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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.

  • MICHAEL SHORT: Zero.

  • So N2 is going to equal 0.

  • But what's the slope of N2 going to be?

  • This is how we can get started solving

  • these graphically without even knowing

  • what the real forms are.

  • So we've already said that at a very short time,

  • N2 is approximately 0.

  • So if that's zero, then that whole term

  • is zero, which means that the slope of N2

  • is approximately lambda 1 N1, just the activity of N1.

  • And hopefully, that follows intuitively

  • because it says for really short times

  • before you get any buildup, the slope of N1

  • determines the value of N2.

  • So if we were to start graphing these--

  • let's just start looking at some limiting behavior-- that's t,

  • and we're going to need some colors for this.

  • Let's stick with the ones on the board.

  • Oh, hey, awesome.

  • Make N1 red, N2 blue, and N3 green.

  • So let's start drawing some limiting behavior.

  • So we know that N1 starts here at N10.

  • And we know it's going to start decaying exponentially.

  • So the slope here is just going to be

  • minus lambda 1 and N1, which is going to be

  • the negative slope of N2--

  • looks pretty similar, doesn't it?

  • So we know N2 for very short times

  • is going to start growing at the same rate that N1 is shrinking.

  • So we already know what sort of direction

  • these curves are starting to go in.

  • How about N3?

  • What's the value of N3 for very short times?

  • Anyone call it out.

  • Well, we've got kind of a solution right here.

  • If we know that N2 is about zero for very short times,

  • what would the value of n' three have to be?

  • Also zero.

  • And what about the slope of N3?

  • Also, about zero.

  • If there's no N2 built up, then there's nothing to create N3.

  • So we know that our end three curve is

  • going to start out pretty flat.

  • Now how do we find some other limiting behavior?

  • Let's now take the case--

  • let's see, I want to rewrite that a little closer here,

  • so we have some room.

  • So that's at t equals about zero.

  • And at t equals infinity, what sort of limiting behavior

  • do you think we'll have?

  • What's the value of N1 going to be at infinite time?

  • Zero-- it will have all decayed away,

  • will equal zero at t equals infinity.

  • How about N2?

  • AUDIENCE: Zero.

  • MICHAEL SHORT: Zero.

  • How about N3?

  • AUDIENCE: [INAUDIBLE]

  • MICHAEL SHORT: N10-- correct.

  • Because of that conservation equation right here.

  • So we know for limiting cases, N1 one is going to be 0.

  • N2 is going to be 0.

  • And N3, it's going to be N0.

  • So we've now filled in all the four corners of the graph

  • just intuitively without solving the differential equations.

  • Now let's start to fill in some middle parts.

  • What other sorts of things can we

  • determine, like, for example, where H2 has a maximum?

  • That shouldn't be too hard.

  • So let's make another separation here.

  • So what if we want to find out when does the N2 dt equal 0?

  • What do we do there?

  • Anyone have an idea?

  • Using the equations we have up here.

  • AUDIENCE: [INAUDIBLE]

  • MICHAEL SHORT: Yeah, well, we can just

  • take this equation right here.

  • We can figure that out in terms of N1.

  • So if D and 2D equals zero, then we

  • know that lambda 1 N1 is going to equal lambda 2 N2.

  • What this says intuitively is that the rate of production

  • of N2 by decaying on 1 equals the rate of destruction of N2

  • by its own decay.

  • So at some point, the N2 is going to have to level off.

  • When that point is depends on the relative differences

  • between those half lives.

  • So we already know if we were to just kind of fill in smoothly

  • what's going to happen, N2 is probably

  • going to follow something roughly looking like this.

  • We already know the solution to N1.

  • I think we can figure that out graphically.

  • It's simply exponential decay.

  • The only trick now is how does N3 shape up?

  • What do you guys think?

  • How would we go about graphically

  • plotting these solutions without solving them?

  • I don't think yet I've given you the full form.

  • It's kind of ugly, and I doubt that if you looked at it

  • you'd be able to tell me exactly what it would do.

  • So this is just the mathematical expression

  • of N10 minus N1 minus N2.

  • So how do we figure out all the stuff about N3?

  • Yeah.

  • AUDIENCE: You could just draw a curve

  • so that you get all three curves and always

  • add it to the same number.

  • MICHAEL SHORT: Yeah.

  • AUDIENCE: [INAUDIBLE]

  • MICHAEL SHORT: Absolutely, that's totally correct.

  • Yeah, if you just take N10 minus N1 and N2,

  • that gives you the value of N3.

  • That's completely correct.

  • So you could do that sort of one point at a time

  • and say, well, maybe around there, maybe around there.

  • It might take a little while, though.

  • So I want to think what's another intuitive way?

  • What would the value or the slope of N3 track--

  • what other variable in the system?

  • Or in other words, how are they directly related?

  • Yeah.

  • AUDIENCE: The slope of n2 are they equal?

  • MICHAEL SHORT: Yeah.

  • The slope of n3 depends directly on the value

  • of n2 and nothing else.

  • So initially, you can see the value of n2 is almost 0,

  • so the slope of n3 is almost 0.

  • As the value of n2 picks up, so should the slope of n3

  • until we reach here.

  • What happens at that point?

  • AUDIENCE: N2 decreases.

  • MICHAEL SHORT: Yep.

  • The rate of production of n2 decreases because the val--

  • I'm sorry-- yeah, the rate of production of n3

  • decreases because the rate of production of n3

  • is just dependent directly on the value of n2.

  • So the maximum slope of n3 has to be

  • right there at which point it has

  • to start leveling off and eventually reaching 0.

  • You're going to see this kind of problem on the homework.

  • You're going to see this kind of problem on the exam.

  • I guarantee you.

  • But it's not going to have this exact form.

  • But what I'll want you to be able to do

  • is follow this example.

  • Let's say I pose you a small set of these first order

  • differential equations.

  • Can you use any method that you want-- intuitive, graphical,

  • mathematical--

  • to predict what the values and slopes of these isotopes

  • are going to be as a function of time?

  • So in order to get nuclear activation analysis right

  • you need to be able to do this.

  • In nuclear activation analysis it's just one twist.

  • I'm going to move this over to add the twist.

  • You're also producing isotope n1 with a reaction rate.

  • By some either isotope and not what you put in the reactor.

  • So if you want to know what your impurities and naught were

  • to undergo what's called nuclear activation analysis,

  • then you can figure out, depending on which one you

  • count, what they could be.

  • So this right here.

  • Let's look at the units of this versus the units of this.

  • First of all, if we're adding them

  • together they'd better be in the same units, right?

  • So we already talked about the units of this decay equation.

  • It's like number of decays per second.

  • So this reaction right here better

  • give us a number of atoms produced per second,

  • or we're kind of messed up in the units.

  • So anyone remember what is the units of--

  • I'll make a little extra piece right here--

  • what's the units of microscopic cross sections or barns?

  • What is that in some sort of SI unit?

  • AUDIENCE: Centimeter squared.

  • MICHAEL SHORT: Yep.

  • It's like a centimeter squared.

  • And what about flux?

  • This one you may not know, but it definitely

  • depends on the number of neutrons

  • or the number of particles that are there.

  • AUDIENCE: Is it barns [INAUDIBLE]

  • MICHAEL SHORT: Almost.

  • So the flux describes how many particles

  • pass through a surface in a given time.

  • So we have how many particles per unit

  • surface, per unit time.

  • Ends up being neutrons per centimeter squared per second.

  • Just like the flux of photons through a space

  • or the flux of any particle through anywhere,

  • it describes how many particles go

  • through a space in a certain time.

  • And then there's the number of particles that are there.

  • If we're going with atoms, it's just atoms.

  • These are all multiplied together.

  • The centimeter squared cancel.

  • And we end up with some sort of a atoms per second produced.

  • We can put in a little hidden unit in the cross section.

  • If there is a reaction going on where in goes a neutron

  • and out goes an atom or something,

  • that should cancel all things out.

  • Let's not get into that now.

  • The whole point is we have the same sort of unit

  • going on here, which is some number of atoms produced

  • per second.

  • Same thing as number of atoms decayed per second.

  • So it's the production-destruction

  • equivalent of each other.

  • So, in that way, we can have a reaction rate that we impose,

  • something artificial, by sticking

  • something in the reactor and controlling its power level.

  • And then follow the decay process

  • which is a natural radioactivity event.

  • And this is one of the simplest governing

  • equations for nuclear activation analysis.

  • Now, one, I might give this to you on an exam

  • and say OK now draw the curves for nuclear activation

  • analysis.

  • And maybe calculate what's the impurity level if you measure

  • this many counts of something.

  • Then you just work backward through the math.

  • But I want to get you guys thinking conceptually

  • right now.

  • What are the real equations for nuclear activation analysis?

  • Let's just do these in terms of n1, n2, n3.

  • That's dt, d, and 3dt.

  • We'll start with the stuff that's

  • up there minus lambda 1 n1.

  • Plus some cross section times the flux times some other atom

  • n0.

  • What other things are we missing?

  • Are there any other methods of production or destruction

  • of isotope n1 that we need to consider?

  • Well, we've got isotope n1 in a reactor.

  • It can decay, or it can absorb one of the neutrons nearby.

  • So how do we write that term-- that destruction term?

  • Yep.

  • AUDIENCE: Flux times the absorption cross section

  • times n0.

  • MICHAEL SHORT: Yeah.

  • So let's say that's the absorption of atom 1 times

  • n what did you say?

  • AUDIENCE: Naught.

  • MICHAEL SHORT: And would it be n0 or would it be n1?

  • If you want to know how quick is n1 being destroyed--

  • AUDIENCE: OK.

  • MICHAEL SHORT: --By absorbing neutrons.

  • So then let's call this absorption of n0.

  • And is it a plus or a minus?

  • If it's a destruction rate.

  • AUDIENCE: [INAUDIBLE]

  • MICHAEL SHORT: It's a minus.

  • Yep.

  • So what's really going on here is

  • you've got some precursor isotope, whatever

  • impurity you want to measure n0, producing n1.

  • And you're looking at n1's decay signature, like its activity,

  • to determine how much was there.

  • But you also have to account for the fact

  • that isotope n1 can be burned in the reactor.

  • So this is like producing.

  • This is decaying.

  • And this is we'll call it being burned.

  • This isn't burned in the sense of creating of fuel--

  • creating energy by burning fuel--

  • but we will refer to this sort of in--

  • colloquially too burning-- because we're then

  • absorbing neutrons by n1 and removing that

  • from the available decay signature.

  • How about n2?

  • How do we modify our equation to account correctly

  • for the production and destruction of n2?

  • And by the way this is not in the book,

  • so I don't expect you to know it off the top of your head.

  • AUDIENCE: It's the same type of thing, flux.

  • MICHAEL SHORT: Yep So let's first take every term

  • that we have up there.

  • We have lambda 1 n1 minus lambda 2 n2.

  • And what else do we have to account for?

  • Yep.

  • AUDIENCE: N2 also being burnt.

  • MICHAEL SHORT: That's right.

  • So n2 is also being burned so we'll

  • have a minus, a flux times the absorption cross section for n2

  • times the amount of n2.

  • How about n3?

  • We'll start with what we had there.

  • Lambda 2 n2.

  • And, just like before, we've got to account

  • for the burning of n3.

  • So then we'll have minus flux times the absorption

  • cross section of 3 times n3.

  • And these equations hold true only for the time

  • that your material is in the reactor.

  • What happens when you take the material out of the reactor?

  • AUDIENCE: You go right back to zero readings.

  • MICHAEL SHORT: You do.

  • Yep.

  • When you come out of the reactor all of the fluxes go to zero.

  • And that's the end of that.

  • Yeah.

  • AUDIENCE: Why don't you account for the production of n2 and n3

  • in the burn rate?

  • MICHAEL SHORT: Ah, so did I necessarily specify--

  • the question was why don't we account for the production

  • of n2 and n3 by the burning.

  • Right?

  • AUDIENCE: Yeah.

  • MICHAEL SHORT: Did we specify that absorbing a neutron

  • is the way to make n2?

  • AUDIENCE: No.

  • MICHAEL SHORT: Oftentimes it's not.

  • So if you burn n1 by absorbing a neutron,

  • then you will make another isotope

  • that has the same proton number and one more neutron.

  • And it may decay by some other crazy way, or it may be stable.

  • Who knows.

  • But by decay-- this could be by beta, positron, alpha,

  • spontaneous fission--

  • not gamma because then you wouldn't

  • have a different isotope.

  • But oftentimes you won't have--

  • the burning process won't produce the same isotopes

  • as the decay.

  • So the situation we looked at on Friday when we said let's

  • escalate things.

  • That was a purely hypothetical situation

  • where isotope n2 could be burned to make n0.

  • I'm not saying it can't happen, but it's not likely.

  • But still we can model it.

  • We can model anything.

  • That just wasn't a realistic situation.

  • This is.

  • This is what you guys are going to have to look at

  • to understand how much impurities there

  • are in each of your materials.

  • So this I would say is the complete description

  • of nuclear activation analysis in the reactor.

  • At which point you then have to account

  • for what happens when you turn the reactor off.

  • So what actually?

  • What physically happens when you turn the reactor off?

  • Yep.

  • Oh Yeah.

  • You've answered a lot.

  • So Chris, yeah.

  • CHRIS: Well, you try to-- you put your control rods all

  • the way in and try to stop as many neutrons

  • as you can to stop the chain reaction.

  • MICHAEL SHORT: Yep.

  • So normally to shut down the reactor

  • you'd put the control rods in and shut down the reactor.

  • Or the easier thing is just pull the rabbit out.

  • Remember those little polyethylene tubes

  • I showed you?

  • This way we can keep the reactor on and remove your samples

  • without changing anything.

  • So it makes the reactor folks--

  • angry would be an understatement--

  • to constantly change the power level of the reactor.

  • Reactors, especially power reactors and research reactors,

  • they're kind of like Mack trucks.

  • If they're moving they want to stay moving,

  • and if they're not moving they don't want to be moving.

  • And it takes an awful lot of effort to change that.

  • It also happens to screw up experiments.

  • If you are irradiating something like I was a couple months ago

  • for 30 days, you want to have a constant flux

  • so that your calculations are easy.

  • You don't want 15 students to come in and turn the knobs

  • all up and down, and then you have

  • to account for that in your data.

  • Which has happened.

  • So you guys are going to be manipulating the reactor

  • power when the experiments are out, and it's at low power.

  • So you're won't be infuriating anyone else on campus

  • like we did last year.

  • So if you didn't account for that, but they still let us in.

  • So they're bad.

  • Whatever.

  • [LAUGHING]

  • Yeah.

  • So after you either shut down the reactor

  • or pull the rabbit out of the reactor then

  • the production and destruction by neutrons is over

  • but the decay keeps going.

  • Which means if you wait too long, like for some

  • of those short isotopes, if you wait more than a day or so,

  • you'll have so little activity left

  • that you won't be able to measure it.

  • So what we're going to be doing is sticking your samples

  • into the reactor for maybe an hour or so,

  • pulling them out, and immediately running them over

  • to the detector, so that we get the most signal per unit time.

  • Because the things are going to be the hottest when they come

  • right out of the reactor, and every second you lose

  • from there you lose signal.

  • Which means you have to account for longer

  • to get the same amount of information

  • with the same certainty.

  • This is a nice segue way to what we'll

  • be talking about Thursday which is statistics, certainty,

  • and precision.

  • How long do you have to count something

  • to be confident within some interval

  • that you've got the correct activity?

  • For background counts-- who here is made a NSC Geiger counter?

  • Hopefully almost all of you.

  • Maybe you guys remember how long you

  • had to count to be 95% sure that your background rate was

  • accurate.

  • It ends up being about 67 minutes or over an hour,

  • and the reason is because the count rate is very low.

  • So I'll do a little flash forward to Thursday

  • since we're talking about it.

  • When you count something with a very low count rate,

  • you have to account for longer to be as confident

  • that your number is correct.

  • So let's say you want to be 95% confident

  • or within plus or minus 2 standard deviations or 2 sigma.

  • You have to count for longer and longer.

  • For something that's really radioactive

  • you can be sure, or 95% sure, that the count

  • rate you measured is accurate for a shorter counting time.

  • So everything in this class seems to come up in trade-offs.

  • Right?

  • You trade off stability for a half-life.

  • You trade off decay constant for half-life.

  • You trade off binding energy for excess mass.

  • You trade off counting time and precision.

  • You trade off exposure and dose. which

  • you're going to get into later.

  • We'll see if anyone wants to use a cell phone

  • or eat irradiated food afterwards.

  • And I do all the time, so that should tell you the answer.

  • So in the last seven minutes or so,

  • I want to walk you through playing around

  • with what happens when you change the values of lambda

  • 1 and lambda 2?

  • So what do they look like when the half-lives are

  • roughly equal and when one is much larger than the other one?

  • So let's set them to be about equal.

  • These are just unitless.

  • So let's set them equal to one.

  • I think the system explodes when we set them exactly equal

  • because that term right there.

  • So let's say that's 1.001.

  • It's about as close as we can get,

  • and let's confirm that we get the same sort of behavior.

  • So isotope n1 just follows exponential decay.

  • There's nothing that changes that.

  • Isotope 2, its slope tracks the value of isotope 1

  • for a little while until you build up enough n2

  • that it starts to decay.

  • You can find when that point is when

  • lambda 1 n1 equals lambda 2 n2.

  • There's one little step that we didn't fill in if you

  • want to find the value of n2.

  • So then you can just rearrange this a little bit,

  • and you'll say n2 would have to equal lambda

  • 1 over lambda 2 times n1.

  • Which is n1 0 e to the minus lambda 1 t.

  • So if you want to find that point right there in time,

  • you can solve this.

  • Then let's look at n3.

  • So n3 when n2 is almost 0, n3 slope is almost 0.

  • It's a little hard to see because--

  • I'll tell you what-- let's make all the half lives longer which

  • kind of expands the graph.

  • Wrong way.

  • Let's make it--

  • Ah, we'll just move that decimal point 0.1.

  • There we go.

  • That's like expanding the graph.

  • Right?

  • So when n2 is almost 0, the slope of n3 is almost 0.

  • And when n2 reaches a maximum, so does the slope of n3.

  • Just like we predicted using our graphical method right here.

  • And then over longer times--

  • let's put the half lives back to the way they were.

  • Over long times n3 trends to n1 0.

  • Don't let that little piece fool you.

  • Again t equals less than 0 is not a valid time for this,

  • so we're not accounting for that.

  • And n3 tracks right to here to the value of n1 0,

  • and n2 and n3 turn to 0.

  • So for this case where you have the half-lives roughly

  • equal to each other, you can expect a pretty big bump in n2.

  • What's going to happen when lambda 1 is extraordinarily

  • big meaning the half-life of n1 is extraordinarily short.

  • What do you guys think will happen?

  • Not mathematically but physically.

  • If n1 just kind of goes ba-boom and instantly decays away.

  • AUDIENCE: There would be a lot of n2--

  • right then.

  • MICHAEL SHORT: Yep.

  • AUDIENCE: [INAUDIBLE]

  • MICHAEL SHORT: Your n1 is just going

  • to turn into n2 right away, and n2

  • is going to take its sweet time decaying to n3.

  • So let's see what that looks like.

  • If lambda 2 is much bigger than lambda 1,

  • let's make the maxima a little different.

  • Change our slider value a bit.

  • So if l2 is big and l1 is small, well,

  • let me change the actual axes to make

  • this a little easier to see.

  • There we go.

  • You can see that much, much more quickly

  • than we have it in this graph right here l1 just decays away

  • right away.

  • L-- I'm sorry-- n1 decays right away.

  • n2 builds up to a much higher relative value

  • because it's produced faster than it's destroyed

  • for short amounts of time.

  • So you can end up with a great spike in n2

  • which slowly decays away to n3.

  • How about the opposite effect?

  • What if l1 lambda 1 is really, really small

  • indicating a very long half-life,

  • and lambda 2 is really, really large

  • indicating a small half-life.

  • Yeah.

  • AUDIENCE: It would basically go from n1 to n3.

  • As soon as it goes to n2, it's going to decay to n3.

  • MICHAEL SHORT: That's right.

  • In this case you've got n2 as soon

  • as its created self-destructs.

  • So let's see what that looks like.

  • So we can just slide l2 to be big, slide l1 to be small,

  • and you can actually graphically see n2 just shrink

  • towards the x-axis.

  • And it's almost like you only have two equations.

  • It's like you just have n1 and n3

  • and n2 basically doesn't exist.

  • Where the slope of--

  • where the slope of n3, except at extremely short times,

  • just tracks the value of n1.

  • And I know in the book they're called

  • secular or transient equilbria.

  • I'm not going to require that you memorize those terms.

  • It's more important to me that I can give you

  • a real physical situation.

  • Say here's these three isotopes, for four isotopes,

  • or six doesn't matter because we can solve these pretty quickly.

  • Tell me what's going to happen based

  • on the relative half-lives as long as they

  • decay in a nice linear chain.

  • I'm not going to give you something where

  • n1 can beget n1 or n4 or n6.

  • Because at that point you can construct the equations,

  • but I don't expect you to be able to graphically solve them.

  • And I may also throw you curveballs

  • like nuclear activation analysis to see well

  • what happens when you turn on or turn off a reactor.

  • I've got an example of that too.

  • We're right at this point here, I guess, t equals 50.

  • Yeah.

  • I've set it up such that you turn off the reactor

  • and n3 is stable right there, but n1 and n2

  • continue to decay.

  • So it's not hard to cad these--

  • to code these sorts of things up.

  • I'll share the links with these equations

  • so you guys can play with them yourselves.

  • Add to them yourselves, and try just getting an intuitive feel

  • for how series radioactive decay happens.

  • So I want to know now that we spent a couple of days on it,

  • would you guys be comfortable setting up

  • sets of differential equations like this?

  • Say yes, no, maybe?

  • I see a lot of up and down shaking heads.

  • That's a promising sign.

  • If not I'm willing to spend a little more time on it

  • on Thursday if folks would like a bit of review.

  • And if you're afraid to tell me, just

  • send me an email anonymous or not.

  • Yeah?

  • AUDIENCE: Do you think maybe Thursday we

  • could do a like a real example?

  • MICHAEL SHORT: Yeah.

  • AUDIENCE: Of a like a series.

  • MICHAEL SHORT: I think so.

  • Yeah, with real example with numbers and everything.

  • AUDIENCE: Yeah.

  • MICHAEL SHORT: Sure.

  • OK.

  • Well, we can make one of those up for Thursday.

  • Cool.

  • And what about the graphical solution

  • method since I don't know whether they teach that

  • in the GIRs, but what I do want to be able to do

  • is look at the limiting cases.

  • In other words, fill in the four corners of the graph.

  • At t equals 0, what are things actually doing?

  • n1 is just decaying at its half-life.

  • There is no n2 yet.

  • So these slopes are equal and opposite.

  • And there's no n3 yet, so that there is no slope of n3.

  • So I would like you guys to try reproducing this.

  • And I will-- again I'll provide pictures of these blackboards

  • so you guys can see, but it would

  • be very helpful for you guys to try to reproduce these graphs

  • as we saw them.

  • Then you can check them here on the graphical calculator,

  • and then play around with the amount of n0, or--

  • I think I just broke it.

  • Let's just call that one.

  • There we go.

  • And play around with sliders or values of n1, n2, or n3.

  • That's an interesting solution.

  • So since it's about four or five of,

  • I want to open it up to any questions you guys may have.

  • Yeah.

  • AUDIENCE: I have a question from the [INAUDIBLE] First

  • time do this.

  • Did you have-- do you know what integrated video because

  • like there are endless possibilities if you

  • just [INAUDIBLE] add up to the right mass number?

  • MICHAEL SHORT: Oh yeah.

  • The questions for a spontaneous fission.

  • What fission products do you choose?

  • I'll say they're all good.

  • As long as you pick something with roughly equalish masses,

  • so you don't pick like it fizzes into hydrogen and something

  • quite smaller, which should be better known

  • as just proton emission.

  • You're going to get roughly the same answer.

  • Yep.

  • AUDIENCE: Would that be something we just like set up.

  • Like the top number is 80.

  • We just look at whatever 40, 42, 38 arms pick a number.

  • MICHAEL SHORT: Roll a D 80.

  • You'll get basically the same result. Roll an 80 sided dice.

  • Hopefully at MIT you could find one, or write a program

  • to make random number between about 10 and 80 or 10 and 70.

  • Let's go to the actual problem set to see what you guys mean.

  • I want to make sure I'm answering the correct question.

  • Problem statement Yep.

  • Yep.

  • So allowable this is for this one

  • allowable nuclear reactions.

  • Yep.

  • So for a spontaneous fission just pick one

  • you think would be likely.

  • You can also look up what sort of isotopes

  • are created when elements fizz.

  • It's not straight down the middle,

  • so like uranium won't often split into two

  • equally sized fission products.

  • They'll have roughly different masses,

  • but which ones you pick?

  • You're still going to get the same general solution.

  • Yep.

  • AUDIENCE: I'm so confused on how to find one.

  • Like a situation where it is unlikely possible because is

  • it spontaneous fission or is it generally possible for heavier

  • elements like transuranic elements?

  • MICHAEL SHORT: So the question is when is spontaneous fission

  • possible?

  • Is it only for heavy elements?

  • There is a difference between energetically

  • possible and observed.

  • That's part of the trick to this problem.

  • If you do out the Q equation to find out for add fission

  • products that you picked, you may be surprised at the result.

  • However, you're right.

  • You don't tend to see spontaneous fission happen

  • until you get to really heavy things like uranium.

  • So there's more to will something spontaneously fizz

  • than does the Q value allow it to happen.

  • So I don't want to give away anymore

  • but I will say if you're surprised at your result,

  • you might be right.

  • Yep.

  • AUDIENCE: On this question for electron capture.

  • In the equation you gave us, you're

  • calculating Q in an electronic capture.

  • Its the massive parents minus the dollars minus the I

  • think binding energy of the electrons is what you wanted.

  • When I try find out what the binding energy of an electron

  • is, it says it depends on the shell that its in.

  • MICHAEL SHORT: Yep.

  • AUDIENCE: So how do we know which

  • electron the nucleus is after?

  • Do we assume its from the innermost shell?

  • MICHAEL SHORT: Yep.

  • So the question was if you're doing electron capture where

  • you have some parent nucleus and you've

  • got a lot of electron shells.

  • The binding energy of every electron is different.

  • Which one goes in?

  • One, you find the data on the nest tables.

  • Two, chances are it will be the closest one.

  • So a roughly 80% of the time these things

  • happen from the K shell with very decreasing probabilities

  • from the outer shells.

  • So you can pick either the k or the L shell,

  • like both things may happen.

  • But I would say for simplicity's sake assume it's an inner most

  • shell electron.

  • And you can look up the binding energy on the NIST tables

  • on the learning module site.

  • So any other questions?

  • Yeah, Luke.

  • LUKE: On graphing the spectrum, the satellite

  • intensity vs. the energy for 4 2.

  • MICHAEL SHORT: For 4 2.

  • Ah yes.

  • So graphing the-- this would be like

  • if you had an electron detector.

  • Is that what I asked for?

  • 4 2, write the full nuclear reactions

  • and draw the energy spectrum you expect from each released form

  • of radiation including secondary ejections of particles

  • or photons.

  • So by a spectrum, I mean, yep, energy versus intensity.

  • LUKE: OK.

  • MICHAEL SHORT: So there you'll have

  • to account for the spectrum like the various range

  • of the betas that can be released,

  • any ejected electrons, any Auger electrons, any photons

  • from X-ray emission from electrons falling down

  • and energy levels.

  • Yeah, Alex, you had a question?

  • ALEX: Yeah.

  • What are the Auger electrons?

  • MICHAEL SHORT: The Auger electrons

  • is that funny case where in our mental model

  • a gamma ray hits an inner shell electron,

  • and it's usually an inner shell electron, shooting it out.

  • Then another electron will fall down

  • to fill that hole emitting an X-ray.

  • And the Auger process can be thought

  • of that second X-ray hits an electron on the way out

  • and fires out the electron.

  • So this here would be the Auger electron.

  • They tend to be particularly low energy.

  • Yeah, Luke.

  • LUKE: If you have that cascade of electrons during an electron

  • capture, are they still Auger [INAUDIBLE] radiation?

  • MICHAEL SHORT: Yep.

  • As long as you have a higher level shell coming down

  • to a lower shell and the ejection of an outer shell

  • electron.

  • That's than Auger electron emission process.

  • Regardless of whether it started with gamma

  • or started with electron capture.

  • Yep.

  • AUDIENCE: How do we know if that happens?

  • MICHAEL SHORT: You can actually sense or detect

  • the energy of those Auger electrons

  • with a very sensitive Auger electron detector.

  • These are in the sort of hundreds or thousands

  • of EV range.

  • AUDIENCE: But for like the context of this question,

  • how do we know--

  • where would we go to find them?

  • MICHAEL SHORT: Oh, to get the Auger electron data?

  • AUDIENCE: Yeah.

  • MICHAEL SHORT: For that you can actually

  • look up the binding energies of an outer shell electron,

  • and you can do that energy balance

  • where it would be E2 minus E1.

  • Whatever the energy of that X-ray

  • is minus the binding energy of the emitted electron.

  • And because there's infinite possibilities.

  • I mean you could eject any electron, just pick one.

  • And say here's an Auger electron,

  • or draw a couple of lines in places.

  • I don't have I don't want you to get every single line.

  • If we asked you to do this for uranium,

  • there's like you know 92 electrons

  • and a lot of different transitions

  • that's not what we're going for.

  • I want to make sure you know the physics.

  • Not that you can draw 92 lines accurately

  • with a fine-toothed pencil.

  • Do you need a question two?

  • AUDIENCE: Yeah.

  • So for 2 1 if we write two possible nuclear reactions

  • for 239 on [INAUDIBLE] the right was only

  • off the case of any decision and that it's

  • a state which decay processes and repeating processes may

  • be possible for each general type of reaction.

  • What exactly does that--

  • I find the answer for number two is alpha.

  • MICHAEL SHORT: Maybe that's the answer.

  • AUDIENCE: Oh, OK.

  • MICHAEL SHORT: And does anything compete with alpha decay?

  • Does anything compete with spontaneous fission?

  • OK.

  • Cool.

  • [INTERPOSING VOICES]

  • AUDIENCE: Is that an indication of [INAUDIBLE]

  • MICHAEL SHORT: Beta decay.

  • Well, is there-- well, for that you

  • can look up the table of nuclides

  • which I've got up here.

  • So let's take a look at plutonium 239,

  • and it precedes by alpha or spontaneous fission.

  • So every year I switch up the isotope

  • and make sure that there's at least a couple of decay modes,

  • and therefore the answers are going to change every year.

  • But the general question doesn't,

  • So this year I happened to pick an interesting one.

  • Yeah, it's kind of a mind game, right?

  • What are you missing?

  • AUDIENCE: Nothing.

  • MICHAEL SHORT: Nothing.

  • Yeah.

  • [LAUGHING]

  • Yeah, go with your physical intuition.

  • Any other questions, and maybe time for one more.

  • AUDIENCE: For 3 2, I could only find one nuclear reaction.

  • The action played after the nuclear reaction.

  • MICHAEL SHORT: Ah Yeah,

  • AUDIENCE: That's very curious you're really

  • [INAUDIBLE] decay.

  • MICHAEL SHORT: Yeah.

  • AUDIENCE: And, so I was wondering where

  • the molybdenum [INAUDIBLE]

  • MICHAEL SHORT: Yeah, so the question

  • is I specifically wrote which nuclear reactions could

  • make 99 molybdenum, despite there

  • being only one natural one.

  • So what could you induce artificially?

  • And if you can do that profitably, I'll guarantee you

  • there's a startup in it for you.

  • So what are all the different particles

  • that something could absorb to create molybdenum 99,

  • and which of those are allowable nuclear reactions?

  • And if they're not allowable, how much energy

  • do you have to put in an accelerator

  • to make that reaction happen?

  • And is the price of electricity in the accelerator

  • worth the Molly 99 that you create?

  • There are actually quite a few startups

  • working on this problem right now.

  • So the answer to this question is be creative.

  • Think about all the different particles you know of,

  • and how they could create Molly 99.

  • And figure out are any of those processes allowed,

  • and if they're not allowed, how energetic

  • do you have to make the incoming particles to allow them?

  • Ah, good question.

  • There's some creativity hiding in these problems.

  • So it's 10:02.

  • I want to cut it off here, and we'll start off Thursday

  • with a numerical example of this stuff.

  • Nuclear activation analysis.

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