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  • MICHAEL SHORT: So I'd like to do a quick two or three

  • minute review of the stuff we did last time

  • to get you back into where we were.

  • We were talking about different types

  • of technologies that use the stuff you'll

  • be learning in 22.01.

  • Everything ranging from nuclear reactors for producing power,

  • and the Cherenkov radiation that tells you-- well,

  • that the beta particles are moving faster

  • than the speed of light in water.

  • It's a neat thing, too.

  • I've actually been to this reactor

  • at Idaho, to the spent fuel pool.

  • Even the spent fuel, once it comes out of the reactor,

  • is still giving off betas and still

  • giving off Cherenkov radiation.

  • And you can tell how long it's been out of the reactor

  • by how dim the glow gets, which is pretty cool.

  • So you can tell how old a fuel assembly is by the blue glow.

  • So remember, I told you guys, if someone says,

  • oh, you're nuclear, do you glow green?

  • You can be like, no, it's blue.

  • That's the right way of things.

  • We talked about fusion energy and got

  • into some of the nuclear reactions involved

  • in fusion energy.

  • And I'll be teaching you more about these today.

  • Why fission and fusion work, it all

  • has to do with the stability and binding

  • energy of the nuclei involved.

  • And that'll be the main topic for today,

  • is excess mass, binding energy, nuclear stability.

  • We looked at medical uses of radiation,

  • from implanting radioactive seeds called

  • brachytherapy seeds in certain places to destroy tumors,

  • to imaging, to X-ray therapy, to proton therapy using

  • accelerators, or cyclotrons, to accelerate protons and send

  • them into people.

  • If you remember, last time we ran the SRIM code

  • of the stopping range of ions and matter,

  • and actually showed that protons all

  • stop at a certain distance in tissue,

  • depending on their energy and what you're sending them into.

  • Let's see-- we talked about brachytherapy.

  • We talked about radiotracers, and this

  • is going to be one of the other main topics for today, is

  • these decay diagrams, and figuring out not only

  • what products are made, but what energy levels do

  • the nuclei have, and how do you calculate

  • the energy of the radioactive decay

  • products and the recoil nuclei, which do take away

  • some of the energy.

  • We talked about one way to get rich.

  • If you guys can figure out one of the ways

  • to solve this moly-99 shortage.

  • Right now, it's mostly made in reactors.

  • The future has got to be accelerators

  • or some sort of switchable device

  • where you don't need to construct a reactor

  • to make these medical isotopes for imaging, and tracers,

  • and such.

  • And finally, we got all the way up

  • to space applications, shielding, crazy, different

  • types of shielding, like electromagnetic shielding,

  • to protect from high energy protons, all the way

  • to radiothermal generators, which use alpha decay

  • to produce a constant amount of energy

  • on the order of one to 200 watts for like 100 years.

  • And finally, to a different configuration

  • of nuclear reactors, where you can

  • design them to produce thrust, not necessarily electricity.

  • And that's where we stopped on Friday.

  • So let's move on to one of the things I'd alluded

  • to earlier, which is semiconductor processing.

  • This is actually a diagram from the MIT reactor,

  • because we have this beam port here.

  • Has anyone got to see the silicon beam

  • port at the MIT reactor?

  • Oh, seven, eight-- OK, about half of you.

  • For those who haven't, who here has not

  • had a nuclear reactor tour?

  • Oh man.

  • OK.

  • Well, you'll get one when you get to control

  • the thing in early October.

  • So you actually get to go down to the control room

  • and see the rest of what's going on.

  • So make sure to ask them, show us the beam port

  • for silicon ingots.

  • And I think I already told you the story about the poor UROP

  • who held the ingots up to their chest,

  • getting about 10 months of dose, which is not dangerous,

  • but it meant that for 10 months, they

  • could have no radiation exposure,

  • and they had to answer the phone.

  • So that's how we ensure safety around here.

  • There's other ones that--

  • applications that you're probably

  • carrying around in your pocket.

  • You can use the fact that charged

  • particles have very finite ranges in matter

  • to separate little bits of that matter from other things.

  • So this is actually how single-crystal sapphires

  • can be separated in little slivers for protective phone

  • covers, because sapphire is one of the most--

  • the hardest, or the most scratch-resistant materials

  • there is.

  • Single-crystal sapphire is exceptionally strong,

  • and optically transparent, and expensive.

  • I know that because on one of our experiments,

  • we use a single-crystal sapphire window to see into reactor

  • conditions at like 150 atmospheres, and 350

  • degrees Celsius, and pretty corrosive chemistry.

  • So you want to use as little as possible.

  • So you can use a big, expensive accelerator

  • to limit the amount of sapphire that you use.

  • And this is actually done here in Boston.

  • There's a facility not far from here that uses an accelerator.

  • And this is their super detailed diagram

  • of what the radiation looks like--

  • yeah, whatever.

  • But what they do is they take-- they accelerate protons.

  • They send them through bending magnets

  • to steer that beam path.

  • And then they send them into a large piece

  • of single-crystal sapphire, which is exceptionally

  • expensive to make.

  • And they can actually lift off a thin sliver

  • with micron precision.

  • The reason for that is the same reason

  • that we showed with that SRIM code.

  • If you have this exact energy of protons

  • going into well-known matter, you

  • know what its range is going to be with an uncertainty

  • or so of about a micron.

  • So you can have things that come out thin, uniformly

  • thick, and smooth, right away.

  • There's some other really wacky products--

  • like has anyone heard of these betavoltaic batteries?

  • No?

  • They rely on beta decay or the direct capture and electricity

  • generation from a radiation source

  • like radioactive tritium.

  • So in this little chip is about 2 curies worth of tritium.

  • You guys will learn in about a week,

  • how to go from activity in curies to mass,

  • or something like that.

  • And so this chip actually contains

  • a lot of radioactive tritium that

  • directly creates electricity.

  • So you can hook into that chip and produce

  • nanowatts for years.

  • So it's one of these batteries that lasts-- well,

  • as long as a couple half-lives of the isotope that's inside.

  • Now there's a trade-off here.

  • The shorter the half-life, the more active

  • a given isotope will be for the same number of atoms,

  • but the shorter it will last.

  • So you can have higher power for lower time, or lower power,

  • higher time.

  • It's the classic energy trade-off--

  • works the same way with irradiation.

  • And so now I wanted to get into some

  • of the more technical stuff, where

  • we'll be talking about nuclear mass and stability.

  • And this is where the nuclear stuff really begins in 22.01.

  • First, I want to make sure that we all agree on notation.

  • So I'll be writing isotopes in this sort of fashion,

  • where we refer to A as the atomic mass, or just

  • the total number of nucleons.

  • This is not the exact mass of a nucleus.

  • It just refers to the sum of the protons and neutrons

  • in the nucleus itself.

  • And a lot of what we'll be talking about today

  • is the difference between this nice integer mass number,

  • and the actual mass of the nucleus,

  • and that difference is given by the binding energy,

  • or the excess mass, which are directly related.

  • Z is just referred to as the atomic number,

  • or the number of protons.

  • It's what makes the element what it,

  • which makes the name kind of redundant.

  • But it's-- humans learned by association.

  • It's easier to remember element names or symbols

  • than which element is which just by the number of protons.

  • So a lot of times we'll use the name, or at least the symbol

  • just so we know what's going on.

  • And anything up here is some sort of a charge.

  • I do want to warn you guys of the dreaded multiple symbol

  • use or multiple use of symbols.

  • I'll try to stick for a lower case q--

  • will be charged.

  • And uppercase Q is going to refer

  • to the Q value or the energy consumed or released

  • by a nuclear reaction.

  • So they're both Q's but we're going to keep

  • one upper and lower case.

  • And like we mentioned before, let's

  • say we were to write a typical nuclear reaction,

  • like the capture of neutrons by boron

  • to produce lithium-7, helium-4, better

  • known as an alpha particle, and some amount of energy.

  • There's two places where we actually use this reaction.

  • One of them is as control rods.

  • A lot of reactors use boron for carbide,

  • or this compound B4C, which is conveniently solid, fairly

  • dense, and contains a whole lot of boron in one place.

  • Specifically, enriched in boron-10,

  • because boron-10 has a high cross section, or probability,

  • for neutron capture.

  • And the other one is in what's called

  • boron neutron capture therapy.

  • Have I discussed this with you guys already, BNCT?

  • Good, because that's what we'll be

  • talking about for a few slides.

  • And to write this whole reaction is the same thing

  • as writing this shorthand nuclear reaction.

  • So this is often how you see them in the reading,

  • and in papers, because it's shorter to do that.

  • But it's the exact same thing.

  • So I have a couple of questions for you guys then.

  • I have this extra Q here.

  • Where does that Q actually go?

  • So let's say boron and neutron absorb,

  • it produces two nuclei with different binding energies.

  • What happens to the excess energy

  • created from the conversion of mass to energy?

  • Yeah, Alex?

  • AUDIENCE: That could be heat.

  • MICHAEL SHORT: Yep.

  • And heat, and more specifically?

  • AUDIENCE: Kinetic energy.

  • MICHAEL SHORT: Kinetic energy of the radiation released.

  • And so that kinetic energy is actually

  • used to our benefit in BNCT, or Boron Neutron Capture Therapy.

  • The way this works--

  • once I hit play--

  • is you can either-- you can use any sort of source of neutrons,

  • either a reactor or an accelerator,

  • through a lower complex chain of events, like this.

  • In this case, an accelerator-- so you don't need a whole

  • reactor--

  • fires a beam of high-energy protons

  • into a beryllium target.

  • Does that sound fairly familiar?

  • Firing something at beryllium, releasing neutrons,

  • like what Chadwick was doing?

  • Except he was firing alpha particles into it.

  • This releases neutrons.

  • What they don't have labeled here is slowing down stuff,

  • or probably hydrogenous material,

  • so that the neutrons slow down to a lower energy.

  • And their probability of capture increases

  • or their cross section increases.

  • And if you don't know what a cross section is,

  • the definition is two slides away.

  • And the idea here is that these neutrons then enter the brain

  • or wherever the tumor happens to be.

  • And we rely on the fact that tumor cells consume resources

  • much faster than regular cells, especially neurons, which

  • after you're about five, don't tend to grow very much.

  • So it's all downhill by the time you enter kindergarten.

  • And we use that to our advantage so that the neutrons coming in

  • will hit the cancer cells, which will preferentially

  • uptake the borated compounds, leaving most

  • of the normal cells intact.

  • And the difference in dose can be a factor of 5

  • or a factor of 10.

  • So that the cancer gets fried while doing as little damage

  • as possible to the remaining brain cells, of which we

  • have fewer and fewer every day.

  • So I say statistically speaking you

  • guys are probably smarter than me

  • if we go by number of neurons in your brain,

  • because I think I'm the oldest person in the room.

  • And so now we can start to explain, how does BNCT work,

  • and why did we make the choices that we did?

  • For example, they use 30 MeV protons in order

  • to induce these neutrons.

  • So we have a nuclear reaction that looks something like this.

  • We start off with beryllium-9 plus a proton--

  • let's just call it hydrogen to stick

  • with our normal notation--

  • and becomes-- well, can someone help me balance this reaction?

  • We know we get a neutron.

  • What else is left?

  • So Monica, what would you say?

  • AUDIENCE: Let's see--

  • MICHAEL SHORT: Even just say number of protons and neutrons,

  • and we'll figure out the symbol later.

  • AUDIENCE: Number of protons should be--

  • MICHAEL SHORT: Sorry, oh, that's a 4.

  • AUDIENCE: The number of protons should be five.

  • MICHAEL SHORT: Yep.

  • Five protons, which means it's boron.

  • And number of neutrons in the nucleus?

  • Someone else?

  • Yeah?

  • AUDIENCE: Nine.

  • MICHAEL SHORT: Nine.

  • So we have boron-9.

  • Not a stable isotope of boron, but it doesn't really matter,

  • because boron-9 almost immediately decays into

  • an alpha, and an alpha, and a hydrogen.

  • But this nuclear reaction right here

  • is what we'll be studying for a little bit.

  • And there'll also be some amount of energy.

  • And this Q can actually be positive or negative.

  • No one said there had to be energy released

  • in a nuclear reaction, because in this case,

  • we actually start off with 30 MeV protons and roughly zero

  • MeV beryllium.

  • If you want to get really exact, it's

  • on the order of about 0.01 eV, which

  • is why we neglect the kinetic energy of beryllium

  • at room temperature.

  • There are other reactions that when

  • you fire a proton into them will produce

  • neutrons, such as the absorption of lithium.

  • But can anyone think of why we'd want to use

  • beryllium instead of lithium?

  • Kristin, what do you think?

  • What could be a bad thing about using lithium?

  • You ever throw in water?

  • AUDIENCE: No.

  • MICHAEL SHORT: OK.

  • Then I should show you what happens

  • when you throw it in water.

  • There's a few bad things about lithium.

  • It does this when you throw it in water.

  • It's one of the alkali metals.

  • It's got an awfully low melting point.

  • It reacts with oxygen to produce an oxide almost

  • instantaneously.

  • So if you ever take a lithium battery apart,

  • which you shouldn't, but if you watch the video of somebody

  • else doing it, you'll see that the lithium foil turns black

  • almost instantly.

  • It also has a pretty poor thermal conductivity

  • and doesn't hold that structural integrity when it melts.

  • So it's not that good of a target to use.

  • Beryllium's pretty cool in that it's the lightest

  • structural material there is.

  • Folks tend to make satellites out

  • of it, because it costs a lot of money

  • to launch things into space.

  • And if you want something that has a high melting point,

  • and is light, and is structural, beryllium's your way to go.

  • It also happens to be a great neutron generator.

  • And then why 30 MeV?

  • In this case, we're going to use a table called JANIS,

  • which I've got open over here.

  • And I just have to clone my screen so you can see it.

  • This is a resource that I think you guys are going to be using

  • quite a lot in this course.

  • We have a link to it on the learning module site.

  • And I'm going to show you how it works right now.

  • So I tend to use the web version because it works

  • on any browser, any computer.

  • And now you can start to pick which nuclear reaction you're

  • looking at.

  • And you can get tabulated cross sections.

  • So I'm going to start by zooming all the way out.

  • We can pick our incident particle.

  • Since in this case, we're looking

  • at the firing of protons into beryllium,

  • I'll pick the incident proton data right here.

  • There's a lot of different databases with sometimes

  • conflicting information.

  • I tend to go with the most recent one you can find.

  • And click on cross sections.

  • And this is, again another table of nuclides, anything

  • in green there's data for.

  • Anything in gray, there isn't.

  • So let's go all the way back to the light nuclei,

  • zoom in, go back down to the light nuclei

  • again until we find beryllium-9.

  • Double-click on that, and let's look for the anything

  • cross section.

  • And this is a pretty wide energy scale.

  • So you can actually change your X minimum and maximum.

  • So let's change it to a minimum and maximum--

  • I don't know-- a maximum of 50 MeV.

  • We don't have to see all of that other stuff going on.

  • 50 MeV and maybe a minimum of 10.

  • If you notice-- actually, I'll go back to 1.

  • And I want to point something out.

  • You can actually get a good yield of beryllium.

  • Let's see-- you can actually get a good yield of neutrons

  • by firing protons at beryllium in lower energies.

  • But I notice there's this interesting feature

  • right around there.

  • The cross section's flatter.

  • And so if you want to get an--

  • ensure that you get the right dose,

  • you might want to deal with a flatter cross

  • section or a flatter probability region,

  • so that you have something more predictable

  • instead of in a really high slope region.

  • But some of these nuclear reactions

  • actually take extra energy in order to move forward.

  • And we'll show you another example pretty quickly.

  • Let's go back to our slides here.

  • Then another question is, how does the boron

  • only get into the cancer cells?

  • Like we mentioned before, cancer cells

  • are actively growing, which means

  • they need a very large and active blood supply.

  • And so it's one way for things to, let's say,

  • not quite cross the blood-brain barrier.

  • If the cancer cells are growing and your neurons aren't, then

  • your cancer cells are going to use more energy, take

  • in more sugar, which might be doped with boron,

  • or some other compound doped with boron,

  • and that's all you can get the boron into the cells

  • that you want.

  • And then why was boron selected for the therapy?

  • Let's think about that.

  • What happens after the neutron is created?

  • And let's write the next stage of the reaction.

  • In boron neutron capture therapy,

  • we rely on doping the patient with boron-10 to release

  • an alpha particle, and lithium-7 and a gamma ray.

  • So now what we can start doing is

  • look at the table of nuclides, which I'm going to teach you

  • how to read now, to figure out-- let's

  • say that this neutron had an energy of about zero eV

  • and the boron nucleus had an energy of about zero eV.

  • And in the end, all this stuff here has gained or lost

  • some sort of energy cue, Q. And today we're

  • going to teach you how to calculate this Q.

  • So I want to skip ahead to how to read the table of nuclides.

  • So there's all-- this is like the poster you'll

  • see in every nuclear building.

  • It's kind of what makes us, us.

  • What you'll notice is that there's a whole lot

  • of nuclei at the lower left.

  • They are the light ones.

  • At the upper right, they are the heavier ones.

  • And they're colored by half-life.

  • In general, the blue ones will be stable,

  • and the further away you get from blue, the less stable

  • they get.

  • So right away, without even delving deeper, what patterns

  • do you guys notice here?

  • Yeah, Alex?

  • AUDIENCE: As they get bigger, heavier, they're more unstable.

  • MICHAEL SHORT: Yeah.

  • There's a whole section where there's no more blue.

  • There are no stable elements.

  • So stability drops off after a certain point.

  • And what about in the region of stable isotopes?

  • Does anybody notice any repeating patterns here?

  • Take a look at every other row.

  • There's a bunch of blues and then one,

  • and then a bunch and then one, and then more and then none.

  • That must be technetium, because that's

  • the only element around there that doesn't have any.

  • And then a bunch of blues.

  • So every other row--

  • and in this case, it's increasing number of protons--

  • has more or fewer stable isotopes.

  • It turns out that the even numbered

  • isotopes have a lot more stable ones, for reasons that we'll

  • get into pretty soon.

  • If you zoom in a little bit, you can

  • see all the different isotopes so you

  • can select which ones you want.

  • And again, if you look really closely, that's--

  • let's say, neon right here has got a few stable ones.

  • Sodium has one.

  • Magnesium as three.

  • Aluminum has one.

  • And this pattern repeats all the way up

  • to the point where you don't really

  • get any more stable isotopes.

  • If you double-click on one of them,

  • you get all the information that you'll

  • need for the next three or so weeks of the course.

  • So in this case, I picked on sulfur-32,

  • one of the stable isotopes of sulfur.

  • So if you notice it doesn't have any decay mechanisms here,

  • but it does say its atomic abundance.

  • So you can know how-- what percentage is normally

  • found in nature.

  • And then there's a few other quantities

  • that is going to be the topic of what's going on here.

  • Let's start with the atomic mass.

  • If you notice, the atomic mass is slightly less than 32,

  • 32 being the mass number, or the total number of protons

  • plus neutrons in the nucleus.

  • The actual mass is a little lower by that amount

  • right there, the binding energy.

  • It might be a little funny because I've

  • given you a mass in AMU, and a binding energy in kiloelectron

  • volts.

  • I want to remind you that these are the same thing.

  • The conversion factor you'll be using over and over again

  • throughout this course, especially on the next p sets,

  • is one atomic mass unit is 931.49 MeV c squared.

  • Yeah-- I'm sorry.

  • Yeah, never mind, put that there.

  • So then, again, one, don't round--

  • because we've had times when folks said,

  • ah, this is about 931.

  • And when you're off by half an MeV,

  • you could be at a totally different decay level

  • or get a positive Q when it should be negative,

  • or vice versa.

  • And let's take a quick look here to say,

  • if this atomic mass is 31.9720707 AMU--

  • this is why I brought a calculator.

  • Normally I do mental, math but since I

  • told you guys don't round, I can't

  • do eight significant digits in my head.

  • So I'm going to get that in there--

  • 0707.

  • If any of you guys want to follow along,

  • I encourage you to.

  • And say minus 32, which is the mass number.

  • So in this case we're taking the actual atomic mass

  • minus the actual--

  • the mass number.

  • In this case, it's 32.

  • In this case it's, 31.9720707.

  • And we end up with minus 0.0--

  • I'm going to put all the digits here--

  • 293 AMU.

  • If we convert this to MeV--

  • times 931.49, we get minus 26.0159 MeV.

  • See this number anywhere on the KAERI table?

  • Right there-- that's the excess mass.

  • And in this case, we usually give this the symbol delta

  • for the excess mass.

  • And these are how these quantities

  • are directly related.

  • The excess mass-- well, actually,

  • what does the excess mass really mean?

  • It's the difference between the actual mass

  • and a fairly poor approximation of the mass.

  • So the excess mass doesn't really

  • have that much of a physical connotation.

  • But it is nice, because if you know very well

  • the tabulated atomic number--

  • I'm sorry, the-- yeah, the mass number and the excess mass,

  • you can figure out--

  • let's see-- yeah, you can figure out

  • what the real atomic mass is.

  • And I want to switch now to the actual table of nuclides

  • and show you one example.

  • If you want to very quickly jump between isotopes,

  • you can type them in right up here.

  • And does anyone know what the gold standard for atomic mass

  • is?

  • And I'll give you a hint, it's not gold.

  • Yep?

  • AUDIENCE: Carbon-12.

  • MICHAEL SHORT: Carbon-12.

  • What do you think the excess mass of carbon-12

  • is going to be without doing any calculations?

  • AUDIENCE: Zero.

  • MICHAEL SHORT: Exactly.

  • Zero.

  • So if we go to carbon-12, because that

  • is set as the standard, the way atomic masses were done

  • was carbon-12 weighs exactly 12 AMU.

  • The excess mass here is zero.

  • And that's why the atomic mass is 12.0 to as many decimals

  • as we care to note.

  • So is everyone clear on what excess mass is?

  • Yep?

  • AUDIENCE: What's the point of c squared for that conversion?

  • MICHAEL SHORT: So mass does not actually equal--

  • oh right, and it's actually on the--

  • where did my chalk go?

  • It's actually down below.

  • The point is that energy is related to mass by c squared.

  • So they're not the same units, but they're

  • directly convertible.

  • AUDIENCE: OK.

  • MICHAEL SHORT: Yep.

  • And so this way, you have an E over a c squared.

  • You get an m, and there we go.

  • I had the units upside down.

  • However, carbon-12 does not have a zero binding energy.

  • Yeah, Luke?

  • AUDIENCE: How come when you did that calculation,

  • you didn't use the c squared?

  • So like, it seems like then that would be 26.0159--

  • MICHAEL SHORT: MeV per c squared.

  • Yeah.

  • AUDIENCE: But they don't say that up there--

  • or it didn't say that on the table.

  • MICHAEL SHORT: Yeah, that's true.

  • AUDIENCE: [INAUDIBLE]

  • MICHAEL SHORT: So it is funny, right?

  • The binding energy is give it in keV, and that's correct.

  • An energy is an energy.

  • An excess mass, it really should say

  • keV per c squared, because if we're talking in units of mass,

  • it's got to be in m.

  • Or in this way, you could say m is an energy per c squared.

  • So this, to me, is a semantic inconsistency in the table.

  • But you guys will know that a mass is always

  • going to be an AMU, or kilograms, or MeV

  • per c squared.

  • And energies will be in MeV, keV, some sort of eV,

  • usually, in this course.

  • The binding energy, though, that's correct.

  • That's in keV, because that's an actual energy.

  • Now then the question is, what does the binding energy

  • actually represent?

  • Does anyone remember from Friday or Thursday?

  • I can refresh your memory, because that's

  • what I'm here to do.

  • The binding energy is as if-- let's say

  • we're assembling carbon-12 from its constituent nucleons.

  • There's going to be 12 of them.

  • Let's say we had six protons and six neutrons.

  • We can calculate the total mass energy

  • of this ensemble of nucleons when they're infinitely

  • far apart from each other.

  • And forgive the little--

  • it's not to scale.

  • But they are infinitely far apart from each other.

  • And we can say that--

  • let's say there is Z number of protons.

  • So we'll say the binding energy is

  • the number of protons times the mass of a proton

  • plus the number of neutrons--

  • A minus Z-- times the mass of a neutron

  • minus the energy of the assembled carbon-12 nucleus.

  • So there's actually a measurable difference

  • in mass between six protons and six neutrons,

  • and the actual mass of a nucleus with atomic number A and--

  • I'm sorry, with atomic number Z and mass number A c squared.

  • So is everyone clear on how we arrived at this formula?

  • It's effectively the energy released

  • when you take the individual nucleons, assemble the nucleus.

  • You don't have as much mass as when you started.

  • Or in some cases, you might have a little more mass

  • than when you started if things are particularly unstable.

  • And you can use the excess mass and binding energies

  • in relative amounts to see, is a nucleus going to be stable?

  • For example, let's look at iron-55 I'm

  • going to jump here, make it a little bigger

  • so the important stuff is easier to see.

  • And if you notice, the binding energy of iron-55--

  • there's quite a bit of it.

  • It's very well-bound.

  • In fact, this is one of the most well-bound nuclei

  • in the whole chart of nuclides.

  • Let's look at something that we know

  • to be particularly unstable.

  • Someone have any idea?

  • Let's just add like 20 neutrons to iron

  • let's see if it even exists.

  • No-- doesn't happen.

  • Let's try adding 10 neutrons to iron--

  • or go even crazier.

  • What about 70?

  • Too small-- all right, let's meet somewhere in the middle--

  • 68.

  • Still a pretty high binding energy,

  • but you can look at it as a difference

  • in binding energy per nucleon.

  • So in this case, the binding energy per nucleon--

  • if you take the binding energy and divide

  • by the total number of nucleons, will give you

  • a relative measure of how tightly bound that nucleus is.

  • Now these are not absolute things.

  • You can't just say, certain binding energy

  • leads to certain stability, but they do give you

  • pretty good trends to follow.

  • And we're actually going to be coming up with--

  • probably on Thursday--

  • a semi-empirical formula to get the rough binding energy

  • of any particular assembly of protons and neutrons.

  • And it follows experimental calculations pretty well--

  • surprisingly so.

  • I want to jump back to here, because I've

  • mentioned cross sections, and I want to actually define

  • what a cross section is, because this is a quantity

  • that you're going to be using everywhere.

  • Let's say that we fired a beam of particles--

  • it doesn't matter what it is--

  • at a target of other particles.

  • Let's say, the beam particles are atom A,

  • and the target particles are atom B.

  • And once these A particles pass through the target

  • B, a little bit fewer of them come

  • out the other side reacted, or unscathed, or unscattered.

  • And some of them are absorbed, or scattered, or bounced off,

  • or scattered backwards, or what have you.

  • We can write the sort of proportionality

  • constant between the change in intensity of our A beam

  • and the thickness of our slab.

  • And we give that proportionality constant

  • this symbol, little sigma.

  • We'll get something going up here.

  • Little sigma, which we call the microscopic cross section.

  • It's in effect, a constant of proportionality

  • that relates the probability of absorbing

  • an atom from this beam I--

  • or from this beam of atoms A through a slab of B.

  • And then if you take this formula,

  • you divide by that delta X--

  • so I'm going to take what's on there

  • and say delta I over delta X equals minus cross section

  • ABn--

  • which refers here to the number density.

  • So I'll keep our table of symbols

  • altogether so it's a little easier to follow.

  • n is our number density, which means the number of atoms

  • per unit volume.

  • Usually, in nuclear quantities, we use centimeters

  • because these are things that are actually fairly measurable,

  • and cross sections are actually in units

  • of centimeters squared.

  • And let me finish that expression.

  • We had the number density of our target

  • B. We had our initial intensity, and that's it.

  • Anyone know how to solve this differential equation?

  • If we take the limit of small deltas,

  • it should start to look like a differential equation.

  • The final answer is up there on the screen.

  • Does anyone remember the method to actually solve

  • this differential equation?

  • This is the easy one--

  • separate the variables.

  • So in this case, we can divide each side by I of of X,

  • multiply each side by X. I'm going to bring this up

  • so I'm not bending down.

  • So we have dI over I equals minus sigma ab n of B times dX.

  • Integrate both sides and we get log of I

  • equals minus sigma ab n of BX, and some integration constant.

  • You can apply an initial boundary condition to say at X

  • equals zero, the intensity of the being

  • x was some intensity I naught.

  • Whatever intensity of the beam that we initially

  • fired at the target.

  • And by combining these two, you end up

  • with the expression you get right there,

  • which is that the intensity of the beam coming out

  • is the initial intensity times e to the minus sigma ab nbx.

  • And we've kind of derived the idea

  • of exponential attenuation.

  • For those who haven't seen that word before,

  • attenuation or the gradual removal of the beam of incident

  • particles by whatever the target happens to be.

  • This quantity right here, we actually

  • have another symbol for it, which we give big sigma.

  • And in this case, big sigma we call the macroscopic cross

  • section.

  • I'll draw a box around these so we

  • know these are our symbols that we're keeping defined here.

  • And so you may see that the microscopic cross section just

  • depends on single reactions between the incoming atoms

  • A and the target atoms B. The macroscopic cross section

  • depends on how much B is there.

  • So if you want to get per atom probabilities of absorption

  • scattering, whatever thing you're looking at,

  • you use the microscopic cross section.

  • And if you have a finite amount of stuff there,

  • and you know the number density of your substance B,

  • you can use the macroscopic cross section

  • to get actual total probabilities of beam

  • attenuation--

  • or to calculate exponential attenuation.

  • We're going to see this again in another form when

  • we talk about designing shielding,

  • and how much shielding do you need

  • to remove how much of the beam?

  • Well, this quantity right here, there's

  • actually tabulated values for a lot of this stuff at the--

  • on the NIST website.

  • And I have links to that as well on the Stellar website,

  • so you can-- instead of having to look these all up on JANIS

  • and multiply number densities, there

  • are some easier graphical functions you can just

  • find the value for.

  • But we'll get back to that in a few days.

  • So anyway, on reading the KAERI table,

  • there's a few quantities right there.

  • We've already defined what the excess mass

  • and the binding energy is.

  • And I want to note right here, if you

  • want to actually calculate binding energies by hand, which

  • I'm going to ask you to do a bit on problem set 2,

  • you'll need to know what the mass of the proton,

  • and the neutron, and the electron

  • are to, again, usually like six or seven digits

  • is the idea behind this course.

  • Notice that they're not exactly one atomic mass unit,

  • because one atomic mass unit, again,

  • was set with that carbon-12 standard.

  • I'm not going to use the word gold standard because that's

  • a misnomer in this field.

  • And so like I said, what does excess mass really mean,

  • physically?

  • Not much, because it's the comparison

  • to an arbitrary standard or a rather poor approximation

  • of the mass.

  • The binding energy actually does represent the conversion

  • of mass to energy when you assemble

  • a nucleus like Voltron-style from its constituent nucleons.

  • So let's try a few examples in class right here.

  • I'd like you guys to follow around and try and calculate

  • the binding energy of each of these three nuclei of sulfur.

  • Let me get a better blank board so we can follow along.

  • And there's a few different ways of calculating

  • that binding energy.

  • You can do it by the excess mass.

  • You can do it by--

  • let's go back to the table of nuclides

  • so I can show you how I would do it.

  • Let's start with sulfur 32.

  • And we'll write up the quantities that we're--

  • that we know.

  • Let's say the excess mass is the actual mass

  • minus the mass number.

  • The binding energy is Z times mass of hydrogen plus A minus Z

  • mass of a neutron minus the actual mass of that nucleus

  • with AZ c squared.

  • And then what we can do is rearrange this excess mass,

  • isolating the mass term right here, and make a substitution.

  • So we can say the mass is actually

  • the excess mass plus A. Stick that in right there,

  • and now we can calculate and confirm the binding energies

  • that we see right here from tabulated excess mass values,

  • atomic number, mass number, and the masses of a hydrogen

  • atom and a neutron, which, for reference,

  • I'll write up here as well.

  • So the mass of a hydrogen is the mass

  • of a proton plus an electron.

  • So 1.0072-- 007276 plus 0.000--

  • make that a little easier to read--

  • how many zeros-- 00054858 AMU.

  • Mass of a neutron, surprisingly close to Chadwick's prediction.

  • 8664 AMU.

  • So now I'll head back to the table of nuclides.

  • And let's see if you guys can follow along.

  • What we want to do is try to confirm this binding

  • energy using the atomic mass, the excess mass,

  • or if we don't even know the atomic mass,

  • we can use the excess mass plus A right there.

  • So let's see-- Z, in this case, for sulfur, is 16 times

  • the mass of hydrogen. This is definitely a calculator

  • moment, because like I said, I don't know about you guys,

  • but I can't do eight significant digits in my head.

  • 0054858-- 1.007855-- probably enough digits--

  • plus 16, because there's--

  • mass number here is 32.

  • The atomic number is 16.

  • That leaves us with 16 neutrons times the mass of a neutron,

  • 08664--

  • minus the excess mass, which in this case is 26.015 MeV--

  • 015 MeV per c squared.

  • So thanks for that-- thank Jared for that question because,

  • indeed, the excess mass, if you want

  • to write it in terms of a mass, should

  • be in MeV or keV per c squared minus A,

  • which is 32 times c squared.

  • So let's do all this out--

  • shouldn't take too long.

  • 007825 plus 16--

  • 1.008664 minus 32 minus 26.015 divided by c squared.

  • It's basically nothing.

  • Gives us on the order of--

  • let's see-- times c squared.

  • What did we get right here?

  • AUDIENCE: Is the 26 negative?

  • MICHAEL SHORT: Ah, let's see.

  • I believe it is, because we have to subtract the mass,

  • and we're substituting in this delta--

  • AUDIENCE: Isn't the delta negative?

  • MICHAEL SHORT: Oh.

  • Good point.

  • There is a negative there.

  • So that's minus negative that.

  • And A is 32.

  • Thank you.

  • Yeah, good point.

  • Let me try this again.

  • Ah, I know what I'm doing wrong.

  • This part right here, we want to convert to AMU.

  • So we can take our minus--

  • thank you-- 26.015 MeV per c squared and divide

  • by our conversion factor, 931.49--

  • let's see-- MeV per c squared per AMU.

  • What does that give us?

  • 26.015 over that.

  • 0.027928 AMU negative.

  • Let's put that in and see how we do.

  • So plus 0.027928 minus 32.

  • And then we get 271.--

  • I'll just say 764 MeV.

  • I think six digits is enough.

  • The actual binding energy, 271.780 MeV,

  • so we're off by 16 electron volts--

  • close enough.

  • Also note that I used a five-digit accurate conversion

  • factor.

  • That might be part of the source of it.

  • Does someone have a question?

  • AUDIENCE: Yeah.

  • In the equation on top, you did the atomic number

  • times the mass of the proton, but in the one on the bottom,

  • you used atomic mass times the mass

  • of the hydrogen including the electron.

  • Is there--

  • MICHAEL SHORT: Oh yeah.

  • I actually added the two.

  • So the mass of an electron, since it's got that extra zero,

  • makes so much of a--

  • so the mass of--

  • oh-- yeah.

  • The mass of hydrogen would be the proton

  • plus the electron right there.

  • AUDIENCE: Right.

  • But why do hydrogen, though, if [INAUDIBLE] just the proton?

  • MICHAEL SHORT: Oh, because there's an electron there, too.

  • Now this can usually be neglected

  • because it's such a small fraction compared

  • to everything else.

  • So now we're talking about-- what--

  • the fifth or sixth decimal place.

  • But just for exactness, I stuck on it.

  • Yeah, in your calculations, you can try with and without,

  • and I think you'll find that it doesn't matter that much,

  • because in the end we get the binding energy that we see

  • on the table to within 16 electron volts for a total

  • of-- yeah--

  • 271 MeV.

  • That's pretty accurate.

  • Yeah.

  • AUDIENCE: If you wanted to calculate

  • the like energy released from a reaction,

  • would you do the binding energy for [INAUDIBLE] reactants

  • that's trapped products for the reactants?

  • MICHAEL SHORT: That's the next slide.

  • We'll get right there.

  • Yeah, so you're catching on to where we're going.

  • So once you can calculate either the excess mass,

  • or the binding energy, or the total mass of any nucleus,

  • you can start to put them together

  • into nuclear reactions.

  • So since you asked, let's take a quick look at them.

  • Where is our nuclear reaction board?

  • Anyone mind if I hide this board for now,

  • so we can go back to our original?

  • OK.

  • Let's take a look at this reaction right

  • here, the actual boron neutron capture therapy reaction.

  • And now we can get towards calculating this Q, what

  • the difference is between the--

  • the total energies of the products and the reactants,

  • and where does that go?

  • So now we can either look up or calculate

  • the binding energy of each of these nuclei,

  • subtract off the energy of the gamma, which I've looked up

  • already, is about 0.478 MeV.

  • And we can figure out what the total Q of this reaction is.

  • So in this case--

  • I'll skip ahead to the slide where

  • I've got it because that way I won't write anything wrong

  • on the board--

  • got everything right up here.

  • We assume that both boron and the neutron

  • have roughly zero kinetic energy.

  • And at the end, they come out with

  • some other kinetic energies as well as this gamma ray.

  • The sum of this energy differences, we refer to as Q.

  • And we can actually confirm this total Q

  • with a few different methods.

  • In this case, it's always conserve something.

  • That's the whole theme of this course,

  • is you can conserve total masses,

  • you can conserve total kinetic energies.

  • We may not know those, but tabulated in the KAERI table

  • are the binding energies of each of these nuclei.

  • So let's try that out right now.

  • So let's look at the binding energies of each

  • of these nuclei and see what the difference is,

  • the total energy released.

  • First of all, what's the binding energy of a lone neutron?

  • Anyone have any idea?

  • I see a lot of these--

  • zero.

  • Yep.

  • You haven't assembled an nucleus out of a lone neutron,

  • so we'll go with the neutron has a binding energy of zero MeV.

  • Boron, not quite the case, but we

  • can go back to the table of nuclides and punch that in--

  • boron-10.

  • We can look up its binding energy,

  • which is about 64.7507--

  • I keep saying about, which is exactly 64.7507 MeV.

  • And then our other two nuclei, helium-4--

  • so you can punch in helium-4 here.

  • It's got a binding energy of exactly 28.295673 MeV.

  • And finally, lithium-7, let's punch that in.

  • I think you guys are going to get

  • very familiar with this table.

  • There's a few versions out there.

  • There's a new slick Java version that I

  • found a little hard to use.

  • So I like the text-only version, because it's just as simple

  • and fast as it gets--

  • 39.244526-- 526.

  • So any sort of increase in total amount of binding energy

  • between the reactants and the products

  • is going to release or absorb energy.

  • Now because boron does capture a thermal neutron, or a neutron

  • with approximately zero eV of kinetic energy,

  • does anyone have any idea whether this would

  • release or consume energy?

  • In other words, do think this is an exothermic or endothermic

  • reaction?

  • Yeah, Alex?

  • AUDIENCE: I'm guessing that heat would

  • be released through the material-- the capture

  • material would be heated up.

  • MICHAEL SHORT: OK.

  • Indeed.

  • If the total Q value is greater than zero,

  • we refer to this as exothermic--

  • kind of like in chemistry.

  • And if Q is less than zero, we refer to this as endothermic.

  • So let's do our binding energy subtraction now.

  • We want to figure out how much excess binding energy is

  • released.

  • So I'm going to take the reactants--

  • I'm sorry-- I'm going to take the product.

  • So helium 295673-- add lithium 244526--

  • subtract boron 0.7507-- subtract the neutron, which is zero,

  • and we're left with 2.79 MeV.

  • And because it's positive, this is an exothermic reaction,

  • which is what we'd expect, because this reaction actually

  • happens.

  • If this was an endothermic reaction,

  • what could you do to make it occur?

  • Yeah?

  • AUDIENCE: Heat up the reactants.

  • MICHAEL SHORT: Like with temperature,

  • or what do you mean?

  • AUDIENCE: Make them have higher kinetic energy or--

  • MICHAEL SHORT: There you go.

  • So actually-- yeah-- you kind of said the same thing twice.

  • Heating things up does give them higher kinetic energy.

  • If you rely on temperature, you'll

  • be imparting eV worth of kinetic energy.

  • But if you accelerate them, or get them

  • from a different nuclear reaction,

  • and you get them up to the MeV level,

  • where whatever this Q value could be might be negative,

  • then you can get the reaction to occur.

  • For example, what is the Q of that reaction?

  • AUDIENCE: Negative 2.79.

  • MICHAEL SHORT: Negative that.

  • So in this case, if you want lithium

  • to absorb an alpha particle, and make boron and a neutron,

  • you would have to accelerate the alphas

  • to that same amount of energy in order to get this to occur.

  • So nuclear reactions do go both ways, just not as easily.

  • Kind of like chemical reactions, you

  • can drive them in different directions

  • by changing the temperature or changing the concentration

  • of the reactants.

  • Here the concentration doesn't matter.

  • But the kinetic energy related directly to the temperature

  • definitely is.

  • And so in this case, it's 2.79 MeV.

  • If I tell you the gamma ray takes off 0.478 MeV of that,

  • we're left with 2.31 MeV between the lithium

  • nucleus and the helium nucleus.

  • Now my next question-- my last question for you today--

  • oh man--

  • is what's the split?

  • I think I don't want to keep you longer, because it's

  • one minute of 10:00.

  • So this is the question that we're

  • going to pick up with on Thursday,

  • which is how much of the energy is taken off by helium?

  • And how much is taken off by lithium?

  • Sorry, I should have kept better track of the time.

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