone have a question I thought I saw?

  • OK, and so now this is where I went wrong last time

  • because I think I was exhausted and commuted in from Columbus.

  • I just assumed right away that C equals zero,

  • but it's not the case.

  • So if we plug-in the condition at t equals 0--

  • and two should equal 0--

  • let's see what we get.

  • That would become a zero.

  • That t would be a zero, which means that we just

  • end up with the equation lambda 1 N10

  • over lambda 2 minus lambda 1 plus c equals

  • zero so obviously the integration constant

  • is not like we thought it was.

  • So then C equals negative that stuff.

  • That make more sense.

  • So you guys see why the integrating constants not zero.

  • So in the end--

  • I'm going t5o skip ahead a little of the math because I

  • want to get into nuclear activation analysis--

  • we end up with N2 should equal, let's see,

  • lambda 1 N10 over lambda 1 minus lambda 1 times--

  • did I write that twice?

  • I think I did--

  • times e to the--

  • let's see, e to the minus lambda 1t minus e

  • to the minus lambda 2t.

  • And so since we know N1, we've found N10.

  • We know N3 from this conservation equation.

  • We've now fully determined what is

  • the concentration of every isotope

  • in this system for all time.

  • And because the solution to this is not that intuitive--

  • like I can't picture what the function looks like in my head.

  • I don't know about you guys.

  • Anyone?

  • No?

  • OK, I can't either.

  • I coded them up in this handy graphing calculator where

  • you can play around with the eye of the concentration N10, which

  • is just a multiplier for everything,

  • and the relative half lives lambda 1 and lambda 2.

  • And I'll share this with you guys,

  • so you can actually see generally how this works.

  • So let's start looking at a couple of cases--

  • move this a little over so we can see the axes.

  • Let's say don't worry about anything before T equals zero.

  • That's kind of an invalid part of the solution.

  • So I'll just shrink us over there.

  • And so I've coded up all three of these equations.

  • There is the solution to N1 highlighted right there.

  • That's as you'd expect simple exponential decay.

  • All N1 knows is that it's decaying

  • according to its own half life or exponential decay equation.

  • And two, here in the blue, which expands, of course,

  • looks a little more complicated.

  • So what we notice here is that N2 is tied directly

  • to the slope of N1.

  • That should follow pretty intuitively

  • from the differential equations because if you

  • look at the slope of N2, well, it depends directly

  • on the value of N1.

  • For very, very short times, this is

  • the sort of limiting behavior in the and the graphical guidance.

  • I want to give you two solve questions like,

  • what's on the exam or how to do a nuclear activation analysis.

  • Is everyone comfortable with me hiding this board right here?

  • OK.

  • So let's say at time is approximately zero,

  • we know that there's going to be N1 is going to equal about N10.

  • What's the value of N2 going to be very, very

  • close to 2 equals 0?

  • AUDIENCE: Zero.