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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. 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 N2 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.
