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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: I wanted to do a
quick review of all the photon interactions, because I've
released problems at 5:00 for you guys. It involves your banana data. So we're going to be taking a
second look at all the banana data and the problem statement
for the lab part is simple. Identify all the peaks. Tell me where they came from. And tell me all the peaks
that should be there that you don't see and why. And that's like a quarter of
the problem set or something. So just to review
the three effects that we talked about sort
of in order of what energies they're important in,
my spelling I'm sure will be slower than usual today. I spent like 3 and
1/2 days listening to Russian presentations with an
English translator microphone. Russian scientific presentation
is really similar to English. All the technical
words are the same, but as soon as you start trying
to talk to a two-year-old, you're just lost. But it's pretty cool. So we went over the
photoelectric effect, Compton scattering,
and pair production. And so in addition to
knowing what these three mechanisms actually are and how
to tell what they would look like on a given
detector spectrum, the other two important
things we wanted to remember are what are the
cross-sections, so what are the relative probabilities
of each happening as a function of their energy
the photon and the material that they're going in and then
filling in this map of if you have energy and z, where are
these effects most prevalent? Does anyone want to kick me off? Does anyone remember the general
form of the cross-sections for any of these effects? Or does anyone remember
what this map looks like? Yeah? Chris. STUDENT: [INAUDIBLE] PROFESSOR: Indeed. STUDENT: [INAUDIBLE] PROFESSOR: That's right. So pair production-- I think we gave it
the symbol kappa to go along with a reading
that you guys have-- was around here. That's because it's not going
to happen below 1.022 MeV, because you need the energy
to make the positron electron pair. And indeed, the more electrons
there are in each atom, the more likely
pair production is, this happens when the photon
gets near the nucleus, which is going to have a higher
charge for higher z and so on. And so pair production,
it was proportional to-- you'll never need to
know the exact things that the cross-sections. That's what books are for. But it was proportional to-- let's see, I think it was
like z to the third or fourth, pretty strong like that. What about Compton scattering? Where does that lie on this map? STUDENT: [INAUDIBLE] PROFESSOR: Yup,
high z, low energy. So in this general region. We'll just give it a C
for Compton scattering. And in that one,
the cross-section was proportional
to something like 1 over the energy or h bar
omega in your reading. That's the same thing
as saying photon energy. And then what about
photoelectric effect? Well, that's the only
place there is left, right? We'll give it the symbol tau. So I'll put these up here. I don't quite know why
they chose those symbols. But I'll just stick to the
notation in the reading. And then the idea here is this
was proportional to something like z to the fifth over-- what is it, like
energy to the like 7/2. So significantly low energy,
significantly high z. And does anyone
remember at what energy does the photoelectric
effect start to kick in? Very close to zero. So here, the energy has got to
be greater than or equal to, but what is the photoelectric
effect physically? STUDENT: [INAUDIBLE] PROFESSOR: Yep. A gamma gets absorbed,
or any photon gets absorbed that
knocks out an electron. So how energetic does it have
to be to knock out the electron? STUDENT: Binding energy
of the electrons. PROFESSOR: The binding energy
of the lowest bound electron, which we give that symbol
phi or the work function. The idea here is that as soon
as you have enough energy to eject the outermost
electron, which is super low for the alkali
metals, like sodium, potassium, cesium, then you can
exceed the work function and get the photoelectric
effect going. And for this one, we
said the energy here has to be greater than or
equal to 2 times the rest mass of the electron c squared,
better known as 1.022 MeV. Is there a minimum energy
for Compton scattering? Photons can scatter. They don't have to have
any energy to scatter. Certainly. And let's see the two
interesting bits of technology we talked about
related to these, one was called a Compton
camera, where you could actually use two detectors. Let's say you're looking for
a tiny source in a big box somewhere. You can have one detector. And you can have
a second detector, so that this source is sending
out gammas in all directions. And let's say one of them
interacts with detector one, bounces off, and
interacts in detector two. At that point,
you've constrained sort of the angle
between these detectors, so that you know what
energy the gamma came from. And you know generally where
it came from physically, which is a cool piece of equipment. I'm going to try
to find pictures of one of these actual
things, because I actually haven't seen one myself. I've just heard it described
physically and it seems to make sense. And the second one
that we touched upon at the very end of last class
had to do with this thing right here. Does anyone remember
thermionic devices? Well, the work function
for some materials, like for cesium,
the work function is a little less than an eV. It's like 0.7
electron volts, which means when you get things
up to about 2,000 Celsius or so, the temperature
of the atoms themselves exceeds the work function,
and the outer electrons just boil off. So if you have two pieces of
material, probably in a vacuum, and one of them is like 2,000 C,
and one of them is, let's say, room temperature, you end up
with this net flux of electrons boiling off the hot
one to the cold one. And this has been one
of the methods proposed to directly convert
heat to electricity for ultra high
temperature applications, like space reactors
or other things that can get super crazy hot. So it's one of those energy
conversion mechanisms-- did anyone ever hear about
this one in high school? Highly, highly doubt that
it would ever be mentioned. One of the professors in our
department, Elias Gyftopoulos was one of the folks that
came up with this whole idea. And in my senior design
course, we actually designed a space reactor
that uses thermionics, and he showed up in the
audience by surprise. And that's probably
the dumbest I've ever looked at a presentation,
explaining something that someone invented. They knew every single mistake
and everything that was wrong. So since then I've kind of boned
up on thermionics knowledge. But that's enough
for the photon stuff. Now we want to start getting
into ion nuclear interactions and in today's reading,
it started off-- I think the first
paragraph went something like, the formula
for stopping power can be expressed as follows. Squared times log of-- squared over mean
ionization energy. And I find this explanation
to be unsatisfactory. I'm not a fan of the kind
of books that just say, here's a formula. For practice, plug
things in and use them. So instead, I'm
going to skip ahead to a little bit of
next week's reading or the [INAUDIBLE] reading
and actually derive it. So when I just throw
up a formula like this, it's like how the hell do
you remember that, right? Well, it's going to make a lot
more sense once we actually derive it. So let's set up this problem. You have a charged particle
with charge little z times e. Little z, we'll
say, is the number of protons in this
nucleus or the charge on an electron if you
want it, times the unit charge of an electron. And it's firing at
some other electron somewhere else in the material. So the basis of any sort
of ion electron interaction has to start with the ion being
either struck or repelled-- I'm sorry, with
the electron being struck a repelled by the ion. And so let's say
that this ion exists. We'll draw kind
of a unit cylinder around this physical situation. And if we draw this
distance right here, it's one of those
rare cases where the nomenclature is the same
in pretty much every reading. There's this distance b, which
we call the impact parameter. It's kind of a
funny name for it, but it just means by how
close does your particle get to that electron
when it undergoes this single interaction. And so this particle is moving
quite fast with some speed v towards and then away
from this electron. And what's going to
happen is there's going to be some
sort of a Coulomb force between this charged
particle and this electron. So let's say you're firing
an electron at an electron. There's going to be
some negative repulsion. Or if you're firing
an ion at an electron, there might be some
positive attraction. But at any rate, there's
going to be some deflection. So let's just say it's a
negatively charged particle. If we draw its
actual trajectory, it's actually going to
go off kind of barely in that direction, right? Two charges passing
in the night. They know each other's there. And they kind of
repel each other. So what we want to
do is figure out what is the total
amount of force in the x and the y direction. Let's just define our axes
to make sure we're all on the same page. And can we resolve that into
a total amount of energy lost per unit distance? This quantity right here was for
referred to as stopping power. Before we launch
into it, does anyone know why I put a negative
sign on this quantity? STUDENT: This is
all [INAUDIBLE].. PROFESSOR: Exactly. Yep. If you're changing the
energy in the particle, unless it's, let's
say, following it to some gravitational field,
which we're not covering today or ever then, it's any
sort of interaction is going to cause the
particle to lose some energy. So this quantity right here
is going to be negative, and so this quantity right
here is going to be positive. We stick a minus
sign in front of it. But let's get back
to the basics then. What is the force between
this charged particle and the electron from 802? This Coulomb force. STUDENT: It's a constant. PROFESSOR: There is a constant. Let's just call it k0, because
the reading calls it k0. STUDENT: And then it'd
be z e [INAUDIBLE].. PROFESSOR: Yes. Yeah. So this is like your
q1 and your q1, right? Your charge 1 and
your charge 2 over? STUDENT: The distance. PROFESSOR: The distance squared. Let's call that the distance
between the two particles. And so now we can say if
this is the distance away in the x direction,
then we know that r is root x squared plus b squared. So we can stick that
in there, and we know that our Coulomb force
would then be this k0 little z e squared over root-- I'm sorry not square root,
just x squared plus b squared. So like we've done
with everything so far in the class-- it's kind of dark in the back. Like we've done with
everything in the class, let's split this up
into x and y-forces. So if we assume that
the electron basically doesn't move, what's
the net amount of force in the x direction
that this particle is going to feel when it
goes from minus infinity, so over here, to plus
infinity like over here. STUDENT: Zero. PROFESSOR: Zero. Why do you say that? STUDENT: Because [INAUDIBLE]. PROFESSOR: Exactly. Whatever force it feels
repelling it from here, as soon as it hits
this midpoint, it gets that same
amount of propulsion in the other direction. So your net force, if you
integrate from minus infinity to infinity, of your x
force as a function of t, that comes out to zero, which
makes our life a lot easier. All we have to worry about
is the total integral of the y force to figure out
how much net deflection do we get in that direction. This integral is also
better known as a momentum. Anyone recognize where
with this comes from? If you take the
integral of a force, it's like the integral of a mass
times an acceleration, which is like mass times the
integral of acceleration, which is like mv, which is a
momentum this is where some of our particle wave stuff
is going to get funky, because we're going to start
throwing in expressions for particle momentums
in wave equations when we start to determine,
well, if this is really an electron. There's some limitations
on how we can treat it, where it kind of loses its
character as a particle. So I just want to warn
you that that's coming up. So now, let's make an
expression for the y force. If we were to say what is the
y momentum imparted, which is an integral of the y
component of the force dt, we already have the
expression for the force, like you guys derived. K0 time as little
ze times e over r squared, which is x
squared plus b squared. And then how do we get
the y component of it? Well, we've got to
define an angle. That's our angle theta. What's the y component
of that force? STUDENT: [INAUDIBLE] PROFESSOR: Yeah, it's just
times cosine theta dt. What's the expression
for cosine theta in this physical situation? STUDENT: [INAUDIBLE] PROFESSOR: Close. b over r. And in this case, r is root
x squared plus b squared. And the last thing we want is
because we have things in terms of x and b, b's a
constant, x is a variable, t's kind of the wrong variable. So we can do a
variable change and say this is equivalent to the
velocity of the particle over-- I'm sorry-- to dx over v. We're
just using this whole like velocity equals, what is
it, distance times time, so our whole, what is it-- yeah. STUDENT: [INAUDIBLE] PROFESSOR: Thank you. Distance equals velocity times. Thinks, [INAUDIBLE]. OK, anyway. So luckily, I had
the expression right and the explanation wrong. So thank you. Was that Luke or Jared's voice? Awesome. OK. So let's put this
whole expression in, keep that little
embarrassment behind us. We have the integral
from negative to plus infinity of k0
little z e squared times b over x squared plus
b squared times the square root of itself. So let's just say x squared
plus b squared to the 3/2, and there's a v
on the bottom dx. This is finally valuable. So we're getting closer. Let's take all the
constants and shove them outside the integral. So we have a k0 z squared eV
over b squared b over velocity times the integral of
just 1 over x squared plus b squared to the 3/2 dx. Not remembering the formula off
the top of my head, I-- yeah? STUDENT: So we can treat
the velocity as a constant even though it's losing energy. PROFESSOR: Yes, that's--
well, we'll call it a crude derivation. But if we're assuming that
the electron basically doesn't change position,
that it changes so little, then we're going to assume that
also the velocity basically doesn't change,
that one collision for a high enough velocity
doesn't lose that much energy. So that's what we're
going with for now. And we'll actually
be able to compare this kind of crude
derivation to one done from quantum mechanics,
and they look pretty similar. There's like an extra
factor of two or something. But as I showed you guys in
preparing for the test, when I said 9 equals about
10, it therefore follows that 1 equals
about 2, and as long as we get the constants and
orders of magnitude right, we're going to gain the
physical intuition for what we're looking at. I'll leave it up there. Anyway, evaluated this integral,
and it came out to something like 2 over b squared. So this just comes out to k0
ze squared b with a 2 over 2 vb squared. Cancel the b's. I don't know where
that 2 came from. Whatever. Yeah, that's what we
have for the stopping power for this sort of one
particle hitting one electron. Now, we have-- well, sorry,
that's the momentum equation. But we're interested in
the change in energy. So what's that equation
we've used before to go from momentum to energy? Our kinetic energy t. STUDENT: Square root
of q [INAUDIBLE].. PROFESSOR: Other way around. So let's do it that way,
right, so p equals root 2 mT. OK, so square both sides. Yeah, you got it. And we have our energy
T is p squared over 2m. So let's take this small
little mess, stick it in here, and then we end up with 4k0
squared little z squared e to the fourth over
2mv squared b. Cool. And so this gives us the little
differential energy change from one electron collision. Yeah? STUDENT: [INAUDIBLE] PROFESSOR: I think we
cancel one of those, right? STUDENT: Yeah, but then
when you square [INAUDIBLE].. PROFESSOR: Oh
yeah, you're right. Thank you. Comes back. b squared b squared. Yep, you're right. Thank you. So now we've only accounted
for the ion hitting a single electron as it moves
through this hollow cylinder of whatever medium
it's going through. So this is when we can kind of
take things back from abstract to reality and say,
all right, it's moving through some actual material. And we have to describe
its electron density in this cylindrical shell. So the electron density in the
cylindrical shell depends on-- well, the number
density of the material itself, just how many atoms
are there times big Z, the number of protons
in that nucleus and therefore the number of
electrons in each nucleus, and the volume of the
cylindrical shell. So what's the expression
for the volume of the cylindrical shell? In differential form? Yeah, I started hearing? I heard a 2. That's correct. Keep going. Well, 2 pi b gives
us the circumference of the circle on the
outside of the cylinder. And if we add a little db there,
some differential thickness element, and we add on a
little dx for some differential distance down the
cylinder, we end up with 2 pi b dv dx,
multiplied by this stuff. And we get some differential
change in energy scales like-- let's say there is
a 4 and a 2 there. So we end up with 4. That's not a 4. pi k0 squared, little
z squared, big Z e to the fourth b dv dx
over mv squared b squared. Now those other b's cancel. We can divide everything by dx. And we've already almost got
our stopping power expression. We're getting pretty close. Anyone see some
similarities between the one I just threw out of my head
and what we've got so far? We've almost got
the makings of it. So now if we want to
account for the fact that our charged
particle is probably not shooting through the center
of a perfect hollow cylinder, but we're just firing it
into like actual matter, we have to account for
every possible impact parameter in every
possible cylindrical shell that it would be moving through. So in this case, we
can integrate this. We've already got
our db right there. That's our integrating variable. And now here's where things
get a little tricksy. Can we actually integrate this
from an impact parameter of 0? And this is not an
easy question actually. What do you guys think? STUDENT: [INAUDIBLE] PROFESSOR: So Luke says no, why? STUDENT: [INAUDIBLE] PROFESSOR: We actually
have an over b. We have a v squared, but that's
not our integrating variable. Yeah, so we have like
a 1 over b looking-- STUDENT: [INAUDIBLE] PROFESSOR: Yeah, that's fine. STUDENT: [INAUDIBLE] PROFESSOR: That's true. There's another more
physical reason though. But you're right mathematically. Can you know precisely the
location of an electron ever? Now I see a lot of
people saying no. Why do you say that? STUDENT: [INAUDIBLE] PROFESSOR: That's right. There's this thing-- the De
Broglie uncertainty principle. It's kind of the punchline of a
lot of quantum mechanics jokes. You never know where
something is going to be or where it's going. We used to say this about
some of the older professors in this department. If you call them and
say, I'm on my way, I'm getting there
as fast as I can, they could be
anywhere in the world. And if they say, don't
worry, I'm three miles away. You don't know how long it's
going to take them to get here. Same thing with me
and getting here. Although I was on
MIT standard time, which means five minutes late. Not bad. So in this case, we have
to ascribe the electron some sort of a wavelength. So in this case, we can't
just treat the electron like a particle whose
position we know. We're going to go with
our original equation for a photon energy, which
looks like hc over lambda. Rearrange that so that we'll
have some lambda wavelength equals hc over E. I'm sorry. This is a momentum thing,
not an energy thing. And what's the momentum
of the electron? From the classical definition? It's just mass times
velocity, right? So we'll just stick in
the mass of the electron times the velocity right there. And this wavelength right
here, the De Broglie wavelength of the electron is as close
as we can specify that impact parameter. And it turns out to be pretty
significant, like on the order of 0.1 to 0.2 angstroms. You can't tell where an electron
is going to be finer than that. So we're going to
have this b minimum. I'll just write that in
there and some b maximum, where are b minimum is
the same as our De Broglie wavelength of the
electron, because we can't define its position
any better than that, which is just Planck's constant
over its mass times velocity. For b max, it comes
out to something like hv over this
quantity, I bar what's called the mean
ionization potential. What this quantity
physically represents is that if your impact
parameter is too large, then the electron will--
or the charged particle will feel so little force,
that it won't eject an electron and won't really be deflected. And the farthest away
it can be corresponds to the minimum amount of energy
to create an average ionization in the material. And this mean ionization
potential scales with something like
this constant k times z, where k is on the
order of like 30 to 35-- think it's like eV. But remarkably tight constant,
so picking a mean value like that is no problem. And there we have
our b min and b max. Those are our limits
of integration. I think I planned this just
to fill up the boards today. So let's write out the
final integral that we have and see what we get. So we have our
stopping power, should be integral from b min h over mv
to b max jv over I bar of 4 pi k0 squared little z squared
big Ze to the fourth over mv squared b db. And like I think it
was Sarah that you mentioned that we'd have a log. I forget who's mentioned it. Sorry. That was Luke, OK? You're right. So it ends up just
being a natural log. It's like all this junk
on the outside times the integral of 1 over b. So this just comes
out to 4 pi k0 squared little z squared
big Ze to the fourth over mv square times the natural
log of dv max over b min. The h's cancel. We get a v squared. And so all this
stuff inside just becomes the natural log of mv
squared over mean ionization potential. And we've arrived basically
at the same equation that we have over
there, that I took care to memorize on the plane. So great that we've gotten
here through the math. Let's actually see
what this means, and we're going to go into
some of the limits of validity like Luke was saying, where you
can't have a natural log of 0. So the stopping power formula
isn't quite going to work at 0. Nor will it work at
super low energies. So if you want to write what
this should be proportional to. I kind of see some constants
here that we don't really care. They don't vary at all. But this looks kind of
like a kinetic energy term, doesn't it? Like kinetic energy terms. So it's kind of proportional
to this function 1 over t times the natural log of t. When you get rid of
all the constants and just express it in
terms of the variables, it looks a whole lot simpler. And so let's see what
this would look like if we started to graph it out. And this is pretty universal
for any charged particle stopping power. So if this was the
kinetic energy t, and this was our stopping power,
we've got this 1 over T term that's going to look
something like this. And we have this natural
log of T term, which is going to look something like
this that actually goes down to infinity. So like Luke was
saying, if these two are multiplied by
each other, we're not going to have negative infinity
as a stopping power, which would physically mean that once
the particle hit zero energy, it speeds up to
infinite speed, and that doesn't make any sense. But we can start to draw
what the curve would look like with this
general envelope. And so at low energies, the
stopping power kind of scales like 1 over e. Let's now start
drawing another graph with a little more
physical intuition, the range of the particle. So while stopping power might
be kind of a new quantity that represents the differential
amount of energy lost as a function of distance--
that's kind of a mouthful-- the range is pretty simple,
just how far it goes. So to get the range
from the stopping power, you can integrate--
let's say you fire a particle into a bunch
of matter at some energy T. So you start off an energy
T, and you want to see how far it gets at distance 0. Well, you can just
integrate the stopping power as a function of T,
or you can switch your limits of integration. Let's see. I'm sorry. I'm not going to switch
those limits of integration yet, which is like saying
from 0 to T of dt dx dt, which is like saying from 0 to T
of dt dx to the minus 1 dx. Much simpler expression. And when you forget all
the crazy constants, and you just take
this kind of form as the variable part of the
expression for stopping power, unless your energy
is really high, and this natural log counts at
all, your range kind of scales like the integral
of just 1 over T. I'm sorry, that to
the minus 1, which is like the integral of T,
which is like T squared, which means that this are pretty
interesting intuitive result, that the range of the
particle increases with the square of its energy. So this gives you
a good hint to say, if I increase the particle
by a certain amount, I'll increase the range by the
square root of that increase. So anyway let's start
drawing this range curve as a function of x. What this says right here is
that if we start our particle at some high energy, and we're
firing into the material, and it's losing
energy as it goes, and we track this value
of the stopping power to figure out how far it's going
to go, change that in a second, for the first little while
as this particle loses more and more energy,
its stopping power stays mostly constant, and it
loses a pretty constant amount of energy as a function of time. As its energy gets
lower and lower, it loses more and more as
a function of position. What this actually means is
that as the velocity goes down or as the particle's
energy goes down, it spends more time in the
vicinity of the electron and gets deflected more. It's just that kind
of a simple argument. Like the more time it
spends near this electron, the more it feels the push. And so it will lose
more and more energy as its energy gets lower and
lower until you hit the point where this curve breaks down. Where do you guys think that is? Even mathematically speaking. Well, what happens if your
natural log term is negative here? Then you get a negative
stopping power, which would be like the
particle picks up energy. That's not quite
physical at all. So in reality, we know
that at some point it's going to taper off,
and the stopping power at 0 should be 0. This maximum right here
occurs around 500 times the mean ionization potential,
which is a pretty low energy, but what this actually says is
that when the particle's moving really slow, it's moving so
slow that once in a while, it will capture one
of those electrons, like if you fire in a proton or
a positively charged heavy ion, if it's moving so slow
that it can feel the pull, it will just
partially neutralize. And that becomes the next
mechanism of energy loss. And so if we keep following
this curve, once we hit some sort of a
maximum, then it's going to lose less and
less and less energy, do less and less damage, and
you end up with the same curve, this kind of brag peak
curve that we did together when we used the SRIM code. Did I go through the stopping
range of ions in matter with you guys? Did I show you
this on the screen? Remember the curve of the-- let's say damage events
per distance or number of implanted ions as a
function of distance. You end up with the
exact same thing. That's what the SR stands
for in SRIM is stopping, what is it, stopping power and
range or something like that. Something range of
ions and matter. Is it the stopping power and
range of ions and matter? I don't know, but all SRIM
is a gigantic stopping power database and a big
Monte Carlo engine. So it takes an equation
just like this one or-- yeah, just like
this one, that one, whichever one you want
and decides, well, how often is the
particle going to lose how much energy depending
on where it happens to be? And that's all there is to it. This point here, we
would call the range or like the average range
at which-- well, that's not quite right. There'd be some average
range around here where the particles
actually stop, when their energy goes to zero. And in reality, because
every one of these processes is random in nature,
the impact parameter is going to be kind of random. Not every particle will
stop at the same place, because all the electrons are
moving around in the atoms. So there's going to be some
sort of a range of ranges, which we call straggling,
which is to say that not all of the particles
end up at the exact same range, but they end up pretty close. I think I want to pause
here for a seconds and see if there's any
questions from this four board derivation or any
intuition questions that you guys might have. Yeah. STUDENT: You didn't initially
have the negative d2 over dx. PROFESSOR: Oh yeah,
where'd that go? Let's trace this through. STUDENT: What happened
to the negative? [INTERPOSING VOICES] STUDENT: But we didn't
actually derive it. PROFESSOR: Yeah, let's see. So the change in
energy should have been a negative change in energy. So if we-- it went
missing there. So there we go. That's the only other place
that seems to be missing. OK, great. Cool. Then if you want to start
looking at the number of damage events that this
particle will incur, we'll call this the
number of ion pairs, which might look suspiciously
familiar if you guys remember the Chadwick paper,
he was talking about this proton of this
energy should make this many ion pairs at this distance. Now you guys actually
have the tools to find out what that
number should be, because it's going to be 1
over some ion pair energy, usually around 30 to 35 eV,
depending on the material times dt vx, which is to say when
the stopping power is higher, you're going to have
more ion pairs produced as a function of distance. So the real label
for this y-axis here should be like ion pairs
or damage or defects or anything like that that refers to the
same kind of thing as damages to the material, either
by ionization or even by similar nuclear processes. And so that's what
results in those SRIM curves that we showed from
before, where you have, let's say, a bunch of protons
entering a material here at some high energy. They don't lose very much
energy when they go in, but as soon as they get
to a low enough energy where they're stopping
power reaches the maximum, they dump most of
their energy in there. And this is the basis behind
proton cancer therapy, which I mentioned to you guys in the
first or second day of class. Now that we know both
exponential attenuation and stopping power, we
can explain theoretically why proton therapy is a
more effective treatment. So let's say this is the
person that contains a tumor. Say it's right there. And you have a choice
between firing in an X-ray or firing in a proton. What is the dose to this
person, not just to the tumor, but through the whole person
going to look like for X-rays? Get another board. So if we look at the number
of ion pairs, and let's say this is the
thickness of the person, and the tumor is in this range. And if you send in
your X-rays, or you send in your protons,
what is the number of ion pairs produced from X-ray
or from proton going to look like in either case? So first of all, who
wants do the X-ray one or tell me what it will be? You guys-- yeah, Luke? STUDENT: Would it
be pretty flat? PROFESSOR: It'll be
fairly flat, but there would be some decay to it. So let's say we defined some
initial intensity, X-rays just get attenuated exponentially. So you're going to do a whole
lot of damage to the person before the X-rays
reach the tumor, which is why when you
do X-ray therapy, you have to send in x-rays from
a bunch of different locations so that the tumor
gives out the most, and the rest of the
person in any location doesn't get that much dose. For proton therapy,
it's quite different. It looks just like this. So you might do a little
bit of damage as you go in, and you tune the energy
of those protons, so that they do all the
damage in the tumor, and then they stop in the
tumor or just beyond it, so that they don't
do any more damage to the rest of the person, and
they do very little going in. And so that's why
proton therapy centers are popping up
all over the world because it's a more
effective treatment. It's also more expensive because
you need a proton accelerator so then, here's a
question for you. This is something they
actually do in the lab. Say, here's your human. There's your tumor. There is your proton
gun at a fixed 250 MeV, firing protons out. How do you change the
range of those protons without changing their energy? STUDENT: The distance
it has to travel? PROFESSOR: Is what? STUDENT: The distance it
has to travel the other way. PROFESSOR: The distance it
has to travel specifically? I mean, if they travel in a
vacuum, do they lose energy? No. So what can you do? STUDENT: [INAUDIBLE] PROFESSOR: You
could deflect them and change their direction. But as we'll get into on
Tuesday, if you deflect them, they're going to
emit lots of X-rays in the form of Bremmstrahlung. So that's probably not
what we want to do. You can put stuff-- and I can't be any more
specific than that-- in between the proton
beam and the patient, because if the stopping power
for 250 MeV protons and 50 MeV protons basically
doesn't change, then you just put
things in the way. So let's say this is the
thickness of the person. You just put some
tissue equivalent stuff, or what they'll call a phantom,
so some tissue equivalent gel or water or some other
stuff to lower the proton energy without deflecting
the beam that much. So as you guys saw in
the SRIM simulation, if you track the 3D
positions of these protons as they enter into the material,
they all go pretty straight, and then they start
getting funny. Computer can fly the
ions faster than I can. But no matter what
it goes through here, while the protons
have high energy, they don't get
deflected much, they don't lose that much energy. And you can very finely tune
the amount of stuff in the way. This would be the stuff section
before entering the person. So this is why it's so useful. So let me check
the time, because I haven't checked at all. The clock's broken. Oh, we have like 10 minutes. So now is a good time
to stop for questions and see if you guys have any
questions from the derivation or the sort of physical meaning
of stopping power and matter. Yeah. STUDENT: What was the
nz sort of [INAUDIBLE]?? PROFESSOR: Yes, the nz, n is
the number density of atoms. So if you're traveling through
some actual block of matter, it depends how many
atoms are in the way. So if the total stopping power
decreases with decreasing density, like if you're
going through tungsten, but it happens to be a tungsten
gas, that tungsten gas will not have nearly as much stopping
power as tungsten metal, because there's just more
tungsten than the weight. Or less, I'm sorry. And the z right here
is the charge per atom, to say if you're firing
electrons into something, the strength of
the Coulomb force that they'll feel, or let's
say, the number of electrons that they can smack into is the
same with the number of protons in that nucleus if we're
not using ionized materials. And we're typically
not firing anything into ionized materials. It's just normal neutral matter. Does that make more sense? Cool. Yeah, Luke. STUDENT: Where did that n go? PROFESSOR: It should
have been there. Thank you. pi n. Should absolutely be there. Anything else? Yeah, Dan. STUDENT: [INAUDIBLE] PROFESSOR: OK. STUDENT: [INAUDIBLE] PROFESSOR: The
charge per atom is big Z. The charge
on the particle is little z, because both
of them actually matter. So little z tells
you the strength of the interaction between the
particle and each electron. Big Z tells you how many
electrons are there per atom. Big N tells you how many
atoms are in the way. And in that way, you have
a complete description of the material. Curious that the mass
of the charged particle is absent from this formula. Isn't it? Yeah. The mass doesn't matter. You will certainly
change the momentum of, let's say, a charged
particle less. But in the end, it's just
non-contact Coulomb forces that determine the
energy transfer between the electrons in there
and the charged particles slowing down in the medium. So that is a curious thing to
look at, but it is intentional. For the case of this ionization
or electronic stopping power, the mass does not enter into
it at least in this formula. There is another version
derived that's in your reading that they just kind of
plop it in front of you that's got the mass somewhere
in the natural long term somewhere where
it really doesn't change much at all, except
for really high energies. So if you want to
think about, well, what do I want you
to know, I would want you to be able to go
through this derivation again, so that I can know you can
go from a intuitive example to an actual equation
you can use, graph what that equation
should look like and talk about where it really matters
and where it breaks down. Like mathematically speaking,
if this natural long term is negative, you're not going to
have a negative stopping power. Something else has got to
occur, and what's happening here is neutralization. And that's why the stopping
power curve diverges for really low energies,
because sometimes electrons get captured. Any other questions
on stopping power? Cool. This is a good place
to stop for now.
