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visit MIT OpenCourseWare at ocw.mit.edu. MICHAEL SHORT: You guys asked
to do some numerical examples of the stuff that we've
been learning today, and I've got a fun one for you. A very real one, because
it happens all the time in reactors all over the world. So to set the stage for
this, hey, suppose I had-- just suppose, I had a
radioactive cobalt-60 source, that was calibrated
in March, 2011 to be approximately
one microcurie. And it's now October 2016. So let's start off
with the easy part. Let's say I posed the question,
how active was this source actually when it was made? Because it just says-- what does it say-- 1 microcurie. Supposedly, this was 1
microcurie on March, 2011. What does that mean the
actual activity could be? There's not a lot of
confidence in that number. Remember significant figures
from high school chemistry and physics. This is when they're
important, because you got to know what you're buying. So what is the actual
activity of this source when it was calibrated? What is our uncertainty on this? Anyone remember
this from sig figs? I hope so. If not, I'll
refresh your memory. It's plus or minus the
next decimal point. So really, this was 1 plus
or minus 0.5 microcuries. So you might have a
1/2 microcurie source. You might have a 1 and
1/2 microcurie source. They specifically
decided to leave out the second decimal
point so they're not liable for a source that's
out of that calibration level. So we supposedly have
a 1 microcurie source. And I've measured it
on-- let's say it's now-- October, 2016. Suppose I went and
made a measurement, and it was 0.52 microcuries. And we want to know, how
active was this source actually when we got it? Where would I begin? AUDIENCE: Look up the half-life. MICHAEL SHORT: OK,
yeah, that's why I've got the table of
nuclides right here. So let's look up the
half-life of cobalt-60. And there it is right
there, 1,925 days. So we know that the half-live
for cobalt-60 is 1,925.4 days, which equals-- and I pre-did the math out so
I wouldn't spend lots of time on the calculator. Activity, how many
seconds is that? 1.66 times 10 to
the eighth seconds. And we also remember
that activity equation, where the activity
as a function of time equals the original activity
times e to the minus lambda t. The only thing missing here
is the initial activity and lambda, so how
do we find lambda? Anyone remember that expression? AUDIENCE: Log of
2 over half-life. MICHAEL SHORT: Yep. And we know that
lambda equals log of 2 over the half-life,
which in this case is about 0.693
over 1.66 times 10 to the eighth, which equals
4.17 times 10 to the minus 9 per second. And so now it's pretty easy. We know what lambda is. We know what our current A is. So we can say that our
original activity is simply our current activity divided
by e to the minus lambda t. What's t? Well, I made an
approximation here. It's been about five
years and seven months. I assumed that a month
has 30 days, on average, which comes out to t-- I remember calculating
this already-- 1.74 times 10 to
the eighth seconds. So we plug in that t right
here, plug in that lambda right there, and we get
our initial activity-- what did I get?-- was 1.07 microcuries. Now hopefully, this result
is fairly intuitive. Because the half-life
of cobalt-57 is 1,925 days, which is
about 5 and 1/4 years. And it's just over 5 and
1/4 years since this source was calibrated. And we have just under half
of the original activity. So hopefully that's an
intuitive numerical example. Now, let's say, how many atoms
of cobalt-60 did we have? Or better yet, what was
the mass of cobalt-60? Where would I go for that? I'll give you a hint. It's on the screen. AUDIENCE: Are you
saying the mass that we have in the pellet? MICHAEL SHORT: Yeah, so for
this plastic check source right here, what's
the actual mass of cobalt-60 that we
put in in the beginning, again, supposing I had one of
these right in front of you? AUDIENCE: Look at
the binding energy. MICHAEL SHORT: You could
look at binding energy, which is a form of mass. Or luckily, we've got the atomic
mass right up there in AMU. So again, this is a quick
review of high school chemistry. We need to find out how many
atoms we have in this pellet. And once we know the
number of atoms-- and we have the molar mass up
there in either AMU per atom or same thing as moles
per gram, pretty much-- we'll know what the
mass of cobalt-60 was. So one important
conversion factor to note is that 1 curie of
radiation is 3.7 times 10 to the 10 becquerel. And remember that 1 becquerel
is 1 disintegration per second. So if we know the initial
activity of our material, using our decay constant, we
should know the number of atoms that we had right there. So we know our initial
activity, A0 is 1.07 microcurie. And anyone remember,
what's the relation between the activity and
the number of atoms present? Just yell it out. I hope you'd know that by now. Does this look
familiar to anyone? Where the activity is directly
proportional to the number of atoms there, times
it's decay constant. So since we now know A0, we
can find N0, because now we know lambda as well. So the number of atoms
we had at the beginning is just, our
activity over lambda. And our activity is
1.07 microcuries. Let's convert up to curies. So we know that there is 1 curie
in 10 to the 6 microcuries. And we'll use that conversion
factor right there, times 3.7, times 10 to the
10 becquerel per 1 curie, divided by our decay constant,
4.17 times 10 to the minus 9 per second. Let's check our units to make
sure everything comes out. I brought a canceling color. We have microcuries on the
top, microcuries on the bottom, curies on the top,
curies on the bottom. And we have a
becquerels, which is a disintegrations per second. And we have a per second
down on the bottom. So the per second goes away
and the becquerels just becomes atoms. And so we actually get, N0 is-- I think it's something
times 10 to 12th. Yeah, 9.5 times 10
to the 12th atoms. That's the way it looks. So the final step, how do we go
from number of atoms to mass? Can anyone tell me? AUDIENCE: Convert it to
Avogrado's number and moles. MICHAEL SHORT: Yep. So the last thing
we'll do is we'll say we have 9.5 times
10 to the 12th atoms, times Avogadro's number, which
is 1 mole in every 6 times 10 to the 23rd atoms. Then we go to the
table of nuclides to get its atomic mass. This is one of those
situations where you don't need to take
the eighth decimal place, because you're not converting
from mass to energy. You're just getting mass. So if you might wonder, where
did all the decimal points go, think of what type of
calculation we're doing. If we started off with a one
significant digit number, do we really care about keeping
the eighth decimal place in the rest of everything? No, definitely not. We're not turning AMU
into MEV, in which case the sixth decimal point could
put you off by half an MEV, or like the rest
mass of the electron. We're just getting masses here. So let's just call that 59.9-- that's enough for me-- 59.9 grams in 1
mole of cobalt-60. And just to confirm, we have
atoms here, atoms there, moles there, moles there. We should get a mass in grams. And this came out to 0.95
nanograms of cobalt-60. Not a lot of cobalt-60
can pack quite a wallop in terms of activity. So even though this has
a pretty long half-life, as far as isotopes go, it
takes very little of it to have quite a bit of
activity, certainly enough for our fairly inefficient
handmade Geiger counter to measure what's going on. Pretty neat, huh? Cool. Now, let's say I ask you another
question, a simpler question. How many disintegrations
per second are coming out of
that cobalt-60 source? Let's say we wanted to find
the efficiency of our Geiger counter. You'd have to know how
many counts you measure. You'd have to know how
far away your source is. And you'd have to know how
many disintegrations per second are happening. So how would I get to the number
of disintegrations per second? AUDIENCE: Convert
it to becquerels. MICHAEL SHORT: That's right. We'll just take our
current activity, which is 0.52 microcuries. And so we'll say, 0.52 times
10 to the minus 6 curies, times 3.7, times 10 to the
10 becquerel in 1 curie is-- well, we have curies here. We have curies there. And this comes out to-- what
was our current activity. It was in the tens of thousands. About 19,000 becquerels. So we know that,
right now, this source is giving off about 19,000
disintegrations per second. Is that how many gamma
rays it's giving off? Do we know that yet? AUDIENCE: No. MICHAEL SHORT: Why not? I heard a lot of no's. What other information
do we need to know? AUDIENCE: Type of radiation. MICHAEL SHORT: Sure. Luckily we've got
that right here. So if you look here,
the mode of decay is beta decay to nickel-60. So cobalt-60 is actually
primarily a beta source. However, it's used for its
characteristic gamma rays. So let's take a quick look. We look at its decay
diagram, it's pretty simple. And let's just say, somewhere
near 100% of the time, it beta decays up
to this energy level and can undergo any number
of transitions like this. The only two we
really tend to see is, like that one and that
one are the most likely ones. So on average,
each disintegration of a cobalt-60 atom is going
to produce two highly energetic gamma rays. So actually, what you'd have
to know for this source is, despite it being 19,000
becquerels of cobalt-60, it's giving off 38,000
gamma rays per second. So that way your
source calculations wouldn't be off
by a factor of 2. Then if you know the
distance between your source and your detector, and you
know how many of those gamma rays are going through
the detector itself-- which we can count calculate
with a solid angle formula, which I'll give you
a little later-- you'll know how
many of them should interact in here, once you learn
photon nuclear interactions. And you'll know how many of
them actually get captured. And that's how you can get the
efficiency of the detector. So we've kind of filled
in half the puzzle. You now know, for
a fixed source, how many atoms there are, how
many disintegrations there are and how many gamma rays
you expect it to give off. And then later in
the course, we'll tell you how to figure
out how many of them make it into the detector
and how many of those should interact in the detector. Is everyone clear on what
we talked about here? Yep. AUDIENCE: Where did you get the
number for 38,000 gamma rays? MICHAEL SHORT: Double the number
of cobalt disintegrations. AUDIENCE: Because we
assume it drops 2? MICHAEL SHORT: Yeah. So I'm ignoring
the 0.2% and 0.6% decays because they're
extremely unlikely. And having looked
this up ahead of time, I know that this transition
and that transition are by far the most likely ones. And if you don't know that-- AUDIENCE: Yeah, does
it say it down there? MICHAEL SHORT: It sure do. Those right there, the
99.9-something intensity ones, it's all there. So I had some
questions on Piazza. What happens when we can't read
the pixelated decay diagram? That's just there for fun. The actual table that will tell
you all the information you need to know is right below. And I just decided, all right,
forget the ones that are 2e to the minus 6 likely or point
0.007, whatever, too many 0's. So you can say that, on
average, to probably two or three significant
digits, that gamma ray, that gamma ray are the
only ones you tend to see. Is everyone clear on how
I made that determination? Cool. Yes. AUDIENCE: So are
disintegrations just a beta, and then for each one of
those, it's two gamma rays? MICHAEL SHORT: That's right. The number of
disintegrations is the number of atoms that leave
this position. So let's take a crazier example. I've already had you
look at americium-241. Sort of bring home the message
that the number of americium disintegrations
does not necessarily equal the number of gamma rays
that you will expect to see. Because any one of
these is possible. There is an 84% likely one. It looks like it goes to the
third level from the bottom, because it's just the third
number I'm seeing here. If you don't believe
that, then check how much you have to zoom
out to see what's going on. Let's see. Intensity, yep. The third most
energetic alpha ray, so the third from the bottom,
is indeed the 84% likely one. Then there's another
one that's 13% likely, so we can't discount
that either. So you're going to hit
anything from, like, the second to the
fourth energy level and any number of those gamma
cascades that come off there. So the intensity of your
source in becquerels or curies does not immediately
tell you how many gamma rays or other
disintegration products you should expect. There are some pretty simple
ones, like the dysprosium one that we saw. Was it 151? I've been here before. It's not purple. No, that's a complicated one. All right, I don't remember
which dysprosium isotope just had a single decay. But the only time
that you would get one particle per disintegration
is if your decay diagram looked like this. That's your parent and
that's your daughter. That's the only time you
can expect one thing. If you have a more complex
decay diagram than that, you'll be looking at more
than one quantum of radiation of some form per disintegration. So is that unclear to anybody? Cool. Now let's get into
a more fun problem. And I'll leave this up
here, since we're going to refer to it a fair bit. So this is a good example of
activity, half-life, mass, number of atom calculations. We're going to use
this information to answer a more
interesting question. How was this made? Anyone have any idea what sort
of intentional nuclear reaction could have produced cobalt-60? And I'll go back to cobalt-60
so we can get its proton number. There it is. Can you say it a little louder? AUDIENCE: [INAUDIBLE] Oh,
I said neutron bombardment of cobalt-59. MICHAEL SHORT: Sure, you can
have a neutron bombardment. So you can have, cobalt-59
absorbs a neutron, becomes cobalt-60
with some half-life. And then we'll decay with
some decay constant lambda, and become-- in this case, it
undergoes beta decay-- to nickel-60. Let me make sure I got
the proton number right. I'm going from memory here. Hooray. OK, how do we set up
the series radioactive decay and production equations
to describe this phenomenon? So we have a physical picture
of what's going on here. How do we construct the
differential equation model? The same way we've been
doing it Friday and Tuesday. So someone, kick me off. What do we start with? AUDIENCE: So the production
of cobalt-59 is 0. Destruction would be sigma
times flux times the amount of cobalt. MICHAEL SHORT: OK, let's make
a couple of designations. Let's call cobalt-59-- I'll use another color for
this, just so it's clear. We'll call cobalt-59 N1. We'll call cobalt-60 N2. We'll call nickel-60 N3, just
to keep our notation straight. And then continuing
our notation pile here, you mentioned that
there's going to be some sigma, some microscopic
cross-section times some flux, the number of neutrons
whizzing about the reactor, times the amount of cobalt-60. So let's stick our DN,
Dt, equals no production minus some destruction. Does this look eerily
familiar to any of you? Forget the fact that
it's a sigma and a flux. Just treat those as constants. What does this
look exactly like? AUDIENCE: The minus lambda D. MICHAEL SHORT: Exactly. It looks just like dN1 dt
equals minus lambda and 1. Yeah, that's exactly it. This, in effect, is like your
artificially-induced decay. So the probability that one
atom of N1 absorbs a neutron, times the number of
neutrons are there, gives you some rate at which
these atoms are destroyed. Just like a lambda gives
you the rate at which those atoms naturally
self-destruct. So you can think of the sigma
times flux like a lambda. Mathematically, they're
treated identically. The only difference is, we're
imposing this neutron flux. So it's like an
artificial lambda, which means solving the
equations and setting them up is exactly the same. So how about N2? What's the production and
destruction rate of cobalt-60? First of all, someone yell it
out, what's the creation rate? AUDIENCE: Lambda N1. MICHAEL SHORT: Which lambda? AUDIENCE: Oh, lambda 1 I mean. MICHAEL SHORT: So what
is lambda 1, this one? AUDIENCE: Sigma, yeah. MICHAEL SHORT: OK, so let's
keep with these notations. I'm going to say this is
like a lambda artificial, but I'm going to keep
with sigma flux and 1. And just so we keep
our notation straight, I want to be able
to cleanly separate the natural and the artificial
production and destruction. And what's the destruction rate? AUDIENCE: Sigma phi N2. MICHAEL SHORT: So you
said sigma phi N2. Is that the physical
picture we have up here? Not quite, because
like we can see here, cobalt-60 self-destructs
on its own, right? Yeah. However, that actually is
a correct term to put in. The other one that we're missing
would be minus lambda N2. So you know what? Let's escalate this a
bit into reality and say, we're going to do this. The actual equation is not
going to be that much harder. So thanks, Sean. Let's do this. Yeah. AUDIENCE: Would those
sigmas be different? MICHAEL SHORT:
Yes, good question. That's what I was getting to. So there is an
absorption cross-section for every reaction
for every isotope. And now I'd like to show
you guys where to find them. On the 22.01 site,
there is a link to what's called the
Janis Database, which is tabulated and plottable
cross-sections of all kinds. So every single database
that we know of, it's updated
continuously by the OECD, so you can trust that
the data is updated. I won't say accurate, because
some of these cross-sections are not very well known. So I've already started it up. In this case, we don't
have neutron capture data for cobalt-60. Let's keep it in
the symbols for now. And then later on, we're just
going to say, it's probably 0. But let's keep it together. So for cobalt-59,
double-click on that, and you are presented with an enormous
host of possible reactions. Like right here, you
have N comma total, which is the total cross-section
for all interactions of neutrons with cobalt-59. And you can see
that this varies, but the same sort of shape
that I was haphazardly drawing. It's low. There are some resonances. And then it continuously
increases at lower energies. Is this the right
cross-section to use? If it accounts for every
possible interaction of a neutron with cobalt-59. No. So what other reactions, besides
the one that we want, which is capture, could this account for? AUDIENCE: Scattering. MICHAEL SHORT: Yep. AUDIENCE: Fission. MICHAEL SHORT: Scattering,
fission, n2n production. Sometimes one neutron
goes in and two come out, like for
beryllium, like we talked about at the
beginning of class. So you've got to know to
choose the right cross-section. And in nuclear
reaction parlance-- that's shorthand right there-- that's N comma total. That accounts for
elastic scattering, inelastic scattering to any
energy level, capture, fission, n2n reactions, sometimes proton
release, sometimes exploding, whatever nuclei do. So let's look at
our other choices. I'll shrink that. We have the elastic
cross-section. Compare that with
a total, and you get a general idea
of how much the total at elastic cross-sections
actually matter. So a lot of those resonances
are elastic cross-section resonances, but there
are other reactions that are responsible for a lot
of the other craziness going on. So let's unselect that. Oh, I'm sorry. I meant MT 1 Let's
compare MT 1 and MT 2. So MT 2, elastic. There we go. OK, that's what
I was hoping for. So right now, the red one
is the total cross-section, and the green one is the
elastic cross-section. And you can see that,
at high energies, the total
cross-section is mostly the elastic cross-section. But at low energies,
especially right around here at the thermal energy of
neutrons in a reactor, there's something
else responsible. That's probably what
we're going after. So let's keep looking
through this Janis Database. Hey, there's the n2n
reaction, if you guys want to see how likely this is. It's another one
of those reactions that-- look at that-- it's
0 until you get to 11 MEV. So what do you guys
think the q-value for n2n production
of cobalt-59 is? AUDIENCE: Very negative. MICHAEL SHORT: How negative? It's on the graph. AUDIENCE: 10. MICHAEL SHORT: Yeah, negative
looks like 10, or maybe 10 and 1/2, MEV. AUDIENCE: 0.454. MICHAEL SHORT: Oh, hey, awesome. [LAUGHTER] AUDIENCE: I think
it moves it around. MICHAEL SHORT: So there you go. Yeah, indeed, it's very
energetically unlikely to fire in one
neutrons and get two. But if you have a
10.454 MEV neutron, you can make it happen. Pretty cool, huh? That's not the reaction
we're going for. What we want-- let's
see if I can find it. Proton plus neutron,
neutron plus deuteron, all the inelastic energy levels. There it is, capture. The gamma reaction here is
what's referred to as capture. And there we go, a nice
normal-looking cross-section. So for cobalt-59, if we
go down to about 0.025 EV here, read it off,
it's about 20 barns. Because you guys asked
for a numerical example. So let's say that our capture
cross-section for cobalt-59 is about 20 barns,
which is to say 20 times 10 to the minus
24th centimeter squared. Let's also put up our
capture one for cobalt-60. So let's go back to our table,
take a look at cobalt-60. I think I know what the
answer is going to be, which is, we don't know. Not in this database,
unfortunately. So for symbolism,
let's keep it there. But we're going to say,
well, we don't know. Yeah. So let's designate these
different cross-sections. Let's call it
sigma-59 and sigma-60. So those will be our
two cross-sections. We'll just call
this one sigma-59, and call this one sigma-60. And we already know the
lambda for cobalt-60. So let's say the lambda for
60 cobalt, from this stuff up here, 4.17 times 10 to
the minus 9 per second. Let's just refer
to that as lambda for ease of writing things down. So we've got a complete set
of reactions for a dN2 dt. What about dN3? What's the production
rate of N3? AUDIENCE: Lambda MICHAEL SHORT: Yep, lambda N2. Anything else? What about the destruction rate? Is what? AUDIENCE: [INAUDIBLE] MICHAEL SHORT: It's
a stable isotope. AUDIENCE: We don't know. MICHAEL SHORT: But you
were on to it, Sean. So what would you add, based
on what you added to N2? AUDIENCE: Sigma
whatever, phi N3. MICHAEL SHORT: Yeah, there's
going to be some new sigma-- let's call it sigma
nickel-60, phi N3. AUDIENCE: Yeah, is the sigma
going to be 0 this time around, or is it actually going to be-- MICHAEL SHORT: Let's find out. Let's go to our tables. There, there's
data for nickel-60. Let's look up its
absorption cross-section. So we'll scroll
down to our Z gamma, our capture
cross-section, plot it. Take a look at
around 1e minus 8. And it's like 2 barns-- not negligible. So our capture cross-section
for nickel-60-- I keep overwriting
myself, and then I remember it's a blackboard. 2 barns, which is 2 times 10
to the minus 24th cm squared. So we can just refer to
that as sigma nickel. For the purposes
of this problem, we don't particularly care how
much nickel-60 we're making. Nickel-60 is a stable
isotope of nickel. Eh, forget it. So for the purposes
of this problem, forget the N3 equation. We don't care how much
stable nickel-60 that we're making, because it's a
lot cheaper to get it out of the ground, probably
something like 10 orders of magnitude cheaper. Yep. AUDIENCE: Where did you pick
that incident energy from, the MEV [INAUDIBLE]
1 times 10/8? MICHAEL SHORT: Yeah, so the
incident energy I picked-- because one thing
we had gone over is that the thermal
energy of a neutron is around 0.025 EV, which
is equal to 2.5 times 10 to the minus eighth MEV. So I took that value,
about 2.5 times 10 to the minus eighth MEV, so
around here, and just went up. And on a log scale, it looks
to be closest to 2 barns. You can always get
the actual value. So if you want to
zoom right in, I'm going to keep zooming into the
10 to the minus eighth region. Maybe not. You can set your
bounds accordingly. Oh, 0 or negative
values-- ah, OK, whatever. Let's just read off
the graph for now. You can use this tool
to get the actual value. But for problems
like this, I think estimating it from the
graph is going to be fine. When you get into
22.05 and you're like, what's the actual flux
in the reactor to within 1% or something, estimating
from the graph is no longer allowable. There are tabulated
values of these things. Oh, yeah, so you can actually
set the plot settings, get the tables. Oh, tabulated-- there we go. So you can read off values
of the cross-sections from a table like that. But for now, since
we want to make sure to get this
problem done, let's just stick with the graph. So we don't care
about the N3 equation. We're also going to say, well,
let's not ignore sigma-60 yet. So how do we solve
this set of equations? Let's make a little
separation here. First of all, the easy one-- what's N1 as a function of time? AUDIENCE: e to the
minus sigma-59, phi. MICHAEL SHORT: e to the minus
sigma-59, phi, and what else? AUDIENCE: N0. MICHAEL SHORT: There is
an N1 0, and there's a t. Because as you're
burning it out, it matters how
much time you have. Doesn't this look eerily
similar to an N equals N0 e to the minus lambda t? Again, it's like exponential
artificial decay, because we're burning
those things out. For a fixed amount
of neutrons going in, the amount that we burn is
proportional to the amount that is there. So that's the burn rate. That's the amount that's there. This is our artificial lambda. So that equation's easy. What we really want
to know is, what is N2 as a function of time? That's the $60 million
dollar question. And I'm not exaggerating there,
because cobalt-60 is expensive. So the question I'm posing
to you guys on the homework-- so for those of you who
have started problem set 4, I'm going to be swapping
out the noodle scratcher problem to this problem that
we're going to begin together in class. And you guys are going to
finish on the homework. So I'm kind of giving
you help on the homework. What is N2, the amount of
cobalt-60 in your reactor, as a function of time? And what is your profit
for running the reactor as a function of time? Assuming a few things-- so
let's set up some parameters. I'm going to say that
the neutron flux is the same as that in the MIT
reactor, which is about 10 to the 14th neutrons per
centimeter squared per second. We already have
all of our lambdas. We have all of our sigmas. I'm going to say that the cost
of running the reactor is-- and I have a quote on this-- $1,000 per day, which is the
same as $0.01 per second. Not a bad rate to stick
something in the reactor, right? It's not bad at all. If you had to build
your own reactor, your daily operating cost
would actually be $1 million a day for a commercial
power plant. So every time a plant goes
down, you lose $1 million a day, plus the lost electricity or
whatever that you have to buy. So let's put that in there. And from the cost of this
hypothetical cobalt-60 source, we know that cobalt-60 runs
about $100 per microcurie, because these sources
run for about $100. And so the eventual problem
that we're going to set up here and you guys are going to
solve on the homework-- and we can keep going a little
bit on Friday if you want-- is, at what point, at what t
do you shut off your reactor and extract your cobalt-60
to maximize your profit? And this is an
actual value judgment that folks that make
cobalt-60 have to make. How long do you keep your
nickel target in there and not hit diminishing returns? Because you're
always going to be, let's say, increasing
the amount of cobalt-60 that you make if your source
is basically undefeatable, until you reach some
certain half-life criterion. But it might not make
financial sense to do so. So let's start getting
the solution to N2. Let's see. So I'm going to
rewrite our N2 equation and we can start solving it. I'm sorry, that's
a dN2 dt, equals sigma-59, flux, N1 minus lambda
N2 minus sigma-60 flux N2. So how do we go about solving
this differential equation? AUDIENCE: Integrating factor. MICHAEL SHORT: Yep, the
old integrating factor. First, we want to
get rid of the N1, because that's another variable. And we've already
decided right here that N1 is N1 0 times e to
the minus sigma-59 flux t. And we haven't
specified, what's N1 0? Let's do that now. The last number we'll put in
is, we started with a 100 gram source of cobalt-59. When we write it in isotope
parlance, it sounds exotic. But that's actually the only
stable isotope of cobalt. So that's just a lump of
cobalt from the ground that we stick in. So we know what N1 0 is. So we can now rewrite
this equation as-- let's just go with N2
prime for shorthand. And we'll put everything on
one side of the equation. So we'll have, plus lambda
plus sigma-60 phi N2, minus sigma 59 phi times e
to the minus sigma-59 phi t, equals 0. So what is our
integrating factor here? AUDIENCE: e to the lambda
plus sigma-60 phi t. MICHAEL SHORT: Yeah. So it is, e to the
integral of whatever is in front of our N2,
lambda plus sigma-60 phi dt, which equals e to the
lambda plus sigma-60 phi t. So we now multiply every
term in this equation by our mu, our
integrating factor. So let's say we have
N2 prime times-- I'm just going to use mu for
shorthand, since it's going to take a long time to write. Plus-- let's see, that right
there is like mu prime, because if we take the
integral of mu times N2-- let's see. Oh, yeah, this is right. So we have lambda plus
sigma-60 N2 times mu, minus mu times sigma-59 phi e
to the minus sigma-59 phi t, equals 0. This stuff right here is like
the an expanded product rule, so we can write it more simply. So we can say, N2
times mu prime-- let's see-- equals mu sigma-59
phi e to the minus sigma 59 phi times t. So next we integrate both sides. And we get N2 times mu equals-- let's see. Let's expand everything out now. So that stuff would
be sigma-59 phi e to the lambda
plus sigma-60 phi, minus sigma-59 phi times t. So if we integrate
all of that, we're going to get sigma-59 phi. I think there's an N0
missing here, isn't there? Let's see. There should be an
N1 0 missing here. Yep. There's an N1 0 that I
dropped for some reason. Let's stick that back in-- N1 0, and N1 0. N1 0 over that stuff,
lambda plus sigma-60 phi minus sigma-59 phi,
times whatever is left, e to the lambda plus sigma-60
phi, minus sigma-59 phi t, plus C. So now we can say, what's
our integration constant C? The last thing we
haven't specified is our initial condition. So let's assume that when
we started our reactor, there was no cobalt-60. That makes for the
simplest initial condition. So we can substitute
that in here. So at t equals 0, N equals 0. So we've got get 0 equals-- if t is 0, then that just
becomes sigma-59 phi N10 over lambda plus sigma-60
phi, minus sigma-59 phi, plus C. And so that
makes things pretty easy, because we know C equals
minus sigma-59 phi N1 0, over that stuff that I keep
saying over and over again. And then we've pretty
much solved the equation. The last thing we have
to do is divide by mu, and we'll end up with
the same solution that we got on Friday
and the same solution that we got on Tuesday. So I realized, the second after
I said it last time, that, oh, we can't just absorb
some e to the something t into our integration
constant C because there's a variable t in it. So I would say, look
at this derivation to know the whole solution. And so finally we end up with-- anyone mind if I erase a little
bit of this stuff up top? OK, I'll erase the decay
diagram because we're not using that anymore. And hopefully everybody
knows our conversion factor. So the end solution, N2
of t, would look like, sigma-59 phi N1 0, over
lambda plus sigma-60 phi, minus sigma-59 phi,
times e to the minus-- what's lambda 2 in this case? Lambda plus sigma-60
phi t, minus e to the minus sigma-59 phi t. And that right there is
our full equation for N2. Now you guys said, let's
make this numerical. OK, we have every numerical
value already chosen. I've already plugged these
into the Desmos thing, so you can see
generally how this goes. So we've solved
it theoretically, so now let's make this numerical
and make some sort of a value judgment, right? We know sigma-59, because
we just looked that up. We know phi. We impose that as 10 to the
14th neutrons per second. There it is. We know our lambda. We don't know our
sigma-60, so we're just going to forget that for now. But the point is, for
everything except time and N2, we have numerical
constants for this. So once we plug it all in, I
modified the Desmos example from last time to
have the actual unit. So you can see that our
fake L1, our lambda-- we'll call it lambda 1--
equals sigma-59 times phi, which is 20 barns. 20 times 10 to the minus 24th
centimeters squared, times 10 to the 14th neutrons per
centimeter squared per second. And the centimeter
squareds cancel. That becomes, to the minus 10. And we get that our lambda 1 is
like 2 times 10 to the minus 9 per second. Our lambda 2, well, we already
have that, 4.17 times 10 to the minus 9 per second. So this is one of those cases
where lambda 1 approximately equals lambda 2. So just like you see
in the book, when you plug in all the numbers,
you get a very similar equation. Which is to say, there's going
to be some maximum of cobalt-60 produced. And in this case,
the x-axis I have in seconds, because
that's the units we're using for everything. The y-axis is number of atoms. So right there, 6 times 10
to the 23rd, that's one mole. So at most, you can
make up about 1/3 of a mole of cobalt-60 out of
100 moles of cobalt-59, which means you're never
actually going to have one mole of cobalt-60. Because of the way that our
natural and artificial decay constants work out, because
they're fairly equal, you're never going to be able
to convert and harvest it all. That's the numerical
output of this. Now I have another
question for you. Is the top of this curve
necessarily the profit point for this reactor? No, good answer. Why do you say no? AUDIENCE: Well, you have
to write another equation for the costs and the profits
to maximize both of them. MICHAEL SHORT: Exactly. That's what you guys are
going to do on the homework. I think we've done the
hard part together here. And so now I want
you guys to decide, given those profit parameters--
and I will write them down on the Pset-- how long do you run
your reactor to maximize your cobalt-60 profit? So this is one of
those examples where we did the whole
theoretical derivation, we decided, yes, let's escalate
the situation for reality. Everything works out just fine. The final answer,
well, you just tack on this extra
artificial bit of decay from the cobalt-60
being in the reactor. But the form of the equation
is exactly the same. There's just a couple other
constants in it for reality. Then if you plug in all the
numbers for the constants, so you pretend like that stuff
is lambda 2 and that stuff is lambda 1, there's lambda
1 again, there's lambda 2, there's lambda 1,
and it's exactly the same equational form as the
original solution that we had. When you plug in
all the numbers, you get something
remarkably similar to what's in the book, just scaled
for actual units of atoms and seconds. Is there a question? Yeah. AUDIENCE: On the
homework, do you want us to just assume
that sigma-60 is 0? MICHAEL SHORT: Sure. AUDIENCE: And have
that all cancel out? MICHAEL SHORT: If you can
find it, that's great. But I couldn't find
it that easily. You can hunt through the
different databases in Janis to try to find it. But I'm not going to penalize
you if you can't find it, if I couldn't find it. So yeah, a lot of
the homework is going to be redoing this
derivation for yourself. Because I want to
make sure that you can go from a set
of equations that models an actual
physical solution. And I guarantee you you'll
have another physical solution on the exam. Solve them using your
knowledge of 1803, get some sort of a
solution, which looks crazy, but it comes from
straightforward math. Then plug in some
realistic numbers and answer an actual question. How long should you
run your reactor to maximize your profit? So it's kind of neat. We've been here
one month together, and you can already start
answering these value judgment questions about
running a reactor. And so again, I don't know who
did, but I'm glad you asked, is this field mostly simulation? And the answer is no. You actually have to use math
to make value judgments if you want to go and make isotopes. And then you go and
make the isotopes. And you sell them
to people like me so I can bring them into class
and scare unwitting members of the public, theoretically. Yeah, hypothetical
source indeed. So any questions on what we
did here, from start to finish? Yep. AUDIENCE: On the equation
right in the middle there, where it says-- in parenthesis,
it has lambda plus sigma-60-- MICHAEL SHORT: This one? AUDIENCE: Yeah. Was the phi drop just like a-- MICHAEL SHORT: Oh, yeah,
that was a mistake. AUDIENCE: OK. I can just pull it back later. MICHAEL SHORT: Yeah,
it probably means I was talking and
thinking at the same time and forgot to write that. But indeed, it's
back everywhere else. Thank you. Yep. AUDIENCE: So the lambda in the
final N2 function equation, is that lambda 1, the
theoretical lambda 1, fake lambda 1, or
is that lambda 2? MICHAEL SHORT: This
lambda right here is the actual lambda
for cobalt-60. Yep. This right here
is just an analogy I'm drawing to say that,
it's almost like that stuff is lambda 1. That's our original
artificial decay constant. And this stuff here
is like our lambda 2, because there's natural
decay and then there's reactor-induced destruction. Yep. AUDIENCE: What is that factor of
dividing by lambda [INAUDIBLE].. Where does that come from? MICHAEL SHORT: That comes
from this solution right here. There's another
interesting bit, too. Did we necessarily say-- let's see. Did we necessarily say that-- no, never mind, that's fine. That comes from--
let's trace it through. So mu contains-- yeah, it
comes from C. That's right. So we had it over
here, because that's part of our solution for-- let's see. Where would we trace it back to? It starts off here in the
differential equation. It starts off here as well. So that's part of
what's inside mu. OK, that's where it came from. So a mu contains this stuff. Yep. AUDIENCE: And then
once you integrate, it just becomes e to the that. It doesn't actually
become lambda plus 65. MICHAEL SHORT: It becomes e
to the lambda plus 60 phi, times t. AUDIENCE: Yeah, but when we
do the integration of that, we don't get any factors of
lambda plus 60 phi coming down. MICHAEL SHORT: We do, actually. There is a mu stuck
in right here. And so I just wanted to say
that, expanding this term comes out to lambda
plus sigma 60 phi, minus lambda sigma-59
phi, times t. So when you integrate
this whole term-- and again, there's an
N0 that should be there. You do bring this whole
pile in front of the t down on the bottom of the equation. So that's where it comes from. Yeah, cool. I don't think there's
any more missing terms. OK, maybe time for a last
question, because it is 10 o'clock. AUDIENCE: This
doesn't really have to do with your
derivation or anything. So I'm pretty sure you also
already explained this. But why can you put
a cross-section, like that's a measure
of probably in units, of centimeters squared. How does that [INAUDIBLE]? MICHAEL SHORT: Ah, so the
cross-section is almost like, if you fire a
neutron at an atom, the bigger the atom appears to
the neutron, the more likely it's going to hit it. So it's kind of a
theoretical construct, to say, if something has
an enormous cross-section, it's like shooting a bullet
at a gigantic target, with a high probability
of impact or interaction. Something with a
small cross-section, there's still only
one atom in the way, but it's like you're shooting
a bullet at a tiny target and have less of a
chance of hitting it. Does that makes sense? Cool. AUDIENCE: So is it just
a theoretical construct, or can you actually relate
it to an actual physical cross-sectional area? MICHAEL SHORT: You
can't relate it to a physical cross-sectional
area, as I know. It's not like a certain
nucleus has a larger cross-sectional area. Otherwise, things
like gadolinium, which has a cross-section
of 100,000 barns, would just be a larger atom. And it's not. And yeah, Sean. AUDIENCE: Are they determined
only experimentally, or do we know some of
way to calculate it? MICHAEL SHORT: Good question. They can be theoretically
calculated in some cases. In the Yip book, Nuclear
Radiation Interactions, he does go over how to calculate
those from quantum stuff. And so you'll get a little
bit of that in 22.02, in terms of predicting the
cross-section for hydrogen and the cross-section for water. And molecular water is not
just the sum of its parts. That's the kind of crazy part. Cross-sections do change when
you put atoms and molecules together, just tend to
be at lower energies, around thermal
energies and such. Let's say, all of
them probably can be theoretically calculated,
just not that easily. But the really simple ones
you can predict theoretically. Predicting the resonances
in those cross-sections, that's tough. Let's look at a simple
cross-section, like hydrogen. AUDIENCE: Can't you calculate
it using simulations? MICHAEL SHORT: Yes. Like if you know, let's say, the
full wave function for a given atom or for all the electrons in
an atom, you should be able to. So let's do N
total for hydrogen. Much simpler, this is the kind
of thing that can be predicted from theory quite easily. In fact, you will be
doing this in 22.02. The other one,
no, I don't expect you to be able to predict this. But you will learn why
the resonances are there and why they take the
shape that they do. So last thing-- we did go
like three minutes late, but everyone's still here. You can go if you
have to, by the way. I can't keep you here. If you want to know, if you
want to make this equation more realistic and account for every
possible energy in the reactor, you can make these
cross-sections a function of energy, and
integrate over the full energy range. And this is actually
how it's done. And you will do
this in 22.05, where you'll be able to take the
energy-dependent cross-section in tabulated or
theoretical form, and then integrate
this whole equation, and also account for the fact
that the flux has an energy component. Usually, it looks something
like-- in a light-water reactor, if this is
energy and this is flux, there'll be a bit
of a thermal spike. There won't be much
going on in the middle. I'm sorry, a fast spike, and
there will be a thermal spike. And knowing how many
neutrons are at every energy level, what's the probability
of every neutron at energy level interacting, and what are the
cross-sections at every energy level integrated over
the full energy range is what gets you the
accurate correct solution. What we've done here is called
the one-group approximation, where we've assumed that
all the neutrons have the same energy,
thermal energy, which is an OK assumption for
thermal light-water reactors. And it'll get you
a good estimate. The more neutrons you have
at different energies, the less good that
estimate becomes. Yeah. AUDIENCE: Wait, so
that thermal energy you gave us, like
0.02 [INAUDIBLE],, that was estimated for the
energy of the neutrons being fired. MICHAEL SHORT:
Let's say you have a neutron at about 298 Kelvin. From the Maxwell-Boltzmann
temperature distribution, you can turn that temperature
into an average kinetic energy. And that average kinetic energy
will give you a velocity. And that velocity is around
2,200 meters per second. And that average kinetic energy
happens to be about 0.025 EV. So thermalized
neutrons, like the ones flying about in the reactor,
are moving quite slowly at just 2,200 meters
a second, compared to the fast neutrons,
which can be moving closer to the speed of
light, not that close, but much, much, much closer. Cool. I'll take it as a good sign
that you all voluntarily stayed a little late. So did you guys find
this example useful?
