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visit MIT OpenCourseWare at ocw.mit.edu. MIKE SHORT: All right. So I am super
excited about today because this is, in my
opinion, the highest point of the apex of
the course where we're going to put together
everything you've done so far and start to explain things
like Bremsstrahlung, radiation damage, X-ray spectra that
you get in a scanning electron microscope, and actually find
a way where cross-sections are areas. Remember before, I told
you guys cross-section is measured in barns, in
centimeters squared? Think of it kind
of like an area. We'll actually be able
to show mathematically that some of them do
derive from actual areas. So it will make
a lot more sense. It's not just an
abstract concept. What I want to do
is a quick review of the ionization and
excitation collisions that we did last time. Remember we had this imaginary
hollow cylinder where we said that there
is some ion traveling in this direction with
charge, let's say, z times the electron
charge and it's colliding, kind of, with an electron
somewhere else separated by some impact parameter b. And let's say this
hollow cylinder head a shell of thickness db. And we started off
with this situation where we wanted to say-- we can find the y momentum as
the integral of the y force. We did something. And one of the
intermediate steps we came up with was that our
energy in part of the electron is p squared over 2 times mass
of the electron, which came out to 2z squared e to the 4th
over the mass of the electron impact parameter squared
velocity squared. Then we multiplied by
the electron density in the material, which was the
number density of atoms times z, the number of
electrons per atom, times the area, the
cross-sectional area of this hollow cylindrical
shell, which came out to 2 pi b, which is the
circumference of that circle right there, times db dx. I'm just going to leave
that there for now, and we'll come back
to it in a second. I will mention, though,
that at the end, we came up with our
stopping power expression. Let's call this
ionizations that came out to 4 pi, this constant
k0 squared, which comes from the Coulomb force
law, little z squared, big Z, e to the 4th over mass of the
electron velocity squared-- I'm running out of room-- times log of mass of
the electron v squared over this mean
excitation energy, which comes out to around,
let's say, 10 to 19 electron volts times z. And what this meant was that
if we graph stopping power as a function of energy, there's
a couple components to it. One of them is this roughly
1 over energy component. So let's say there's
a component of it-- yes, I love when I can do that-- that actually-- that
follows 1 over e. And there's this logarithmic
component that goes that way. And if we sum them together,
we ended up with a much higher stopping power at low energies. But around this local
minimum of about 3 times the mass of the
electron c squared, it starts to increase again. And that's due to this. How many excitations can you
make as a function of energy? So if you think about it, this
is kind of like an energy term. How much energy do you have
divided by how many ionizations can you make? So it's kind of an intuitive
result in that it's how much energy you've got
versus how much energy it takes for a single unit process. And that's all mediated
by this 1 over e term. The last thing that
we said is that this curves back down here. And that's because at around
500 times this mean excitation energy, which as you can see
is around 1 keV times z-- it's not a very high energy-- you start to get
charged neutralization. The reason for that is that
for-- in order for this formula to work, we have to assume
that the deflection is really, really, really small,
infinitesimally small. If it's not, and if the ion
manages to catch that electron, it's not going to lose energy
by undergoing a Coulomb interaction. It's going to lose energy
by absorbing that electron and neutralizing. And so for really
low energy, the ions are moving so slowly
that they can capture some of those electrons. And it gets less and less
effective at stopping the material. And this led to-- let's see-- this,
we're distance, and this, we're dt dx
times the energy required to make an ion pair, which
we'll just call this i. We ended up with a curve that
looked something like this. This right here
we call the range. We just gave that the symbol r. So that's the
review of last time, but I left some extra space for
some reasons which will become clear in about half an hour. Right now I want to take
you through a little bit of a whirlwind in terms
of Bremsstrahlung. I'm not going to derive
anything about it. We're just going to go over what
the cross-sections and stopping powers look like, why they
take that form intuitively. But we're not going to go
through a rigorous derivation because, well, we
simply don't have time. But I do want you to know
what sort of things exist. So for Bremsstrahlung, better
known as braking radiation-- who's actually heard
of this before? Does anyone know roughly-- let's say if we were to
say some cross-section for Bremsstrahlung. Let's call it
cross-section radiative. Because in this
case, we're actually talking about the charged
particle radiating away photons. So let's say we had some
nucleus of charge big Z, and we had some ion-- we'll look at a
different color-- some ion of charge little
Z moving towards it. It will either be
attracted or repelled depending on what
the sign looks like. And sometimes you might
get radiation of a photon. Let's call it of energy
E equals hc over lambda. That'll become
important in a sec. So this radiative
cross-section-- I don't care about--
particularly care about the exact form, but
what is it proportional to? What sort of
factors do you think would make the emission of that
photon more or less likely? Yeah? AUDIENCE: The energy of
the charged particle. MIKE SHORT: Sure, the energy
of the charged particle. So you can-- there's one
expression I like you. Can only take my money
until you take it all. Well, the same thing goes
for the energy of a photon. You can't radiate any
more than the energy of the particle
coming in, so there's going to be some
maximum energy which is going to correspond to
some minimum wavelength, which is-- let's say if we just put
our initial energy in here, that gives us our minimum
wavelength right there. That's true. And it can radiate at any
energy smaller than that or any wavelength
larger than that. I'm basically saying
the same thing. Because if this particle
started off further away and felt less of a deflection,
it could still emit a photon, but of a longer wavelength. Don't have room to draw
enough wavelengths, but I want to make sure
that's physically accurate. So it's actually going
to be also proportional to z of the large nucleus. The stronger the pull, the
more of that breaking radiation you're going to see. It's actually
proportional to z squared. What this says is that heavier
z materials produce a lot more of this breaking radiation. And it's also proportional
to the inverse of the mass squared. What this says intuitively is
a heavier particle deflects less and emits less of
this breaking radiation. Hopefully this makes a
lot of sense to you guys, that the stronger the pull
of the nucleus, the more deflection you're going to get,
and the more Bremsstrahlung you'll get. The larger the mass
of the incoming ion-- let's say that's mass m-- the less deflection you'll
get because the less-- what is it? The less momentum transfer you
can apply with the same force if you've got a
heavier particle. Does this make
sense to everybody? So what this says is it's really
important for high z materials. And we'll see a little bit why. If I actually write the full
expression for stopping power-- and I promise to
you I'm not going to derive it because we don't
really have the time for that. It's proportional to
the number density. You should always think there
will be a number density and stopping power, because
the more atoms there are, the more they stop things. It's just directly proportional. Times that kinetic
energy plus mec squared. And again, this is not something
I want you to memorize, but it is something
I want you to be able to decompose and explain
why the parts are there. Let's see. Times some radiative
cross-section where this sigma radiative is
some constant cross-section. This ends up being about
1/500 barns times z squared times this parameter b, which
if you see in the reading is actually given just-- this b scales roughly
with the atomic-- I'm sorry-- yeah, with the
proton number of the material. So you can see that in
here, in the stopping power is actually directly
the cross-section. So the components of
a stopping power-- there's going to be some
probability of interaction, and there's going to be
some energy transfer part. This is an interesting
result to show. This is why I wanted to just
write the Bremsstrahlung stopping power. Because in here, actually,
is the cross-section times some other stuff. Pretty neat result.
And so now I want to show you how our
cross-section is actually contained in the
ionization stopping power. So let's bring this
back down for a sec. You can think of the likelihood
that the ion comes off with any particular energy
to be directly related to this impact parameter. Because as we saw, the final
expression for stopping power is directly related
to this parameter b. We ended up integrating
over all possible b to get the total stopping
power in the material. But if we didn't
do that integral-- if we stopped, let's say,
at this stage in the game, and we said, all right,
well, the stopping power at some fixed b actually
depends on the probability that that particle enters into
this cross-sectional area right here, this 2 pi b db,
that right there is the cross-section for
scattering as a function of the incoming energy
and the outgoing energy. This is one of the
coolest parts, I think, is that there is an
actual area right here. It's the area of a hollow circle
is the actual cross-section for scattering with a
given ingoing and outgoing, outcoming energy. And then the rest
of this stuff-- if you pull all this together,
if you take a microscopic cross-section times the number
of particles that are there-- because this is your
atomic number density, and these two together are
your electron number density. Like we talked about
before with reaction rates and cross-sections,
this thing right here is your macroscopic
cross-section for an incoming and an outgoing energy contained
directly in the stopping power formula, because
then we integrated over all possible
cross sections, which means all possible outgoing
energies for a given incoming energy. And so the last bit, the way
to link these two together, which is why I left a little
bit of space right here-- we know right now, because
I wrote it up there, that the scattering
cross-section as a function of the ingoing an
outgoing energy per unit energy is just the area of
that hollow circle. So let's divide
everything by dt. We end up with the total
formula for cross-- for the scattering
cross-section. We don't know what
this b db dt is. What's the differential
probability between impact parameter
and outgoing energy? We don't quite know,
but we can express this as a change of variables
that we do know. We do know-- if we have a
certain impact parameter, we know what the scattering
angle is going to be. There is a well-known
relation for that. And there's a
derivation in the book and in another book that
I want to point out to you guys by Gary Was called
Fundamentals of Radiation Material Science on page 32. If anyone wants to see
the derivation from which this result came from,
you can head right there, and it's free on MIT Libraries. Meanwhile, we do know our
relation between this impact parameter and the
angle, and we do know a relation between this
angle and the outgoing energy. And this is where some of the
hard sphere collision stuff comes in. What's the maximum
amount of energy that a particle can
impart to another particle in some sort of a hard,
sphere-like collision? Let's take the easy example. If the two particles
have equal mass, how much energy can one
particle impart to another? AUDIENCE: All of it. MIKE SHORT: All of it, right? It can impart a
maximum of, let's say, this incoming energy ei. As those mass ratios change,
you can impart a maximum-- let's say your
maximum becomes what's called gamma ei where
this gamma right here is 4 times those two
masses multiplied over the sum squared. The full expression-- I'm
going to use a different color because I'm running
out of space here. The full expression for t is
actually gamma ei over 2 times 1 minus cosine theta. The two intuitive limits from
this is if theta equals pi, then this here equals 2,
and t is our t maximum, just gamma times ei. If theta equals 0, that
whole thing equals 0. And in our case of
forward scattering, no interaction occurs, and
the energy imparted is 0. But the important
part here is we have a direct relation
between t and theta and the angle, which
we can put in here. And we actually have a direct
relation between b and theta, which I wrote down
so I wouldn't forget. So we actually have--
our impact parameter is classical radius
of the electron times cosine of angle over 2. You don't have to know
where these came from, but the point is we have
a relation between B and the angle. We have a relation
between angle and energy. So we can just do a
change of variables to get our final cross-section. And it ends up being, I think,
pi times radius of the electron squared over gamma. So in this way, we can go from
known relations between each of the variables and an
actual physical cross-section that has units of real area
down to an energy-dependent form for this cross-section,
which I think is pretty cool. And then this is
the one that you would see tabulated
in the JANIS tables, like the energy-dependent
cross-section for the S reaction or the scattering. So this is one of my favorite
parts of this course, because you can see how
cross-sections really do follow directly from areas. So now for the other
part, now that we've got this Bremsstrahlung
stopping power and we've got our
ionization stopping power, it's useful to find out which
one is more important when. To do that, we just
look at their ratios. So if we look at the dt dx
from ionizations over the dx dx from radiative energy
transfer or Bremsstrahlung-- let me make sure I
get this one right. It's proportional to z times
mass of the electron over m squared times t over 1,400
rest mass of the electron. So what this tells
you here is that-- I'm sorry. I think I have those
backwards, because radiative should get more important
at higher energies. So what this tells you is
for higher z materials, Bremsstrahlung
becomes more dominant, and for higher
energies, Bremsstrahlung becomes more dominant. So if we want to generalize our
stopping power curve from just ionization to everything-- so I know I had another color. No. I ran out of colors. I need a fourth one. There's going to be some
Bremsstrahlung component that starts to get more
and more important with increasing energy. And so then if you
extend this curve, you're going to radiate
more and more and more power the higher energy you go-- not from this component or
that component of ionization, but from radiation
or Bremsstrahlung. So this has some pretty
serious implications to answer questions like,
how do you shield beta rays? Does anyone have any idea? Based on this
formula right here, what would you use to
shield beta particles and not irradiate the person
standing behind the shield? Let's ask a question
everyone knows. What do you use to shield
photons really well? Lead, tungsten,
something with high z. Because as we saw from before-- I'm going to steal a little
bit of the Rutherford stuff. If you graph the energy
versus the mass attenuation coefficient, you get a
curve that looks like this, but everything increases
with increasing z. You get more mass attenuation
with increasing z. And also, denser materials
tend to be higher z. Is that what you want to
do for beta particles? You say-- so Monica,
you're saying no. How come? AUDIENCE: Don't you
just want something with a low cross-section? MIKE SHORT: That's right. Well, you don't
necessarily want something with a low
cross-section, or else it might not shield at all, but
you are on the right track. You can actually look
at the difference between these stopping
powers, and cross-sections are embedded in there. So I think that answer
is pretty much correct. But also, you're going to
get more Bremsstrahlung or more breaking radiation
in higher z materials. So if we actually look at what
thresholds does this become important in lead, this ratio
is about 1 at around 10 MeV, which means that you lose
an equal amount of energy to Bremsstrahlung as ionization
at 10 MeV for electrons. In water, this ratio
is about 1 at 100 MeV. So what this says is if you
want to shield electrons or beta particles safely,
you actually have to use lower Z
materials because they won't make much Bremsstrahlung. But because, like Monica
said, then the cross-section is lower, you actually
have to use more. So you don't have a choice. You can't just use
less high z material. Because while you will
stop more of the electrons, they will create more
x-rays in the process. And those x-rays are
highly penetrating, as we know from these
mass attenuation curves. Once you get to high
energy, this is-- these are logarithmic
scales, so let me correct those and
say these are log of e and log of mu over p. It gets millions of
times less effective at shielding high
energy photons. So that's one of those really
important things to note is if you're designing
shielding for something, and there are electrons involved
that are even around 1 MeV or so, you can't just use high
z materials to shield them, or you will create more
problems than you solve. That's a pretty
important implication. It's quite important for what's
called betavoltaic devices. It's kind of a
sidetrack, so I'm going to stick it on a board
that'll be hidden soon. Has anyone heard of
a betavoltaic device? Anyone? What are they? AUDIENCE: It's
like a beta source that emits electrons onto a
semiconductor [INAUDIBLE].. MIKE SHORT: Yeah,
it's a beta battery. All it is is, let's say,
some pieces of silicon, some circuit that grabs the
power, and a beta emitter. And these beta particles
directly hit the silicon, and the movement of those
betas constitutes a charge. And it's direct-- it's direct
conversion of radiation to electrical energy. They're not very high power, but
they last for a very long time. How long? Around a few half lives
of that beta decay. So for most of these
beta emitters that have half lives in the realm
of, like, 10 to 1,000 years, you can make a microwatt battery
that could last for millennia. This could be pretty useful. Let's say if you wanted to
have some secret sensors in a naughty country
like North Korea, you could drop these tiny
little beta particles that would just-- betavoltaics that would
just trickle charge a battery, make a measurement of-- I don't know-- radiation level,
or weight of the dictator, or whatever you happen
to want to measure, and send that off once
a month or once a year with no need for
external monitoring. Or let's say you're designing
a mission to land on a comet, like the Rosetta Philae Lander,
and your radiothermal isotope generator is going to burn out
in, let's say, 10 or 20 years. You might not need
that much power just to measure temperature,
or light levels, or something else, or
a gas that you might want to know what's there. But you have to choose
your beta isotope wisely. If you want to make these
things in a little chip-- and they actually have been
commercialized in a chip that's about that
actual size using about two curies of tritium. Anyone have any idea why
one would choose tritium? AUDIENCE: It's got
a short half-life. MIKE SHORT: Yeah, it's
got a short half-life, so you can get a lot
of power out of it. That's one of the
two correct reasons. And what is the other one? Lets see who's memorized
there KAERI table of nuclides. What do you think
its beta decay energy would have to be for this not
to blast anyone in the vicinity? AUDIENCE: Low. MIKE SHORT: Very low. Why do you say that? AUDIENCE: [INAUDIBLE] they
don't penetrate all the way through the [INAUDIBLE]. MIKE SHORT: That's true. Their range is much smaller. But the range of all betas
is pretty low in materials. But the answer lies right here-- less Bremsstrahlung. Lower energy betas give
most of their energy off in ionization rather than
by radiating Bremsstrahlung. So you can have a device
with two curies of tritium, which if that's released to the
outside world, that's bad news. That's something that
you might have to report. But as long as it stays
contained in this device, it does not have enough
energy to produce many x-rays from Bremsstrahlung. And therefore, it does not
require an enormous amount of shielding. So you can't just
pick a 1 MeV beta emitter which you might
get a lot of power out of, because it's also
going to be a big, crazy X-ray source that you wouldn't want
in a cell phone or a sensor or some other device you
might put in your pocket, or even 20 feet from you. Cool. So that's the idea
behind Bremsstrahlung. There's a little bit more
I want to tell you about, and I'll save that for
the sidetrack board. We use Bremsstrahlung in a
lot of really interesting applications, including
cyclotron, one of which we just took delivery of here
at MIT, or a synchrotron. And I'll just briefly
explain how these work. In a cyclotron, you've
got two D-shaped magnets. They actually call
them dees because we're so creative in
naming these things. You inject some source
of charged particles, and there is some electric
field lines across these two dee magnets. And what this says is that
in between the magnets, the particle accelerates. And inside each magnet,
the path curves. And it accelerates some more. And it's moving even faster,
so it takes longer to curve. Than it moves even faster,
and it takes longer to curve, and so on and so on, until it
finally shoots out the side. And so this is one way that you
can have an extremely compact-- and I'm talking like
garbage-can-sized-- accelerator that brings
things up to about 13 MeV. That's the one that we've got
in the basement of Northwest 13. The problem is every
time these particles bend, they send
off photons, what's known as cyclotron radiation. And the higher energy
that is, the more intense that cyclotron radiation gets. So you've got this
garbage-can-sized device with a little hole right here, and
it's just blasting out photons in all directions
in this one plane-- let's just call it
the plane of death-- which you don't
want to be in, which is why this the thing is behind
4 feet of concrete shielding, and in the middle
of a room, to help-- that 1 over r squared
keeps your dose down. But we actually use this plane
of death in a synchrotron. What it is is it's a
circular accelerator. It's not quite circular,
so let me correct my drawing a little bit. There are straight segments,
and there are slightly curved segments. But it pretty much
looks like a circle if you look at it
from high up enough. In each of these
curved segments, there is a bending magnet. That's my best
drawing for a magnet. And what this does
is it continuously changes the path of these
charged particles going through usually electrons. And you end up
with intense beams. Let me use a different color. You end up with intense
beams perpendicular to the original path before
it went in that bending magnet of synchrotron radiation. So it's kind of like
a gigaelectron volt spinning ninja star of
death, except at the end of every one of these
stations, you have what's called a beam line. Because there's 60 or
80-odd of these beam lines coming off with,
let's say, 80 kv and below Bremsstrahlung
x-rays, you can use those for a whole lot of
different analysis techniques. You can simply irradiate things. You can send those x-rays
through a monochromator to select only one wavelength,
and then use that wavelength to probe the structure of
matter down to the atomic level. There's actually one of these
just down in Long Island. About a 2-and-1/2-hour
drive from here, there's Brookhaven National Lab. And they just opened up the
National Synchrotron Light Source, or NSLS version
2, where they can actually measure distances with
single nanometer precision. So inside this beam
line is a bigger room which is encased in
another room which is encased in another room. And the whole point of that is
for vibration and temperature isolation. So they maintain
this entire room to within a speck
of 0.1 Celsius. And it's the least vibrating
place, probably, in the US. I don't know about
on the planet. But it's got basically
no vibration. So the atoms are
effectively standing still except for their normal
vibrations in the material. But there's no source
of external vibration. And the cooling has
to come in through these convoluted channels so
as not to blow on the sample, so as not to make any convection
currents or temperature changes. And they can actually probe
the structure of matter with single-nanometer precision
using these synchrotron x-rays all produced by Bremsstrahlung. So it's not all bad. You can use
Bremsstrahlung for good. Then there's a little bit-- I have to hijack
a little more area from Rutherford scattering. You might think
about, well, what is the actual spectrum
of this Bremsstrahlung. Well, you can look to see
what's the probability that an atom enters into any
of these concentric, hollow circles. It looks to be less and
less likely that you're going to enter through
one of the center rings and more and more
likely that you're going to enter through
one of the outer rings. If you start farther away,
there's less of a pull to change the path of
that ion or electron, and the Bremsstrahlung is
going to be lower in energy. This is actually
described by what's called Cramer's law, which
says that the intensity of the Bremsstrahlung as
a function of wavelength scales with some constant k,
and that constant scales with-- surprise, surprise--
the atomic number of the material
times some lambda over lambda minimum minus 1
times 1 over lambda squared. And what this says
is that there's some minimum lambda
or some maximum energy that you can impart to
this Bremsstrahlung, which again, you can only take some
energy before you take it all. And there's going to be some
sort of a fixed minimum lambda if we draw this intensity. And I graphed this on Desmos
just before coming here, so I know it looks something
like this where that right there is lambda minimum. It's taking more area. If you then change variables
from lambda to the angular frequency where,
if you remember, the energy of the photon is just
h bar times that frequency-- so it's kind of like converting
into energy with just a tiny, little constant in front. And I mean really, really
tiny little constant. You end up with
an energy relation that looks like some
maximum angular frequency or some maximum energy. And this is kind of a simple,
linear-looking relation, this 1 over energy relation minus 1. So if we graph energy
versus the intensity of the Bremsstrahlung,
you end up with a curve something
like this where your max energy is the same as
your incoming particle energy. Now, who here has done any
sort of X-ray or SCM analysis before? You have. So can you tell me, is this
the Bremsstrahlung spectrum that you tend to see? AUDIENCE: Well, I've
done [INAUDIBLE] analysis with imaging. MIKE SHORT: OK. Have you ever gotten a
regular, old X-ray spectrum to see what elements are there? Can you draw what
one looks like? AUDIENCE: Maybe. MIKE SHORT: You want to try? They're all the same. So if you remember any
particular one, you're correct. Yep. There's some peaks. And then what does this
background stuff look like? Yeah. There's some noise and junk
on the back of it, right? So this is actually correct. Thank you. And what you actually
see here is a bunch of characteristic peaks. These will maybe be like
the L lines and the K lines for one element or another,
these characteristic X-ray peaks, on top of
the Bremsstrahlung, the breaking radiation
which constitutes the background here. And what you actually see-- I'm just going to draw
the background curve under Julia's curve here-- looks something like this. What happened to
the real spectrum? Why don't we observe
what actually exists? There are a couple of reasons. Does anybody have an idea? So let's take this
to the extreme. Why don't you think you
would observe physically-- and this is when we actually
get into the real world-- any x-rays with energy
in, let's say, the eV range if you were to try and
observe any x-rays at all? This is where we
actually get into what do these detectors look like. So there will be some active
piece of your material if this is your detector. This is most definitely not
Rutherford scattering anymore. And there's got
to be some window. We can make it as thin
as we possibly can. And they make it out of the most
X-ray-transparent structural material that they can,
which tends to be beryllium. So beryllium has got
an atomic number 4. It's the first and
lightest element that you can make
structural anythings out of. So if you want to protect
your detector from, let's say, air or something-- if
this were full of air, it would absorb the
x-rays, so you want there to be pretty much nothing. You can put a very thin,
seven-micron beryllium window in front. But the problem is
we've already got one of these mass
attenuation curves. And when you get down
to these energy levels, you attenuate everything. So the lower energy
your Bremsstrahlung is, the less likely you're
going to see it. So even though this is the
actual Bremsstrahlung spectrum, this is what we observe. And I haven't finished
grading the tests yet. But I like I promised, for
the two folks who do the best, I'm going to ask you
to bring something in for elemental analysis. This is precisely what
we're going to see. You're going to see this
Bremsstrahlung which is not the actual spectrum
coming out, but this has to do with the
absorption of x-rays in the detector window, as
well as some self-shielding. If we're using a scanning
electron microscope, which is nothing more than
an electron gun, and you're firing electrons to
some distance in the material where they'll then interact
and send off x-rays, you've also got this part of
the material to contend with, some self-shielding. So not only do the x-rays
all have to get through the detector window
to be counted-- so the high-energy
ones, which we'll have with small
wavelength, get through, but the low-energy or
long-wavelength ones might get stopped. You also have to get out
of the material itself. The electrons don't
just produce x-rays in the outer atoms
of the material. They go down a micron or two. And then the x-rays that are
produced in those interactions have to get back out again. So it's interesting. It's kind of like the inverse
photoelectric effect, right? In the photoelectric
effect, photon comes in, electrons come out. In a scanning electron
microscope, electrons come in. Photons come out. Many of them are these
characteristic x-rays. Because now if we
start to review what sort of interactions are
possible when we fire electrons into material, we've just
gone over Bremsstrahlung. And we know that with higher
and higher energy electrons, you're going to get more
and more Bremsstrahlung. But you're not going to see
the actual x-rays produced at low energies no matter
what, because this isn't just a system on paper. It's real life. And you're going to get
characteristic x-rays that come from energy transitions. So if you fire in
an electron, and you happen to undergo one of
these ionization collisions, you might just knock
an electron out. So let's say an electron comes
in, knocks an electron out. Then another electron
fills that shell, giving off-- in this case, it
would be a k alpha or a shell 2 to a shell 1 X-ray
the way I've drawn it, which is why Julia
has got everything right on the spectrum here. There's the Bremsstrahlung,
and then there's these characteristic peaks. The background is due to
radiative stopping power, and these characteristic
peaks give away some of the ionization
stopping power. And so all in one spectrum, you
can see just about everything going on in this material. The last thing that you can't
see that I would be remiss if I didn't talk about it as
a radiation material scientist is radiation material
science, which really is concerned with
mostly Rutherford scattering, Rutherford or hard
sphere scattering. This is the last of
the major interactions between charged particles
and matter that concern us. It's not really in your
reading except for, I think, being mentioned once,
because they didn't seem to think it's important. But I happen to think
it's extremely important, because this is the basis
behind radiation damage. In all of these
collisions right here, you have some sort of
displacement of electrons. And those electrons
can get ionized, and other ones will fill
them back in the holes and whatever they'll do. But at no point were
nuclei displaced. You can transfer a lot of
energy without moving any atoms around. But when your energy
starts to get lower, you end up with a new
kind of stopping power-- let's call it nuclear-- which scales with, as always,
a number density times pi. Let's see. Little z, big z, e the 4th-- everything looks pretty
similar so far, except for now we've got the energy of
the incoming material, and now we have a mass ratio. Because in this
case, you're actually undergoing some sort
of a hard sphere collision between one
atom and the other times the natural log. This is going to look
awfully familiar. It ends up being some
energy term over-- actually, let's
just go with-- yeah, gamma ei over some new energy. This thing right here is called
the displacement threshold energy. And it ranges from about 25 to
90 eV, but it's usually 40 eV. And what that is--
it's the max-- the minimum amount
of energy that has to be imparted to a nucleus
smack head on in order for it to move from its
original atomic position. And that's what's known as a
hard sphere type collision. Or in this case,
it's just like all the other
q-equation-looking scenarios that we looked at before. So let's say the little
nucleus goes off, and the big nucleus goes off. This should look familiar
by now because I've harped on it probably too much. Now what I want you to
consider is this big nucleus had a position. It liked where it was, and
now it's been knocked away. What's left over is
an atomic vacancy, which is the most basic building
block of radiation damage. So sometimes it's neat
to look at the ratio of-- let's see. Make sure I get
this ratio right. Ionization on top. To look and see when is
ionization versus radiation damage actually important. And the ratio scales with 2
times the mass of the nucleus over MeZ times their respective
natural log threshold things. And that's the ionization
potential and log gamma ei over ed. So what this says is
that for higher energies, ionization is more important,
and for lower energies, nuclear stopping power
or radiation damage is more important. If we graph these two, let's
say, on a log e graph-- and let's say we have
our nuclear stopping power in blue and our ionization
stopping power in green-- we end up with curves that
look something like this. That's our nuclear. That's our electronic or ionic. So what this actually says is
if you fire high-energy neutrons or high-energy protons
into a material, it's ionization that does most
of the damage at high energies until you slow down to
around like 10 to 100 keV level, curiously very similar
to this 500 times i bar, the mean ionization potential,
at which point Rutherford scattering or hard
sphere scattering becomes the dominant mechanism. And so what this says is if we
want to draw a picture of what radiation damage looks like-- let's say we had a proton that
we're firing into a material, and it hits some
atom that we're going to call the PKA or
Primary Knock-on Atom. That PKA then becomes-- let's say it was nickel. It's like a nickel plus
26 ion because you've knocked the nucleus out of its
electron cloud, effectively, and it's now flying out
through the material. That proton might go off to
do more damage somewhere else, but it's not
actually the protons or the incoming particles
that do the bulk of the final radiation damage. The radiation damage is
mostly self-ion radiation. Even though it all starts
with the incoming particle, nothing would happen if the
incoming particle didn't show up. Most of the final
results of the damage are from these heavy
ion collisions. And so we actually talked
a little bit about-- I think we talked about
when this ionization starts to pick up for electrons
versus heavy ions. If you think about when
electrons start to radiate away most of their energy, taking
it away from radiation, it's like 10 to 100 MeV. What would be the case for a
heavy ion, like even a proton? Well, what's the only
thing that changes when you change from an
electron to a proton here? AUDIENCE: The charge. MIKE SHORT: Well, yeah, the
charge is the opposite sign, but of equal strength. But what else in this formula? AUDIENCE: [INAUDIBLE] MIKE SHORT: That's right. So for heavy ions, for
even things like protons, you need to go at
approximately 1,837 squared more energy
than the electron. So we're talking in
the gigaelectron volt to teraelectron
volt range for ions. So this is why
Bremsstrahlung is not important for any sort
of ion interactions unless you are a
high-energy physicist and you're working in
the GeV or gigaelectron volt in the upper range. So we like to say in the
radiation damage field if you want to know
the total stopping power from all
interactions, you have to take into account
the ionizations. I'll just make that a
minus sign for the symbols. The nuclear and the radiative. For most radiation
damage processes except for high-energy electron
radiation, we neglect that. The reason is that the
radiative to ionization stopping power is pretty close to zero. It's like-- even at 10 NeV,
it's like 1 over 2,000 squared-- or 1 over, I guess, 4,000,000. It doesn't matter at all. With heavier ions, it
becomes even less of an issue because you can deflect
a heavier ion less with the same Coulomb. And so what ends
up only mattering is the ionization stopping power
and the nuclear stopping power. It's this nuclear stopping
power that leads to collisions. And it's like two of five of. So I want to stop here
and answer any questions. And I'll hijack a bit of the
neutron discussion on Thursday with some review
of this and filling in the last gaps of
radiation damage. So anyone have any
questions from today? Yeah? AUDIENCE: Can you
repeat what you just said about why the
radiation term goes away? MIKE SHORT: Yeah. The radiation term goes
away because of that. AUDIENCE: And that's under
the assumption you're working with a proton or a heavier-- MIKE SHORT: Yeah. If you're working
with an electron, then it actually does matter. If you're firing 10 MeV
electrons into something, you must account for the
radiative stopping power, because there's a lot of it. At 10 MeV, there's as much
radiative as ionizing, and there's basically
no nuclear yet. But for anything heavier-- even muons, which are
approximately 237 times heavier, or protons, which
are approximately 1,837 times heavier, it totally doesn't
matter because it scales with the mass ratio squared. It might be 267 for muons. I forget that middle number. But still, 267 squared
is a pretty big number. Was there another question here? I thought I had seen a hand. So remember, you
guys had said you want to see some radiation
material science or radiation damage. This is where it comes from. This is why I love
teaching graduate radiation damage and 22.01
at the same time-- because they're the same thing. Except you guys get
the derivations, and in the grad class, I
say I assume they know it. And then in the homework,
I find out they don't. But it doesn't matter
because they're supposed to. At least you guys will,
so you've got the power. And knowledge brings
fear, as I like to say. OK. I'll see you guys
on Thursday when we'll wrap up a little
bit more radiation damage, because I can't resist. And then we'll start moving
into neutron interactions, which is kind of taking a
step down from here, because there aren't really
any electronic interactions. But because we can deal
with enormous populations of neutrons, things
are going to get messy. Have you seen the
equation shirts that we have here, the neutron
transport equation shirts? Yeah. We're going to derive
that on Thursday.
