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visit MIT OpenCourseWare at ocw.mit.edu. MICHAEL SHORT: And to try
something out a little real, I took a detector that you all
have, as well, my cell phone. And this morning I
went down with EHS to one of the very
radioactive cobalt-60 sources, a 10 millicurie source. If you note, the source that
we were playing around here was one microcurie, so this is
a 10,000 times stronger source. It was actually able to show the
difference between a background count of my phone. You shouldn't see much going
on, except for that one malfunctioning pixel,
because not much is going on. And when I put the phone
over the source itself, things look a little different. You guys see all that digital
noise, or snow in the video? Every one of those white
flashes that you see is a gamma interaction with the
semiconductor in the cell phone camera, with one or
more pixels in your CCD, or charged couple device,
or your CMOS detector, whichever one it happens to be. So I thought this
was pretty cool. You can actually use your cell
phone as a radiation detector. We're going to understand why,
and what sort of radiation it could detect by virtue of its
size and its composition today. Anyone ever try this before? You have? AUDIENCE: Yeah. MICHAEL SHORT: Cool. AUDIENCE: [INAUDIBLE] MICHAEL SHORT:
Probably more intense than this if you were
making neutrons, right? AUDIENCE: Yeah. MICHAEL SHORT: Awesome. OK, cool. So let's just first
figure out, well, where is this
radiation coming from? This is the link between
the first part of the course and what we're going to be
doing over the next month. As we've seen from
the decay diagrams-- and I think I've harped on
potassium-40 as an example for a reason-- it can undergo electron
capture or positron release. And if it undergoes electron
capture by this likely route it gives off a
1.461 MeV gamma ray as the only possible
transition here. It also undergoes
beta decay, which you don't want to forget
about if you're calculating the activity of potassium-40. But today we're
going to be focusing on what does this
gamma ray actually do when it encounters matter? Or what are the possible
things that can happen? I'm going to introduce
them conceptually today, and we're going to go through
the math of the cross sections and the energetics
more tomorrow. So I'm doing a context first,
theory second kind of approach. There's three main things
that gamma rays will do in matter depending
on their energy and the actual matter itself. There is one called the
photoelectric effect, where a gamma ray simply ejects
an electron from the nucleus. So let's say we've got
our potassium-40 atom, have a bunch of
electron shells-- I'm not going to draw
all the electrons, but I'll draw a few inner
and outer ones here. One of the things that
the gamma ray can do is just eject that electron,
have it come firing out. And the energy balance
for that isn't that hard, because this gamma has
some energy E-gamma. This electron had some
binding energy, E-binding. And the kinetic energy, let's
call it t, of the electron is simply the gamma ray in minus
the binding energy back out. It's just however
much energy it takes to remove that electron,
that's what it takes. And so you end up
with, if we go back to our banana
spectrum, what we call a photo peak, or a photoelectric
emission peak, right here. If you trace down
this line, it's awfully close to
1,469 KeV, Or 1.46-- what was it? 1.461 MeV. It won't be exactly
at that energy, because it does take
a little bit of energy to remove that electron. Anyone have a guess on
what order of magnitude that might be? Yeah. KeV all the way down to eV. So this photo peak
will typically be extremely close,
but not exactly equal, to the energy of the
gamma ray coming out. For most detectors that don't
have that good resolution, you can pretty
much assume they'll be in the same channel,
or the same energy bin, because you're detector will
have some sort of resolution. It may have 1,024 or
2,048 channels that span the full energy range. And you might not be able
to tell the difference between 1,460 MeV-- or 1.46 MeV-- and
that minus a few eV. Potassium, in particular, has
quite a small work function. And we'll get into why
that is in a second. The next thing you
can do is what's called Compton scattering, in
the general case, which means that there is an electron here. A gamma ray comes
in with E-gamma. And then it bounces off with
some energy, E-prime-gamma. And then the electron goes off
with some other kinetic energy. Then the last one is what's
called pair production. Just like in the
Q equation, if you have anything
related to positrons you have to first create them. So pair production doesn't
happen below about 1.022 MeV. And it happens with
increasing probability as the energy of
the photon goes up, kind of like in
radioactive decay. There's a lot of parallels here. Is you can make an
electron-positron pair at 1.022 MeV. It's just not very likely. And what we're going
to find out tomorrow is why the most likely
photon effect is shown in these different regions. Anyone have any idea? Why do you think the
photoelectric effect would be most likely at
low energies and high Z? You just have to give
an intuitive guess. Yeah, Luke? AUDIENCE: Because
the binding energy isn't very large for the
outer electron shells. MICHAEL SHORT: That's right. So that explains
the low energy idea. So it doesn't take very much. In fact, does anyone know
what the minimum energy you need to make the
photoelectric effect happen? Well, what's a typical
order of magnitude for binding energy of the
lowest, or the outermost, electron shell, or the
lowest bound electron? AUDIENCE: Work function? MICHAEL SHORT: Yeah. That's called the work function. Anyone know an order of
magnitude, guess what it is? It's in the single eV range. In some cases, it can
even be slightly lower. And that, we're talking
about visible light. So green light,
even yellow light, can eject electrons via
the photoelectric effect. And then the reason that
goes more likely with higher and higher Z,
we'll get into that when we look at the different
cross sections of interaction. Pair production is much more
likely at higher energies, because at higher
energy you're more likely to create a positron. And in addition, pair
production happens when a photon interacts with
either the electron cloud, or the nucleus there. And that gets more
and more likely, let's say, the denser
the electron cloud is or the higher charge
there is on the nucleus. So first, the simplest one,
the photoelectric effect, this is actually what Einstein
won the Nobel Prize for. Not E equals mc
squared, which has been the bane of our
existence for the last month. That's not what he got
the Nobel Prize for. It was demonstration of
the photoelectric effect, where if you start firing
photons of an energy times Planck's constant
times its frequency-- and in the next page I'll give
you a quick photon math primer, in case you don't know
what those quantities are-- there will be no
photoelectric emission until you hit that
work function. Yeah, like Julio was
saying, that lowest bound electron energy. And then, emission
will simply go up. And so this was demonstrated
by applying a voltage to two different plates,
two different metal plates, and then sending in
light via this window and seeing when the current
actually became non-zero. So the way you detect
photoelectric emission is, if you've got electrons
boiling off of one surface to the other, that's
the movement of charge, and that's a current. And so you can measure a
current with an ammeter. That's-- it actually
is that simple. But very elegant experiment,
back for the 1910s or 1920s. And as a quick primer
on photon quantities, so you know what all of
these different symbols mean, the photon energy we
give as Planck's constant times its frequency. And I gave you Planck's constant
right here, for reference. I do recommend that you
guys try checking out all the units to make
sure that they work out. Because if you ever
forget, is it h times nu, or is hc over lambda,
you can always check the units of your
expression to make sure they come out to an energy. Which in SI units is what? AUDIENCE: Joules? MICHAEL SHORT: Joules. And then how about in these
sorts of things, the most reduced SI units? AUDIENCE: eV? MICHAEL SHORT: Is what? AUDIENCE: eV? MICHAEL SHORT: eV is
another unit of energy similar to the joule,
like 1.6 times 10 to the minus 19 joules. But what about in meters,
kg, seconds, other SI units? [INTERPOSING VOICES] MICHAEL SHORT: Yep. Kilogram meters squared
per second squared. Indeed. Per meter squared,
per second squared. Yeah. So just make sure sure
you remember that, because if you're just
looking for joules and you don't remember
what a joule is, it's going to make unit
balance kind of hard. And also, we can describe the
momentum, or p, of the photon as Planck's constant over
lambda, its wavelength. This is going to get
real important when I ask you guys to
do a derivation, much like the q equation one
that we were doing before. But instead of me just doing it
at the board and you copying it down, I want you guys to
try working through it. And it's going to be
another energy and momentum conservation thing,
just like before. And this way you know
what the energy is and you'll know what
the momentum is. So now onto this work function. What is it, actually? There's a great
paper by Michaelson-- I did not look up whether this
is the Michaelson of Michaelson Interferometry, but I
wouldn't be surprised. I'm going to check into that. But I did dig out this
paper that actually shows the different patterns
in the work functions of different elements. So what do you guys notice
in terms of patterns, here? First of all, which elements
are all the way to the left or have the lowest
work function? AUDIENCE: Group one. MICHAEL SHORT: The what? AUDIENCE: The group one metals? MICHAEL SHORT: The
group one metals, like sodium, lithium, potassium. Why do you think that is? AUDIENCE: I mean, they have-- I don't know. MICHAEL SHORT: They've
got one electron in their outermost shell. So it looks like my potassium
picture is not quite accurate. I'm going to draw another
shell, and put one lone electron in that for accuracy. And so that electron
is extremely unbound. That's the same reason
that these elements are so chemically reactive. They want to ditch that electron
to have a filled outer shell. So you may also expect the
work function of noble gases to be extremely high. I don't know if any
are plotted here, but you do see the next row
over, like barium, strontium, calcium, magnesium, has a
slightly higher work function. And as you move this way
through the periodic table, to the left, until you hit the
transition metal craziness, it follows a pretty
regular pattern. And so you can have
a good guess of what the work function of something
will be depending on it's Z, and depending on which-- what is it? Which column it's in
in the periodic table. Now, can anyone
tell me, why do you think that work functions tend
to increase with decreasing Z? Yeah. AUDIENCE: For smaller Z,
the outermost electron is closer to the nucleus,
so it's more tightly bound. MICHAEL SHORT: Indeed. Yep, exactly. For the smaller Z, that first or
second shell is a hell of a lot closer to the nucleus. Even though it has a lower
total charge in the nucleus, it's much more tightly
bound, being much closer. So, like the outermost electron
in caesium is quite far away, it does not feel as
much coulomb attraction. Yeah, good point. So now onto Compton scattering. I'd say though the
most difficult, conceptually, to
understand the energetics. But the kinematics, or what
actually physically happens, should look strikingly
similar to what we've spent the last month on. Instead of two
particles colliding, it's a photon colliding
with an electron. Does anyone remember
what we read in that first day of class,
with the Chadwick paper? When he said, hey, maybe
this quantum of energy is done in a process analogous
to Compton scattering. Well, this is
Compton scattering. His analogous process
was maybe an electron hits a proton and
something happens, which is not actually
what happens. And in this case, you have
a photon with energy h nu, and momentum h nu
over c, striking an electron with rest
mass m of electron c squared, or 0.511 MeV. And afterwards, the photon
leaves at sum angle theta, and the electron leaves
at sum angle phi. So I'm going to show you guys
some of the Compton scattering energetics relations, like
what is the wavelength shift. Which means that if
this photon comes in with a certain
wavelength, lambda, and it gives some of its
energy to the electron, it comes out at a
different wavelength. Is it going to be lower
or higher wavelength, do you think? [INTERPOSING VOICES] MICHAEL SHORT: I
heard a bit of both. So who says lower? AUDIENCE: Me. It can be wavelength. MICHAEL SHORT: So
lower wavelength. Let's go back to
the photon formula. Would a lower wavelength
result in a lower or a higher photon energy? AUDIENCE: Higher. MICHAEL SHORT: OK. So in a Compton
scatter, you start off with an electron
kind of at rest. They're definitely
not actually at rest. But compared to the
energy of the photon, they're at rest enough. And then you give some of
that energy to the electron. That energy has got to go down. And because these two
quantities here are constants, the wavelength has
got to increase. And hopefully this
makes intuitive sense. The photon does what
we call a redshift. It shifts closer to the red
end of the visible spectrum than the blue end. And as you guys know,
the high energy light in the visible spectrum hits
towards the ultraviolet. That's what tans you, or
gives you skin cancer. Red light, or infrared light,
doesn't do much of anything at all. And so this is that on
the extreme scale, where when we say redshift,
we don't necessarily mean the photon is visible. But we do mean that
it's shifting to a lower energy, or a higher wavelength. And so this wavelength shift
is always going to be, well, is it going to be
positive or negative? AUDIENCE: Positive, right? MICHAEL SHORT: Is what? AUDIENCE: Positive? MICHAEL SHORT: So you
say the wavelength shift is going to be
positive, which would mean an increase in wavelength? There you go. Yep. Because it's got to be--
it's got to lose energy. So I'm not going to go through
the derivation of these, because I want you guys to
go through the derivation. But we're going to do it
in the exact same way, and I'll help kind
of kick you off. Where, in this case, what
are the three quantities we can conserve in every
physics, everywhere? AUDIENCE: Mass,
energy, and momentum? MICHAEL SHORT: Mass,
energy, and momentum. The trick here is, what
is the mass of the photon? Massless. So we've got energy
and momentum. And we've got, let's say, some
wavelength shift to determine, which is some change in energy. And we've got two
angles to deal with. That's three unknowns. We need three equations. So we know that our
initial energy coming in is going to be h nu
plus approximately 0 becomes h nu bar, and the
kinetic energy of the electron. So that's our energy
conservation relation. And then what do we
do about the momentum? What do we do last time? AUDIENCE: Split it into x and y. MICHAEL SHORT: Exactly. Split it up into
x and y momentum. So the x momentum
of the photon-- so I'll just label
this as energy-- put the x momentum of the
photon is h nu over c. And there was no x momentum
of the electron to begin with. So then we're going to say
this has outgoing momentum h nu prime over c times
cosine theta plus whatever the electron momentum is, let's
say m electron v, or root 2 m electron, T
electron, cosine phi. And then how about
the y momentum? What's the y momentum of
the system at the beginning? AUDIENCE: Zero? MICHAEL SHORT: Yep. Nothing for the photon,
nothing for the electron. And at the end we've got h
nu over c sine theta minus, because it's in the negative
y direction, momentum of the electron sine phi. I'm going to stop my
part of the derivation there, because I don't want to
steal away your whole homework problem. But you're going to start it
out exactly in the same way as we were doing kinematics
of two particle collisions. Because what is a
particle, but a wave? They're all the same thing. It's modern physics. And then-- here's an
interesting bit, here-- this maximum wavelength
shift, if you want to figure out what is the-- well, look. Let's say, we call it
the Compton wavelength. So if you were to decide what
is the maximum wavelength shift, where would that be? At what angle? Did you have a question, or did
you say what was the question? AUDIENCE: Yeah. MICHAEL SHORT: Oh, OK. Is what? AUDIENCE: Pi over 2. MICHAEL SHORT: Is
that angle pi over 2? Because at that point cosine
of pi over 2 equals zero. Yep. And so then you get
this interesting result. No matter what the incoming
energy of the photon is, you get this
0.238 MeV shift. And that's actually going to
help explain, to jump back to our banana spectrum, what the
distance is between our photo peak-- which is our
photoelectric peak, which is pretty close to the
energy of the photon-- and this part right here,
which we call the Compton edge. Which would mean the
maximum scattered energy of that photon, in this case. Or no, I'm sorry. That would be the maximum
energy imparted to the electron. Almost misspoke there. And no matter what this
energy of the photon is, that distance right there,
that's the Compton wavelength. Interesting quirk
of physics, huh? Because in the end,
all that matters is if the angle's all the same,
everything else cancels out and you just get a
bunch of constants. Let me jump back to there. So now we'll take another
look at our detector spectrum and start identifying
some of these peaks. If you notice, this
0.238 MeV looks just like what it does on the graph. This is the kind of cool
thing, like you guys threw some bananas in
a detector last week. We got a spectrum
yesterday morning, and how well-timed it was. We're actually going to
start explaining it today. There's a whole lot more going
on in this banana spectrum. Part of what we'll be
explaining tomorrow is, why do you get this kind of
bowl shape, this Compton bowl? And it turns out that there's
a different probability of scattering at
every different angle, or what we call a
differential cross-section. A d theta over d omega. Because the probability
of that photon scattering off in any
direction is not equal. But if you know what direction
the photon scatters off in, you know what energy it has,
or you know what sort of energy it gives to the
electron, because that's a one to one relation. And that's why you end
up with this very smooth, almost cosine-ish
looking kind of curve. You guys will actually get
to derive that yourselves. So then onto the wavelength
and energy shift. By looking at the
electron recoil energy and this wavelength
shift, from that you can actually get some
sort of an energy shift. You can arrive at what is the
recoil energy of that electron. And so here's one of the topics
that's usually hard for folks to understand, but I want
to stress it right now. When you send gamma
rays into a detector-- let's draw an
imaginary detector. In fact, let's draw the real
one that we used in our banana counting experiment. So we had these copper walls,
we had our bag of bananas, and we had our high
purity germanium detector. Let's say we had a
good shield on top, and then a good
shield on the bottom. That right there is
our active detector, and this banana is sending off
gamma rays into that detector. The way a detector works is
not by counting the energy of the gamma ray directly. It can't actually do that. In this germanium
detector you've got a huge voltage
applied across it. I think what Mike
Ames actually said was the one we used
was about 2000 volts. What happens here
is, let's say-- I'll actually need
three colors for this-- let's say a gamma ray comes in-- that's our gamma ray-- and interacts in the detector. That gamma ray will redshift--
let me get a redder color, because that'll be like
physically accurate-- that gamma ray is going
to hit an electron, go off at a different angle,
and redshift, or get lower in wavelength. Meanwhile, that electron that
it hit actually goes flying off and in the other direction,
we're going to call it a hole. A defect missing one
electron of some sort in this semiconductor. Normally if there was
no voltage applied here these two would just find
each other and annihilate, and you would have nothing. So what's to count? But by applying a
gigantic voltage, let's say this voltage
was really plus and this voltage
was really minus, this electron keeps on
moving and this hole keeps on moving
to the electrode. Instead of recombining
in the detector, they're actually then
sent through where they're counted in some sort of ammeter
or some sort of energy pulse counter. And what we're
actually measuring is the recoil spectrum
of the electrons that the photons make. You're not directly
measuring photon energy, you're measuring the
electron effects. Part of that is because
chances are photons just go through everything. This is why I wasn't
so worried this morning standing with my face over a
10 millicurie cobalt source. Because while I was
getting billions of gammas per second
flying through my brain, most of those billions just
flew out the other side. It's literally in one
ear, out the other. And so most of these gammas,
if they interact at all, will escape again. The electrons, however,
because they're charged and very low mass,
have a very low range in the detector. So chances are
the electrons that are made are going
to stay there, unless you happen to make
one, like, right here at that surface few
atoms, and it escapes. That almost never happens. So forget that. This was, last year, a
huge source of confusion to say, why are we seeing
some of the other peaks that I'll be explaining
in five or 10 minutes? Or, why aren't we
seeing a 0.238 MeV peak? Because what you're seeing
here is a photon losing it's energy minus 0.238
MeV in its maximum energy transfer, which is
given to that electron. Then what actually happens
next is this electron slams into a bunch of
other ones, and that slams into a bunch of other
ones until all the energy is lost in the detector. And all of those electrons
get sucked into the electrode by this very high voltage. And then, the way you count
the energy of an interaction is by how many electrons you
get in a certain little amount of time. And so that's why, for
example, for the photo peak, that's the kind of
simplest reaction. A gamma goes in, a really
high electron comes out, it smashes into tons of
other electrons imparting all of its kinetic energy
in the detector, which is all summed up
in a nanosecond, or however long we collect for. And then we say that we saw
an energy blip containing about 1,460 KeV of energy. It all came from
that first gamma. And then it was all given
to that first photo peak electron, which then slammed
into a whole bunch of others. And they slammed into
a bunch of others. And there's this what's called
this ionization cascade, where a whole bunch of electrons
make a whole bunch more until all of them
have too little energy to ionize anything else. And then they're just collected. So that's what we mean
by a pulse in a detector. It's not exactly an
intuitive concept, because it's not like
the gamma goes in and we just count its energy. There's more things that
physically happen in here. But it's important
for you guys to know, especially when we start
to look at pair production. You guys remember some of
this stuff from the positron annihilation spectroscopy? Well, the way we actually know
that positron annihilation spectroscopy, or PAS,
works is by measuring photons, or their
eventual electron recoil, that can only be possible
from this process. So as a quick review, let's
say you had a positron source, like sodium-22, which naturally
undergoes radioactive decay, and forms a positron
along with a gamma ray from a very short
isometric transition, or IT. Then that positron bounces
around in the material until it reaches an electron. And once it hits that electron,
because the positron-- let's see, the rest mass
of the positron is the same as the rest mass of the
electron, which is 0.511 MeV-- once the two of
these combine, they annihilate, producing two
511 KeV or 0.511 MeV photons. And it's those photons at
this exact energy all the time that really give it away. Because there's not
many other processes that produce a huge amount
of exactly that photon. Now that we've talked a little
bit about momentum and energy conservation, does
anybody know why you get what's called a
blueshift or a redshift in positron annihilation
spectroscopy? I'll give you a hint. It goes down to
conserving the same thing that we're doing all the time. Yeah, Kristen? AUDIENCE: I was going to
say, is that something to do with wavelengths? MICHAEL SHORT: You're close. I mean, technically
you're close, if you treat electrons as
waves, which you totally can. The electrons themselves
do have a non-zero momentum as they're flying about in the
atom or around the nucleus. And when a electron
collides with a positron, if that electron already
has some momentum associated with it, then the cell system
of mass is not at rest. It's moving at some small speed. So this little minus delta
energy and plus delta energy accounts for the initial
momentum of the electron, which means not only can you
tell from the lifetime how many electron looking
defects there are, but you can probe electron momentum by
looking at the slight energy changes as positrons
collide with electrons. It's a really cool
and powerful technique that uses only 22.01
concepts to probe matter at its deepest level. So what's happening on
the atomic scale is, let's say a photon
made a positron, and the positron bounces about
what's called thermalizes, or just slows down
via collisions, via other types of collisions
that we'll go into soon, and then gets trapped in a
defect, which is a relatively electron-poor place. But it doesn't mean there's
no electrons, In every space everywhere, there's
a probability that there's an electron there. In a defect not
containing an atom that probability is
lower, but not zero. And so by figuring out
how long they last, and when those 511 KeV
gammas are emitted, you can tell, let's say,
what size defect that was. But now let's talk about
what happens to these 511 KeV gammas. What evidence do we have
that positron pair production actually exists? So before I reveal
the labels, can anyone tell me what on
this graph suggests that positrons are happening? And there's actually two things. What do you think? Yeah. AUDIENCE: There's
a peak 511 KeV. MICHAEL SHORT: That's right. That's exactly right. There's a peak at 511 KeV
that if I trace that up, I went one over. Yeah, right there. 511 KeV. Is it exactly 511 KeV? What do you guys think? So forget the fact that
it came from a positron, let's say a 511 KeV
gamma came in somewhere. How would it then release
electrons to be counted? It then undergoes
photoelectric emission. So the actual
energy of this would be 511 KeV minus the work
function of the material. This is one of those
tricky questions that you might not even
see it on the spectrum, but I want you to physically
understand what happens here. It's not like 511 KeV
positron photons magically get counted at 511 KeV. They then have to eject
an electron, somehow. And for those electrons
to be counted, they have to interact in
exactly the same way as all the other electrons. There's no difference. What else? Oh, yeah. Luke, you have a question? LUKE: So, from the banana. Is a gamma ray coming from a
banana, and then that gamma undergoes pair production? And then the gamma from
the pair production-- I guess, where are the
gammas coming from? MICHAEL SHORT: That was
my next question to you. So let's think about
this a little bit. We'll start off with gammas
being emitted in all directions from our bag of banana ashes. Now the question is,
where do these 511 KeV photons come from? If the gamma ray interacts
with the detector by any mechanism
including pair production, what are the possible
things that could happen? There's three
different scenarios. Let's pick a 511 KeV color. Well first of all, it might
just undergo pair production. And it'll release
two 511 KeV gammas. Let's see, those are
our 511 KeV gammas. And because they're gammas,
and they interact with almost nothing, they can get out. So you might end up-- your energy that you
detect in the detector might be the energy of your
gamma ray minus 2 times 511 MeV. This is what we refer
to as double escape. Close the quotes like that. So if this gamma ray right
here came in at 1,460 KeV, and the double escape peak--
if it undergoes pair production in the detector and both of
those 511s elevens escape, because a lot of them do-- where would you expect
there to be a double escape peak on this spectrum? AUDIENCE: Add that minus 11.022. MICHAEL SHORT: Yeah. Let's say, add that minus-- so we're at 1,460 minus 1.022. That comes out to about 450 KeV. 450 KeV right here, not
much going on, is there? You're not going to see
it in every detector. Especially the larger
the detector is, the less likely both of those
photons are going to escape. So this is where the
concept of detector size can tell you whether
or not you're going to see every peak
that's physically happening. So in this case, the germanium
detector is pretty big, it's pretty expensive. So chances are a lot
of those 511 KeVs, even though they're produced
in pairs, one of them didn't quite get out. Yeah, Luke? LUKE: So, the gamma from the
banana goes into the detector. And then it produces
pairs, and then those pairs are annihilated,
and that produces the vibration. MICHAEL SHORT: That's right. That's right, why don't we
write that down in steps for, let's call this pair
production in the detector. So step one would
be gamma emission. Step two would be
electron-positron creation. Step three would
be annihilation. Annihilation in the detector. And then step four would
be somewhere between zero to two photons escape. So we have, actually,
three scenarios that could happen here for pair
production inside the detector. One of them we just described. Where pair production
happens, you get annihilation in a
very short time frame, like tens of picoseconds
or hundreds of picoseconds. Both the gammas get out. That would have
produced a 460 KeV peak, which it might be there. But I can't tell if that's
a peak or if that's noise. So we don't really know. And chances are, the
reason that didn't happen is because the detector was big. So our next possibility. What if one of those photons
gets out and one of them doesn't? It then interacts via
Compton scattering, or photoelectric effect, or
any of the possible mechanisms. Then you'll end up
with an energy counted equal to energy of the
gamma minus only one of those things getting out. And we call that single escape. At what energy would that
single escape peak be? Oh. AUDIENCE: 511? MICHAEL SHORT: It would be-- that peak would be at the energy
the gamma 1,460, minus 511 KeV. So roughly 900 KeV or so. There we go, there it is. That's the second
bit of evidence that there is pair
production going on. Not only do you have a peak
at 511 KeV, which we have not explained yet, but you also
have the single escape peak, which is the energy of
your gamma minus one escape from a 511 KeV photon. Yeah? LUKE: When you mean
escape, do you mean escapes through the detector,
or what is that? MICHAEL SHORT: Yes. I mean-- when I escape, I
mean it escapes the detector and is no longer counted. So it might go and
drop somewhere else, but your detector
doesn't know it. So what's the third
scenario that could happen? What if zero of
these photons escape? What energy will you count? AUDIENCE: [INAUDIBLE]. MICHAEL SHORT: Exactly. So all that's going
to happen is it's going to look like the
photoelectric effect. In reality, you'll have
slightly, slightly lower energy, because
you have three work functions to subtract off from
the three photons doing stuff. But I would count
that as correct. It's going to look just like
the photoelectric effect. First, you get that
energy minus 1.022 MeV. And then both of
those 511 KeV photons interact in the
detector by, probably, photoelectric emission. And you just get another count
at this channel, right here. Now the last question
I want to ask you guys, where did this peak come from? Under what circumstance
would the detector just count 511 KeV? I'll give you a hint. There's a reason I drew gammas
going off in every direction. AUDIENCE: So they
don't hit the detector. MICHAEL SHORT: Yeah. So most of the gammas
don't hit the detector. But let's say you had a gamma
that went into anything else, like the copper shielding, and
it underwent pair production. And one of those gammas made
it through the detector. I'm sorry, one of those photons
made it through the detector. That's actually where these
things are coming from. Because most of
those gammas are not heading towards the detector. This is a very
small, solid angle. But surrounding the
rest of the detector is this really dense copper,
and these high energy gammas in this relatively
high Z material undergoes a lot of
pair production, so it's firing out 511 KeV
photons in all directions. And some of those
enter the detector when nothing else
enters the detector. And that's why you get this
511 KeV peak right here. So we haven't explained
every peak on this graph. Does anybody have
any ideas where-- what's that about? Or that? Or those? AUDIENCE: Cosmic radiation? MICHAEL SHORT: Yeah. Could be cosmic rays. That's probably
what's contributing to a lot of the noise,
here, as well as thermal noise in the detector. But what else haven't
we accounted for? Now, to bring this a
little more into reality, we ran an experiment
where we burned bananas. We didn't put a potassium-40
certified source in. We put bananas in. What else could be going on? AUDIENCE: Other isotopes? MICHAEL SHORT: Other isotopes. That's right. But you can identify
them quite easily, one, by checking to see where
you expect the photo peak. So just from the
decay diagram, you'll expect to see some interactions
or photoelectric effect interactions, at these
transition levels. Luckily, you know they're
not due to potassium, because potassium has
only got one of them. In addition, you should see
some very similar features. So if you have a
photo peak here, you would expect to see another
Compton edge 0.238 MeV away-- and it's kind of hard to
tell if it's going on, because that's a rather
weak photo peak-- and you would expect, then,
for the high energy gamma rays to see another single escape
peak-- maybe right there-- and add to the 511 KeV peak,
because those are all the same. So when you take the
spectrum of a real thing, and you have to deconvolute
it, or take it apart in terms of its
constituent interactions, it's important to know what all
these possible interactions are so that you can take them
apart and say, start off with a photo peak,
which should tell you what elements are there. And then you can subtract
off the expected amount of Compton scattering,
the expected amount of single
escape peak, and then see what's left over, what other
isotopes may there be that you haven't accounted for yet. The last thing I want us to
try, as a mental exercise, is to draw two spectra. Let's say, this will be
energy versus intensity. And for this I want you
to imagine that, at first, your detector is very small. And then I want you to
imagine that your detector is very large. And I'm going to keep
this visible so you can have this as a mental model. If we had just one
isotope, potassium-40, what do you think
the spectra would look like for an
extremely small detector and for an extremely
large detector? So where do we start? AUDIENCE: [INAUDIBLE] MICHAEL SHORT: That's right. And will there be any
difference between the two? Probably not. So small detector,
maybe a large detector is going to have a
larger intensity. But for the same
type of detector, you're going to have
pretty much the same thing. What's next? AUDIENCE: Compton edge? MICHAEL SHORT: Compton edge. So there's going
to be some energy that Compton scattering
is going to start out, and then it's going
to proceed up, thusly. Is there going to be any real
effect of the detector size? Probably not, because as soon
as you release that Compton electron, that electron slams
into all the other ones, and nanometers or
microns of material, and all the energy is collected. What's the real
difference going to be? AUDIENCE: The 511 peak? MICHAEL SHORT: That's right. So the 511 peak, and the
other associated ones. So for a really,
really small detector we have the possibility
for a double escape peak, a single escape peak,
and just more photo peak. What's the most likely scenario? AUDIENCE: Double escape? MICHAEL SHORT: Double escape. So if we go down here, let's
say if this difference is 1.022 to MeV, you would expect
a larger double escape peak. And what would you expect
your single escape peak to be? AUDIENCE: Smaller? MICHAEL SHORT:
Significantly smaller. So let's say this difference
right here is 511 KeV. How about for a large detector? AUDIENCE: Opposite. MICHAEL SHORT:
Quite the opposite. You might expect a tiny
or even nonexistent double escape peak, maybe a
larger single escape peak. But most of the time
you're just going to add on to your photo peak,
depending on the resolution of the detector. Because in this case,
for a small detector, if you have an interaction
inside that volume chances are most of those 511s get out. For a large
detector, chances are most of them stay in and undergo
their own Compton scattering, or photo peak reactions. So let's say that all
these detectors will also have a 511 KeV. We'll just mark that off. Let's just give them
the same height. What else are we missing,
if this is an ideal scenario with no noise? Well what are those five-- what
can those 511 KeV photons do? Can they make pair
production of their own? No. They're not high enough energy. In fact, they're half
the required energy. Can they undergo
photoelectric effect? Sure. That's probably where
we're getting those 511s. Can they undergo
Compton scattering? Why not? There's no minimum
energy to scatter. So what you're going to end up
with, then, is 238 KeV away. You should have
another little Compton edge at a distance of 238
KeV away from the 511 KeV. Now in reality, you
probably won't see it because you're going
to have other X-rays, you'll have
bremsstrahlung, which we'll talk about tomorrow, which
is that breaking radiation. You'll have
background radiation. And it might be hard to
see, but technically it should be there, because
any photon of any energy is going to have that same
sort of Compton edge shape. The shape changes
just a little bit, depending on the
energy of the photon, but you're always
going to have an edge. You're always going to
have some sort of a bowl. Just how big the edge
is compared to the bowl, well, we'll get
to that tomorrow. So it's a little
after 5 of 5 of. I think this is a
good place to stop, because it's the full conceptual
explanation of the ways that photons can
interact with matter. So I want to ask you guys if
you have any questions based on what we've done today? Yep? AUDIENCE: So 511 KeV's
the rest mass of electron? MICHAEL SHORT: Yep. AUDIENCE: So that's
just the energy you assume it has when
you have pair production? MICHAEL SHORT: That's right. So the electron and the
positron annihilate, turning their mass into energy. Since the rest mass of
each of those is 511 KeV, the photons come off at 511 KeV. AUDIENCE: OK. Got it. MICHAEL SHORT: Yep? AUDIENCE: So when you say
the electron and the positron annihilate, is the
positron just a hole? MICHAEL SHORT:
Ah, good question. The positron is not a hole. So like here, we were talking
about an electron hole pair. A hole would be, let's say, an
atom with a missing electron. A positron is a particle
itself of antimatter that has the same mass,
but the opposite charge, as the electron. And so every particle's got
its antimatter component, like there are antiprotons
and antineutrons, that if they find their regular matter
selves, do annihilate. Yeah? AUDIENCE: If the detector
doesn't pick up gamma rays directly, how do you measure-- like, why would a small
detector see double escape? MICHAEL SHORT: So
a small detector would see double escape,
because at first-- let's say a gamma ray interacts
and undergoes pair production. And so it's going
to, let's say, create an electron-positron pair. And it's going to give them
a whole lot of extra energy. So they're going to knock
around and ionize things. And that's going to
count up to the energy of the gamma minus MeV. Then, when they
annihilate, if it's a small detector chances are
those gammas just get out. We're going to be
going over why soon, when we get into mass
attenuation coefficients, or cross sections or
interaction probabilities. But as the energy
of a gamma goes up, it's interaction
probability goes way down. And this is a fairly
high energy photon, compared to, like,
the easier KeV X-rays that you tend to see. So chances are, these photons
get made from annihilation, but they don't stay
in the detector. Then the bigger the detector
is, the more mass there is in the way, and more
likely they get counted. But all of this happens,
well, at the speed of light. At least the photon part. And so it's so fast that
the detector picks it up as that sum of all the different
processes of energy in one time interval. Like I said, this is the harder
stuff, because it's not direct. It's a multi-step process
with different possibilities. But it's important to know where
the single and double escape come from, where
the 511s come from, which is outside the detector. Yes, have a question? AUDIENCE: Yes. Would you say the detector
can-- the detector itself can measure the
energy of a photon? Is the measurement of 511
KeV, is that due to the fact that it will hit an electron and
cause the-- what is it called? MICHAEL SHORT: Like
ionization cascade? Exactly. Yeah so if a 511 KeV
photon enters the detector, the detector does not know
until an electron interaction happens. So most of the photons
that enter this detector leave the detector. That's why if you actually look
at the banana stuff, which I'll pull up right now,
at the efficiency, check out those values, there. Efficiency is in the realm
of 10 to the minus 4 or 10 to the minus 3, which is to
say that out of every 1,000 or 10,000 photons that enter
the detector, one of them undergoes an
electron interaction and the other 9,999 just
goes screaming on through, and the detector does not
know that they're there. The way that Mike Ames
got these efficiencies is by putting a source
of known activity in, calculating how
many gammas the detector should have picked
up, and taking that divided by the number that
it actually picked up. And so that way, you know how
many gammas really went in, and how many gammas it saw. And that's how you get
the detector efficiency. And you will have to
account for this when you do this on the homework problem. So the only quantities
you're going to need is how many gammas that you
get, what's the efficiency, and then back that out. So you'll have to calculate
the activity of the bananas, and then figure out how much
a banana weighs, and then you should be able to calculate
the radioactivity of one banana in curies, or
becquerels, or microcuries. It's all good. So good question. So it's three of, so I'm
going to let you guys go. But I'll see you again
tomorrow, and we'll review a little bit of this stuff. And we'll get into more of
the math of the cross sections and why Compton scattering
and pair production take up the energies that they do.
