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visit MIT OpenCourseWare at ocw.mit.edu. MICHAEL SHORT: So I'd like
to do a quick two or three minute review of the
stuff we did last time to get you back
into where we were. We were talking
about different types of technologies that
use the stuff you'll be learning in 22.01. Everything ranging from nuclear
reactors for producing power, and the Cherenkov radiation
that tells you-- well, that the beta particles
are moving faster than the speed of
light in water. It's a neat thing, too. I've actually been
to this reactor at Idaho, to the
spent fuel pool. Even the spent fuel, once
it comes out of the reactor, is still giving
off betas and still giving off Cherenkov radiation. And you can tell how long
it's been out of the reactor by how dim the glow gets,
which is pretty cool. So you can tell how old a fuel
assembly is by the blue glow. So remember, I told you
guys, if someone says, oh, you're nuclear,
do you glow green? You can be like, no, it's blue. That's the right way of things. We talked about
fusion energy and got into some of the nuclear
reactions involved in fusion energy. And I'll be teaching you
more about these today. Why fission and
fusion work, it all has to do with the
stability and binding energy of the nuclei involved. And that'll be the
main topic for today, is excess mass, binding
energy, nuclear stability. We looked at medical
uses of radiation, from implanting
radioactive seeds called brachytherapy seeds in certain
places to destroy tumors, to imaging, to X-ray therapy,
to proton therapy using accelerators, or cyclotrons,
to accelerate protons and send them into people. If you remember, last
time we ran the SRIM code of the stopping range
of ions and matter, and actually showed
that protons all stop at a certain
distance in tissue, depending on their energy and
what you're sending them into. Let's see-- we talked
about brachytherapy. We talked about
radiotracers, and this is going to be one of the
other main topics for today, is these decay diagrams,
and figuring out not only what products are made,
but what energy levels do the nuclei have, and
how do you calculate the energy of the
radioactive decay products and the recoil
nuclei, which do take away some of the energy. We talked about one
way to get rich. If you guys can figure
out one of the ways to solve this moly-99 shortage. Right now, it's mostly
made in reactors. The future has got
to be accelerators or some sort of
switchable device where you don't need
to construct a reactor to make these medical isotopes
for imaging, and tracers, and such. And finally, we
got all the way up to space applications,
shielding, crazy, different types of shielding, like
electromagnetic shielding, to protect from high
energy protons, all the way to radiothermal generators,
which use alpha decay to produce a constant
amount of energy on the order of one to 200
watts for like 100 years. And finally, to a
different configuration of nuclear reactors,
where you can design them to produce thrust,
not necessarily electricity. And that's where we
stopped on Friday. So let's move on to one
of the things I'd alluded to earlier, which is
semiconductor processing. This is actually a diagram
from the MIT reactor, because we have
this beam port here. Has anyone got to
see the silicon beam port at the MIT reactor? Oh, seven, eight--
OK, about half of you. For those who haven't,
who here has not had a nuclear reactor tour? Oh man. OK. Well, you'll get one
when you get to control the thing in early October. So you actually get to go
down to the control room and see the rest
of what's going on. So make sure to ask them,
show us the beam port for silicon ingots. And I think I already told you
the story about the poor UROP who held the ingots
up to their chest, getting about 10 months of
dose, which is not dangerous, but it meant that
for 10 months, they could have no
radiation exposure, and they had to
answer the phone. So that's how we ensure
safety around here. There's other ones that-- applications that
you're probably carrying around in your pocket. You can use the
fact that charged particles have very
finite ranges in matter to separate little bits of
that matter from other things. So this is actually how
single-crystal sapphires can be separated in little
slivers for protective phone covers, because sapphire
is one of the most-- the hardest, or the most
scratch-resistant materials there is. Single-crystal sapphire
is exceptionally strong, and optically transparent,
and expensive. I know that because on
one of our experiments, we use a single-crystal sapphire
window to see into reactor conditions at like 150
atmospheres, and 350 degrees Celsius, and
pretty corrosive chemistry. So you want to use as
little as possible. So you can use a big,
expensive accelerator to limit the amount of
sapphire that you use. And this is actually
done here in Boston. There's a facility not far from
here that uses an accelerator. And this is their
super detailed diagram of what the radiation
looks like-- yeah, whatever. But what they do is they
take-- they accelerate protons. They send them through
bending magnets to steer that beam path. And then they send
them into a large piece of single-crystal sapphire,
which is exceptionally expensive to make. And they can actually
lift off a thin sliver with micron precision. The reason for that
is the same reason that we showed with
that SRIM code. If you have this exact
energy of protons going into well-known
matter, you know what its range is going
to be with an uncertainty or so of about a micron. So you can have things that
come out thin, uniformly thick, and smooth, right away. There's some other
really wacky products-- like has anyone heard of
these betavoltaic batteries? No? They rely on beta decay or the
direct capture and electricity generation from a
radiation source like radioactive tritium. So in this little chip is about
2 curies worth of tritium. You guys will learn
in about a week, how to go from activity
in curies to mass, or something like that. And so this chip
actually contains a lot of radioactive
tritium that directly creates electricity. So you can hook into
that chip and produce nanowatts for years. So it's one of these
batteries that lasts-- well, as long as a couple half-lives
of the isotope that's inside. Now there's a trade-off here. The shorter the
half-life, the more active a given isotope will be for
the same number of atoms, but the shorter it will last. So you can have higher power
for lower time, or lower power, higher time. It's the classic
energy trade-off-- works the same way
with irradiation. And so now I wanted
to get into some of the more technical
stuff, where we'll be talking about
nuclear mass and stability. And this is where the nuclear
stuff really begins in 22.01. First, I want to make sure
that we all agree on notation. So I'll be writing isotopes
in this sort of fashion, where we refer to A as
the atomic mass, or just the total number of nucleons. This is not the exact
mass of a nucleus. It just refers to the sum
of the protons and neutrons in the nucleus itself. And a lot of what we'll
be talking about today is the difference between
this nice integer mass number, and the actual mass
of the nucleus, and that difference is
given by the binding energy, or the excess mass, which
are directly related. Z is just referred to
as the atomic number, or the number of protons. It's what makes the
element what it, which makes the name
kind of redundant. But it's-- humans
learned by association. It's easier to remember
element names or symbols than which element is which
just by the number of protons. So a lot of times we'll use the
name, or at least the symbol just so we know what's going on. And anything up here is
some sort of a charge. I do want to warn you guys of
the dreaded multiple symbol use or multiple use of symbols. I'll try to stick
for a lower case q-- will be charged. And uppercase Q
is going to refer to the Q value or the
energy consumed or released by a nuclear reaction. So they're both Q's
but we're going to keep one upper and lower case. And like we mentioned
before, let's say we were to write a
typical nuclear reaction, like the capture of
neutrons by boron to produce lithium-7,
helium-4, better known as an alpha particle,
and some amount of energy. There's two places where we
actually use this reaction. One of them is as control rods. A lot of reactors use
boron for carbide, or this compound B4C, which
is conveniently solid, fairly dense, and contains a whole
lot of boron in one place. Specifically,
enriched in boron-10, because boron-10 has a high
cross section, or probability, for neutron capture. And the other one
is in what's called boron neutron capture therapy. Have I discussed this with
you guys already, BNCT? Good, because
that's what we'll be talking about for a few slides. And to write this whole
reaction is the same thing as writing this shorthand
nuclear reaction. So this is often how you
see them in the reading, and in papers, because
it's shorter to do that. But it's the exact same thing. So I have a couple of
questions for you guys then. I have this extra Q here. Where does that Q actually go? So let's say boron
and neutron absorb, it produces two nuclei with
different binding energies. What happens to
the excess energy created from the conversion
of mass to energy? Yeah, Alex? AUDIENCE: That could be heat. MICHAEL SHORT: Yep. And heat, and more specifically? AUDIENCE: Kinetic energy. MICHAEL SHORT: Kinetic energy
of the radiation released. And so that kinetic
energy is actually used to our benefit in BNCT, or
Boron Neutron Capture Therapy. The way this works-- once I hit play-- is you can either-- you can use
any sort of source of neutrons, either a reactor
or an accelerator, through a lower complex
chain of events, like this. In this case, an accelerator--
so you don't need a whole reactor-- fires a beam of
high-energy protons into a beryllium target. Does that sound fairly familiar? Firing something at
beryllium, releasing neutrons, like what Chadwick was doing? Except he was firing
alpha particles into it. This releases neutrons. What they don't have labeled
here is slowing down stuff, or probably
hydrogenous material, so that the neutrons slow
down to a lower energy. And their probability
of capture increases or their cross
section increases. And if you don't know
what a cross section is, the definition is
two slides away. And the idea here is that these
neutrons then enter the brain or wherever the
tumor happens to be. And we rely on the fact that
tumor cells consume resources much faster than regular cells,
especially neurons, which after you're about five,
don't tend to grow very much. So it's all downhill by the
time you enter kindergarten. And we use that to our advantage
so that the neutrons coming in will hit the cancer cells,
which will preferentially uptake the borated
compounds, leaving most of the normal cells intact. And the difference in
dose can be a factor of 5 or a factor of 10. So that the cancer gets fried
while doing as little damage as possible to the remaining
brain cells, of which we have fewer and fewer every day. So I say statistically
speaking you guys are probably
smarter than me if we go by number of
neurons in your brain, because I think I'm the
oldest person in the room. And so now we can start to
explain, how does BNCT work, and why did we make the
choices that we did? For example, they use
30 MeV protons in order to induce these neutrons. So we have a nuclear reaction
that looks something like this. We start off with
beryllium-9 plus a proton-- let's just call it
hydrogen to stick with our normal notation-- and becomes-- well, can someone
help me balance this reaction? We know we get a neutron. What else is left? So Monica, what would you say? AUDIENCE: Let's see-- MICHAEL SHORT: Even just say
number of protons and neutrons, and we'll figure out
the symbol later. AUDIENCE: Number of
protons should be-- MICHAEL SHORT: Sorry,
oh, that's a 4. AUDIENCE: The number of
protons should be five. MICHAEL SHORT: Yep. Five protons, which
means it's boron. And number of neutrons
in the nucleus? Someone else? Yeah? AUDIENCE: Nine. MICHAEL SHORT: Nine. So we have boron-9. Not a stable isotope of boron,
but it doesn't really matter, because boron-9 almost
immediately decays into an alpha, and an
alpha, and a hydrogen. But this nuclear
reaction right here is what we'll be studying
for a little bit. And there'll also be
some amount of energy. And this Q can actually
be positive or negative. No one said there had
to be energy released in a nuclear reaction,
because in this case, we actually start off with 30
MeV protons and roughly zero MeV beryllium. If you want to get
really exact, it's on the order of
about 0.01 eV, which is why we neglect the
kinetic energy of beryllium at room temperature. There are other
reactions that when you fire a proton
into them will produce neutrons, such as the
absorption of lithium. But can anyone think
of why we'd want to use beryllium instead of lithium? Kristin, what do you think? What could be a bad thing
about using lithium? You ever throw in water? AUDIENCE: No. MICHAEL SHORT: OK. Then I should show
you what happens when you throw it in water. There's a few bad
things about lithium. It does this when you
throw it in water. It's one of the alkali metals. It's got an awfully
low melting point. It reacts with oxygen to
produce an oxide almost instantaneously. So if you ever take a
lithium battery apart, which you shouldn't, but if
you watch the video of somebody else doing it, you'll see that
the lithium foil turns black almost instantly. It also has a pretty
poor thermal conductivity and doesn't hold that structural
integrity when it melts. So it's not that good
of a target to use. Beryllium's pretty cool
in that it's the lightest structural material there is. Folks tend to make
satellites out of it, because it
costs a lot of money to launch things into space. And if you want something
that has a high melting point, and is light, and is structural,
beryllium's your way to go. It also happens to be a
great neutron generator. And then why 30 MeV? In this case, we're going
to use a table called JANIS, which I've got open over here. And I just have to clone my
screen so you can see it. This is a resource that I think
you guys are going to be using quite a lot in this course. We have a link to it on
the learning module site. And I'm going to show you
how it works right now. So I tend to use the web
version because it works on any browser, any computer. And now you can start to pick
which nuclear reaction you're looking at. And you can get
tabulated cross sections. So I'm going to start by
zooming all the way out. We can pick our
incident particle. Since in this
case, we're looking at the firing of
protons into beryllium, I'll pick the incident
proton data right here. There's a lot of different
databases with sometimes conflicting information. I tend to go with the most
recent one you can find. And click on cross sections. And this is, again another
table of nuclides, anything in green there's data for. Anything in gray, there isn't. So let's go all the way
back to the light nuclei, zoom in, go back down
to the light nuclei again until we find beryllium-9. Double-click on that, and
let's look for the anything cross section. And this is a pretty
wide energy scale. So you can actually change
your X minimum and maximum. So let's change it to
a minimum and maximum-- I don't know-- a
maximum of 50 MeV. We don't have to see all of
that other stuff going on. 50 MeV and maybe
a minimum of 10. If you notice-- actually,
I'll go back to 1. And I want to point
something out. You can actually get a
good yield of beryllium. Let's see-- you can actually
get a good yield of neutrons by firing protons at
beryllium in lower energies. But I notice there's
this interesting feature right around there. The cross section's flatter. And so if you want to get an-- ensure that you
get the right dose, you might want to deal
with a flatter cross section or a flatter
probability region, so that you have
something more predictable instead of in a really
high slope region. But some of these
nuclear reactions actually take extra energy
in order to move forward. And we'll show you another
example pretty quickly. Let's go back to
our slides here. Then another question
is, how does the boron only get into the cancer cells? Like we mentioned
before, cancer cells are actively
growing, which means they need a very large
and active blood supply. And so it's one way for
things to, let's say, not quite cross the
blood-brain barrier. If the cancer cells are growing
and your neurons aren't, then your cancer cells are going
to use more energy, take in more sugar, which
might be doped with boron, or some other compound
doped with boron, and that's all you can get
the boron into the cells that you want. And then why was boron
selected for the therapy? Let's think about that. What happens after the
neutron is created? And let's write the next
stage of the reaction. In boron neutron
capture therapy, we rely on doping the patient
with boron-10 to release an alpha particle, and
lithium-7 and a gamma ray. So now what we
can start doing is look at the table of nuclides,
which I'm going to teach you how to read now, to
figure out-- let's say that this neutron had
an energy of about zero eV and the boron nucleus had
an energy of about zero eV. And in the end, all this
stuff here has gained or lost some sort of energy
cue, Q. And today we're going to teach you how
to calculate this Q. So I want to skip ahead to how
to read the table of nuclides. So there's all-- this is
like the poster you'll see in every nuclear building. It's kind of what makes us, us. What you'll notice is
that there's a whole lot of nuclei at the lower left. They are the light ones. At the upper right, they
are the heavier ones. And they're colored
by half-life. In general, the blue
ones will be stable, and the further away you get
from blue, the less stable they get. So right away, without even
delving deeper, what patterns do you guys notice here? Yeah, Alex? AUDIENCE: As they get bigger,
heavier, they're more unstable. MICHAEL SHORT: Yeah. There's a whole section
where there's no more blue. There are no stable elements. So stability drops off
after a certain point. And what about in the
region of stable isotopes? Does anybody notice any
repeating patterns here? Take a look at every other row. There's a bunch of
blues and then one, and then a bunch and then one,
and then more and then none. That must be technetium,
because that's the only element around
there that doesn't have any. And then a bunch of blues. So every other row-- and in this case, it's
increasing number of protons-- has more or fewer
stable isotopes. It turns out that
the even numbered isotopes have a lot more stable
ones, for reasons that we'll get into pretty soon. If you zoom in a
little bit, you can see all the different
isotopes so you can select which ones you want. And again, if you look
really closely, that's-- let's say, neon right here
has got a few stable ones. Sodium has one. Magnesium as three. Aluminum has one. And this pattern
repeats all the way up to the point where
you don't really get any more stable isotopes. If you double-click
on one of them, you get all the
information that you'll need for the next three
or so weeks of the course. So in this case, I
picked on sulfur-32, one of the stable
isotopes of sulfur. So if you notice it doesn't
have any decay mechanisms here, but it does say its
atomic abundance. So you can know how-- what
percentage is normally found in nature. And then there's a
few other quantities that is going to be the topic
of what's going on here. Let's start with
the atomic mass. If you notice, the atomic
mass is slightly less than 32, 32 being the mass number, or
the total number of protons plus neutrons in the nucleus. The actual mass is a
little lower by that amount right there, the binding energy. It might be a little
funny because I've given you a mass in AMU, and a
binding energy in kiloelectron volts. I want to remind you that
these are the same thing. The conversion factor you'll
be using over and over again throughout this course,
especially on the next p sets, is one atomic mass unit
is 931.49 MeV c squared. Yeah-- I'm sorry. Yeah, never mind,
put that there. So then, again,
one, don't round-- because we've had
times when folks said, ah, this is about 931. And when you're
off by half an MeV, you could be at a totally
different decay level or get a positive Q when
it should be negative, or vice versa. And let's take a quick
look here to say, if this atomic mass
is 31.9720707 AMU-- this is why I
brought a calculator. Normally I do mental,
math but since I told you guys don't
round, I can't do eight significant
digits in my head. So I'm going to
get that in there-- 0707. If any of you guys
want to follow along, I encourage you to. And say minus 32, which
is the mass number. So in this case we're taking
the actual atomic mass minus the actual-- the mass number. In this case, it's 32. In this case it's, 31.9720707. And we end up with minus 0.0-- I'm going to put all
the digits here-- 293 AMU. If we convert this to MeV-- times 931.49, we get
minus 26.0159 MeV. See this number anywhere
on the KAERI table? Right there-- that's
the excess mass. And in this case, we usually
give this the symbol delta for the excess mass. And these are how
these quantities are directly related. The excess mass--
well, actually, what does the excess
mass really mean? It's the difference
between the actual mass and a fairly poor
approximation of the mass. So the excess mass
doesn't really have that much of a
physical connotation. But it is nice, because
if you know very well the tabulated atomic number-- I'm sorry, the-- yeah, the mass
number and the excess mass, you can figure out-- let's see-- yeah,
you can figure out what the real atomic mass is. And I want to switch now to
the actual table of nuclides and show you one example. If you want to very quickly
jump between isotopes, you can type them
in right up here. And does anyone know what the
gold standard for atomic mass is? And I'll give you a
hint, it's not gold. Yep? AUDIENCE: Carbon-12. MICHAEL SHORT: Carbon-12. What do you think the
excess mass of carbon-12 is going to be without
doing any calculations? AUDIENCE: Zero. MICHAEL SHORT: Exactly. Zero. So if we go to
carbon-12, because that is set as the standard, the
way atomic masses were done was carbon-12 weighs
exactly 12 AMU. The excess mass here is zero. And that's why the atomic mass
is 12.0 to as many decimals as we care to note. So is everyone clear
on what excess mass is? Yep? AUDIENCE: What's the point of
c squared for that conversion? MICHAEL SHORT: So mass
does not actually equal-- oh right, and it's
actually on the-- where did my chalk go? It's actually down below. The point is that energy is
related to mass by c squared. So they're not the
same units, but they're directly convertible. AUDIENCE: OK. MICHAEL SHORT: Yep. And so this way, you have
an E over a c squared. You get an m, and there we go. I had the units upside down. However, carbon-12 does not
have a zero binding energy. Yeah, Luke? AUDIENCE: How come when
you did that calculation, you didn't use the c squared? So like, it seems like then
that would be 26.0159-- MICHAEL SHORT:
MeV per c squared. Yeah. AUDIENCE: But they don't
say that up there-- or it didn't say
that on the table. MICHAEL SHORT:
Yeah, that's true. AUDIENCE: [INAUDIBLE] MICHAEL SHORT: So
it is funny, right? The binding energy is give it
in keV, and that's correct. An energy is an energy. An excess mass, it
really should say keV per c squared, because if
we're talking in units of mass, it's got to be in m. Or in this way, you could say
m is an energy per c squared. So this, to me, is a semantic
inconsistency in the table. But you guys will know
that a mass is always going to be an AMU,
or kilograms, or MeV per c squared. And energies will be in
MeV, keV, some sort of eV, usually, in this course. The binding energy,
though, that's correct. That's in keV, because
that's an actual energy. Now then the question is,
what does the binding energy actually represent? Does anyone remember
from Friday or Thursday? I can refresh your
memory, because that's what I'm here to do. The binding energy
is as if-- let's say we're assembling carbon-12
from its constituent nucleons. There's going to be 12 of them. Let's say we had six
protons and six neutrons. We can calculate the
total mass energy of this ensemble of nucleons
when they're infinitely far apart from each other. And forgive the little-- it's not to scale. But they are infinitely
far apart from each other. And we can say that-- let's say there is
Z number of protons. So we'll say the
binding energy is the number of protons
times the mass of a proton plus the number of neutrons-- A minus Z-- times
the mass of a neutron minus the energy of the
assembled carbon-12 nucleus. So there's actually a
measurable difference in mass between six
protons and six neutrons, and the actual mass of a nucleus
with atomic number A and-- I'm sorry, with atomic number
Z and mass number A c squared. So is everyone clear on how
we arrived at this formula? It's effectively
the energy released when you take the individual
nucleons, assemble the nucleus. You don't have as much
mass as when you started. Or in some cases, you might
have a little more mass than when you started if things
are particularly unstable. And you can use the excess
mass and binding energies in relative amounts to see, is
a nucleus going to be stable? For example, let's
look at iron-55 I'm going to jump here,
make it a little bigger so the important stuff
is easier to see. And if you notice, the
binding energy of iron-55-- there's quite a bit of it. It's very well-bound. In fact, this is one of
the most well-bound nuclei in the whole chart of nuclides. Let's look at
something that we know to be particularly unstable. Someone have any idea? Let's just add like
20 neutrons to iron let's see if it even exists. No-- doesn't happen. Let's try adding 10
neutrons to iron-- or go even crazier. What about 70? Too small-- all right, let's
meet somewhere in the middle-- 68. Still a pretty high
binding energy, but you can look at
it as a difference in binding energy per nucleon. So in this case, the
binding energy per nucleon-- if you take the binding
energy and divide by the total number of
nucleons, will give you a relative measure of how
tightly bound that nucleus is. Now these are not
absolute things. You can't just say,
certain binding energy leads to certain stability,
but they do give you pretty good trends to follow. And we're actually going
to be coming up with-- probably on Thursday-- a semi-empirical formula to
get the rough binding energy of any particular assembly
of protons and neutrons. And it follows experimental
calculations pretty well-- surprisingly so. I want to jump back
to here, because I've mentioned cross sections,
and I want to actually define what a cross section is,
because this is a quantity that you're going to
be using everywhere. Let's say that we fired
a beam of particles-- it doesn't matter what it is-- at a target of other particles. Let's say, the beam
particles are atom A, and the target
particles are atom B. And once these A particles
pass through the target B, a little bit
fewer of them come out the other side reacted,
or unscathed, or unscattered. And some of them are absorbed,
or scattered, or bounced off, or scattered backwards,
or what have you. We can write the sort
of proportionality constant between the change
in intensity of our A beam and the thickness of our slab. And we give that
proportionality constant this symbol, little sigma. We'll get something
going up here. Little sigma, which we call
the microscopic cross section. It's in effect, a constant
of proportionality that relates the
probability of absorbing an atom from this beam I-- or from this beam of atoms
A through a slab of B. And then if you
take this formula, you divide by that delta X-- so I'm going to
take what's on there and say delta I over delta
X equals minus cross section ABn-- which refers here to
the number density. So I'll keep our
table of symbols altogether so it's a
little easier to follow. n is our number density, which
means the number of atoms per unit volume. Usually, in nuclear
quantities, we use centimeters because these are things that
are actually fairly measurable, and cross sections
are actually in units of centimeters squared. And let me finish
that expression. We had the number
density of our target B. We had our initial
intensity, and that's it. Anyone know how to solve
this differential equation? If we take the limit
of small deltas, it should start to look like
a differential equation. The final answer is up
there on the screen. Does anyone remember the
method to actually solve this differential equation? This is the easy one-- separate the variables. So in this case, we can
divide each side by I of of X, multiply each side by X.
I'm going to bring this up so I'm not bending down. So we have dI over I equals
minus sigma ab n of B times dX. Integrate both sides
and we get log of I equals minus sigma ab n of BX,
and some integration constant. You can apply an initial
boundary condition to say at X equals zero, the
intensity of the being x was some intensity I naught. Whatever intensity of the
beam that we initially fired at the target. And by combining
these two, you end up with the expression
you get right there, which is that the intensity
of the beam coming out is the initial intensity times
e to the minus sigma ab nbx. And we've kind of
derived the idea of exponential attenuation. For those who haven't
seen that word before, attenuation or the gradual
removal of the beam of incident particles by whatever
the target happens to be. This quantity right
here, we actually have another symbol for it,
which we give big sigma. And in this case, big sigma
we call the macroscopic cross section. I'll draw a box
around these so we know these are our symbols that
we're keeping defined here. And so you may see that the
microscopic cross section just depends on single reactions
between the incoming atoms A and the target atoms B.
The macroscopic cross section depends on how much B is there. So if you want to get per atom
probabilities of absorption scattering, whatever
thing you're looking at, you use the microscopic
cross section. And if you have a finite
amount of stuff there, and you know the number
density of your substance B, you can use the
macroscopic cross section to get actual total
probabilities of beam attenuation-- or to calculate
exponential attenuation. We're going to see this
again in another form when we talk about
designing shielding, and how much
shielding do you need to remove how much of the beam? Well, this quantity
right here, there's actually tabulated values for
a lot of this stuff at the-- on the NIST website. And I have links to that as
well on the Stellar website, so you can-- instead of having
to look these all up on JANIS and multiply number
densities, there are some easier graphical
functions you can just find the value for. But we'll get back to
that in a few days. So anyway, on reading
the KAERI table, there's a few
quantities right there. We've already defined
what the excess mass and the binding energy is. And I want to note
right here, if you want to actually calculate
binding energies by hand, which I'm going to ask you to
do a bit on problem set 2, you'll need to know what
the mass of the proton, and the neutron,
and the electron are to, again, usually
like six or seven digits is the idea behind this course. Notice that they're not
exactly one atomic mass unit, because one atomic
mass unit, again, was set with that
carbon-12 standard. I'm not going to use the word
gold standard because that's a misnomer in this field. And so like I said, what
does excess mass really mean, physically? Not much, because
it's the comparison to an arbitrary standard or
a rather poor approximation of the mass. The binding energy actually
does represent the conversion of mass to energy
when you assemble a nucleus like Voltron-style
from its constituent nucleons. So let's try a few examples
in class right here. I'd like you guys to follow
around and try and calculate the binding energy of each of
these three nuclei of sulfur. Let me get a better blank
board so we can follow along. And there's a few different
ways of calculating that binding energy. You can do it by
the excess mass. You can do it by-- let's go back to the
table of nuclides so I can show you
how I would do it. Let's start with sulfur 32. And we'll write up the
quantities that we're-- that we know. Let's say the excess
mass is the actual mass minus the mass number. The binding energy is Z times
mass of hydrogen plus A minus Z mass of a neutron minus the
actual mass of that nucleus with AZ c squared. And then what we can do is
rearrange this excess mass, isolating the mass term right
here, and make a substitution. So we can say the
mass is actually the excess mass plus A.
Stick that in right there, and now we can calculate and
confirm the binding energies that we see right here from
tabulated excess mass values, atomic number, mass number,
and the masses of a hydrogen atom and a neutron,
which, for reference, I'll write up here as well. So the mass of a
hydrogen is the mass of a proton plus an electron. So 1.0072-- 007276 plus 0.000-- make that a little
easier to read-- how many zeros-- 00054858 AMU. Mass of a neutron, surprisingly
close to Chadwick's prediction. 8664 AMU. So now I'll head back to
the table of nuclides. And let's see if you
guys can follow along. What we want to do is try
to confirm this binding energy using the atomic
mass, the excess mass, or if we don't even
know the atomic mass, we can use the excess
mass plus A right there. So let's see-- Z, in this
case, for sulfur, is 16 times the mass of hydrogen. This
is definitely a calculator moment, because like I said,
I don't know about you guys, but I can't do eight
significant digits in my head. 0054858-- 1.007855--
probably enough digits-- plus 16, because there's-- mass number here is 32. The atomic number is 16. That leaves us with 16 neutrons
times the mass of a neutron, 08664-- minus the excess mass, which
in this case is 26.015 MeV-- 015 MeV per c squared. So thanks for that-- thank
Jared for that question because, indeed, the excess
mass, if you want to write it in terms
of a mass, should be in MeV or keV per
c squared minus A, which is 32 times c squared. So let's do all this out-- shouldn't take too long. 007825 plus 16-- 1.008664 minus 32 minus
26.015 divided by c squared. It's basically nothing. Gives us on the order of-- let's see-- times c squared. What did we get right here? AUDIENCE: Is the 26 negative? MICHAEL SHORT: Ah, let's see. I believe it is, because we
have to subtract the mass, and we're substituting
in this delta-- AUDIENCE: Isn't
the delta negative? MICHAEL SHORT: Oh. Good point. There is a negative there. So that's minus negative that. And A is 32. Thank you. Yeah, good point. Let me try this again. Ah, I know what I'm doing wrong. This part right here, we
want to convert to AMU. So we can take our minus-- thank you-- 26.015 MeV
per c squared and divide by our conversion
factor, 931.49-- let's see-- MeV per
c squared per AMU. What does that give us? 26.015 over that. 0.027928 AMU negative. Let's put that in
and see how we do. So plus 0.027928 minus 32. And then we get 271.-- I'll just say 764 MeV. I think six digits is enough. The actual binding
energy, 271.780 MeV, so we're off by 16
electron volts-- close enough. Also note that I used a
five-digit accurate conversion factor. That might be part
of the source of it. Does someone have a question? AUDIENCE: Yeah. In the equation on top,
you did the atomic number times the mass of the proton,
but in the one on the bottom, you used atomic
mass times the mass of the hydrogen
including the electron. Is there-- MICHAEL SHORT: Oh yeah. I actually added the two. So the mass of an electron,
since it's got that extra zero, makes so much of a-- so the mass of-- oh-- yeah. The mass of hydrogen
would be the proton plus the electron right there. AUDIENCE: Right. But why do hydrogen, though,
if [INAUDIBLE] just the proton? MICHAEL SHORT: Oh, because
there's an electron there, too. Now this can
usually be neglected because it's such a
small fraction compared to everything else. So now we're talking
about-- what-- the fifth or sixth
decimal place. But just for exactness,
I stuck on it. Yeah, in your calculations,
you can try with and without, and I think you'll find that
it doesn't matter that much, because in the end we get the
binding energy that we see on the table to within 16
electron volts for a total of-- yeah-- 271 MeV. That's pretty accurate. Yeah. AUDIENCE: If you
wanted to calculate the like energy released
from a reaction, would you do the binding energy
for [INAUDIBLE] reactants that's trapped products
for the reactants? MICHAEL SHORT: That's
the next slide. We'll get right there. Yeah, so you're catching
on to where we're going. So once you can calculate
either the excess mass, or the binding energy, or the
total mass of any nucleus, you can start to
put them together into nuclear reactions. So since you asked, let's
take a quick look at them. Where is our nuclear
reaction board? Anyone mind if I hide
this board for now, so we can go back
to our original? OK. Let's take a look at
this reaction right here, the actual boron neutron
capture therapy reaction. And now we can get towards
calculating this Q, what the difference is between the-- the total energies of the
products and the reactants, and where does that go? So now we can either
look up or calculate the binding energy of
each of these nuclei, subtract off the energy of the
gamma, which I've looked up already, is about 0.478 MeV. And we can figure out what the
total Q of this reaction is. So in this case-- I'll skip ahead
to the slide where I've got it because that way
I won't write anything wrong on the board-- got everything right up here. We assume that both
boron and the neutron have roughly zero
kinetic energy. And at the end,
they come out with some other kinetic energies
as well as this gamma ray. The sum of this energy
differences, we refer to as Q. And we can actually
confirm this total Q with a few different methods. In this case, it's always
conserve something. That's the whole
theme of this course, is you can conserve
total masses, you can conserve total
kinetic energies. We may not know those, but
tabulated in the KAERI table are the binding energies
of each of these nuclei. So let's try that out right now. So let's look at the
binding energies of each of these nuclei and see
what the difference is, the total energy released. First of all, what's the binding
energy of a lone neutron? Anyone have any idea? I see a lot of these-- zero. Yep. You haven't assembled an
nucleus out of a lone neutron, so we'll go with the neutron has
a binding energy of zero MeV. Boron, not quite
the case, but we can go back to the table of
nuclides and punch that in-- boron-10. We can look up its
binding energy, which is about 64.7507-- I keep saying about, which
is exactly 64.7507 MeV. And then our other
two nuclei, helium-4-- so you can punch
in helium-4 here. It's got a binding energy
of exactly 28.295673 MeV. And finally, lithium-7,
let's punch that in. I think you guys
are going to get very familiar with this table. There's a few
versions out there. There's a new slick
Java version that I found a little hard to use. So I like the text-only version,
because it's just as simple and fast as it gets-- 39.244526-- 526. So any sort of increase in
total amount of binding energy between the reactants
and the products is going to release
or absorb energy. Now because boron does capture
a thermal neutron, or a neutron with approximately zero
eV of kinetic energy, does anyone have any
idea whether this would release or consume energy? In other words, do think this
is an exothermic or endothermic reaction? Yeah, Alex? AUDIENCE: I'm guessing
that heat would be released through the
material-- the capture material would be heated up. MICHAEL SHORT: OK. Indeed. If the total Q value
is greater than zero, we refer to this as exothermic-- kind of like in chemistry. And if Q is less than zero, we
refer to this as endothermic. So let's do our binding
energy subtraction now. We want to figure out how
much excess binding energy is released. So I'm going to
take the reactants-- I'm sorry-- I'm going
to take the product. So helium 295673--
add lithium 244526-- subtract boron 0.7507-- subtract
the neutron, which is zero, and we're left with 2.79 MeV. And because it's positive,
this is an exothermic reaction, which is what we'd expect,
because this reaction actually happens. If this was an
endothermic reaction, what could you do
to make it occur? Yeah? AUDIENCE: Heat up the reactants. MICHAEL SHORT: Like
with temperature, or what do you mean? AUDIENCE: Make them have
higher kinetic energy or-- MICHAEL SHORT: There you go. So actually-- yeah-- you kind
of said the same thing twice. Heating things up does give
them higher kinetic energy. If you rely on
temperature, you'll be imparting eV worth
of kinetic energy. But if you accelerate
them, or get them from a different
nuclear reaction, and you get them up
to the MeV level, where whatever this Q value
could be might be negative, then you can get the
reaction to occur. For example, what is
the Q of that reaction? AUDIENCE: Negative 2.79. MICHAEL SHORT: Negative that. So in this case,
if you want lithium to absorb an alpha particle,
and make boron and a neutron, you would have to
accelerate the alphas to that same amount of energy
in order to get this to occur. So nuclear reactions do go
both ways, just not as easily. Kind of like chemical
reactions, you can drive them in
different directions by changing the temperature
or changing the concentration of the reactants. Here the concentration
doesn't matter. But the kinetic energy related
directly to the temperature definitely is. And so in this
case, it's 2.79 MeV. If I tell you the gamma ray
takes off 0.478 MeV of that, we're left with 2.31
MeV between the lithium nucleus and the helium nucleus. Now my next question-- my
last question for you today-- oh man-- is what's the split? I think I don't want to keep
you longer, because it's one minute of 10:00. So this is the
question that we're going to pick up
with on Thursday, which is how much of the
energy is taken off by helium? And how much is
taken off by lithium? Sorry, I should have kept
better track of the time.
