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  • PROFESSOR: OK, guys.

  • We're actually slightly ahead of where

  • I thought we'd be at this point, so I'm only

  • going to spend about half of today's lecture

  • finishing up some new material on mass parabolas

  • and stability.

  • I also got a comment in through the anonymous box

  • that said please leave a little bit of time

  • after class for questions.

  • So you can get them out right away, because I'm usually

  • running off to teach some other class at, like,

  • the IDC or some other building.

  • So from now on, I'll try and leave

  • about five minutes at the end of class

  • for questions on today's material,

  • and we'll make up with the second half-hour or 25 minutes

  • of this class being for all the questions on the material

  • so far in the first two weeks.

  • But first I wanted to give a quick review of where

  • we were Wednesday and launch back into mass parabolas, which

  • are ways of looking at nuclear stability in relative numbers

  • and even or oddness of nuclei.

  • So you saw last time we intuitively

  • derived the semi-empirical mass formula

  • as a sum of volume, surface, coulomb, asymmetry,

  • and pairing, or whether things are even-even or odd-even terms

  • with the coefficients in MeV gleaned from data,

  • and the forms of the-- and the exponents right here gleaned

  • from intuition.

  • Here we assume that the nucleus can

  • be thought of like a big drop of liquid with some charged

  • particles in it.

  • And so the droplet should become more stable

  • the more nuclei there are-- or the more nucleons there are.

  • But then you have some more outside

  • on the surface that aren't bonded to the others.

  • All of the protons are repelling each other over linear length

  • scales, because the radius of this liquid drop

  • would scale like A to the 1/3.

  • There's an asymmetry term, which means if the neutrons

  • and protons are out of balance, there's

  • going to be some less binding energy.

  • And then there's this extra part that

  • tells you whether the nuclei are even-even or odd-odd.

  • And this works pretty well.

  • If you remember, we looked at theory versus experiment

  • where all the red points here are theoretical predictions,

  • and all the black points are experimental predictions.

  • And for the most part, they look spot-on.

  • It generates the classic binding energy per nucleon

  • curve that you see in the textbook

  • and can predict from the semi-empirical mass formula.

  • Zooming in and correcting for, let's say,

  • just getting absolute values of errors,

  • you can see that, except for the very small nuclei

  • and a few peaks, which we explained

  • by looking even closer, the formula, well,

  • it predicts nuclear stability quite well on average.

  • Again, this line right here, if there's

  • a dot that lies on this blue line,

  • it means that theory and experiment agree.

  • And a deviation by a few MeV here and there, not too bad.

  • But we started also looking at different nuclear stability

  • trends, and we noticed that for odd mass number nuclei,

  • there's usually only one or sometimes none stable isotopes

  • per Z, whereas for even ones, there's quite a few more.

  • And we're going to be now linking up

  • the stability of nuclei versus what mode of decay

  • they will take in order to find a more stable configuration.

  • We looked quickly at the number of stable nuclei with even

  • and odd Z and noted that these places right here where

  • there are no stable nuclei correspond to technetium

  • and promethium.

  • There's no periodic table on the back of this wall,

  • but behind my back on the other wall,

  • there's that periodic table where

  • you can see the two elements that are fairly

  • light with no stable isotopes.

  • That's what those correspond to.

  • And the peaks correspond to what we call magic numbers

  • or numbers of protons or neutrons

  • where all available states at some energy

  • are pretty much filled.

  • And this goes for both protons and neutrons.

  • So something with a magic number for both n, number of neutrons,

  • and Z, number of protons, is going

  • to be exceptionally stable.

  • And we'll see how that's used as a tool

  • to synthesize the super heavy elements that we believe

  • should exist.

  • And finally we got into these mass parabolas.

  • I found this to be a particularly difficult concept

  • to just get mathematically.

  • If you remember, we wrote out the semi-empirical mass formula

  • and said if you take the derivative with respect to Z,

  • as we did it, you would get the most stable z for a given A.

  • And we started graphing for a equals 93

  • where niobium is stable.

  • That was just the one I had on the brain from some failures

  • in lab earlier this week.

  • We started plotting where those nuclei--

  • what is it-- the relative masses are for a fixed A. So

  • let's regenerate that one right now,

  • because we were a little fast at the end of last lecture.

  • Then I want to generate one for a equals 40.

  • And you'll see something kind of curious.

  • I'm going to leave this up here for a sec.

  • If you notice, for odd A nuclei, there's

  • only one parabola, whereas for even A, there are two.

  • Why that is, we're going to see when we

  • look at the table of nuclides.

  • But notice this nucleus right here

  • can decay by either positron emission or beta emission

  • to get to a more stable form.

  • And there are many real examples,

  • and I'm going to show you how to find them.

  • So let's start off by going back to the table of nuclides,

  • finding niobium-93.

  • Just go up one more chunk.

  • And there we are.

  • Niobium-93 is a stable isotope.

  • And if you want to see where it came from,

  • you can scroll down a little bit and see its possible parent

  • nuclides right here.

  • So let's say that niobium--

  • we'll draw it right there--

  • is stable.

  • We'll put it at the bottom of this parabola.

  • And let's work down in Z. So we'll move to zirconium.

  • Zirconium-93 ostensibly has a very similar atomic mass.

  • But if you remember that 93 AMU is a rather poor approximation

  • for the actual mass of all nuclei with A equals 93.

  • In fact, if you look very closely at the atomic masses,

  • zirconium-93 is 92.90.

  • Niobium-93, well, it looks like we

  • have to go all the way to another digit there.

  • 92.906.

  • You have to go to, like, even more digits 92.906375,

  • 92.906475.

  • So we go down in Z, and we've actually

  • gone up in mass by looks like the sixth or seventh digit

  • in AMU.

  • If we go up in mass, we go down in binding energy.

  • That tells us that there's something that's less stable.

  • And if you notice, we went down by a very small--

  • or we went up by a very small amount of mass.

  • Notice also that its beta decay energy is really, really small,

  • 91 kiloelectron volts.

  • So why don't we put zirconium just above?

  • And we note that that decay will happen

  • by beta, where the beta, let's say

  • if we have an isotope with mass number A, protons Z,

  • and let's just call it symbol question mark.

  • In beta decay, we have the same A.

  • We'll have a different Z. We're going

  • to have to give these symbols.

  • Let's call this parent, and we'll call that daughter.

  • Plus a beta, plus an electron antineutrino.

  • And what has to happen to that Z in order for everything

  • to be conserved?

  • It's the same reaction that we've got here for--

  • I'm sorry-- for zirconium.

  • So you'd have to have one fewer proton to release one electron.

  • And so that becomes same A but different Z.

  • And this is the beta decay reaction.

  • Let's go back a little farther.

  • We'll look at the possible parent nuclide for--

  • did anyone have a question?

  • AUDIENCE: Don't you need one more proton [INAUDIBLE]??

  • PROFESSOR: Let's see.

  • Which direction are we going in Z here?

  • That goes to yttrium.

  • It actually looks like it's going down.

  • Oh, yeah, for beta decay.

  • I'm thinking-- I have the reaction backwards.

  • Sorry.

  • I need one more proton to account

  • for the extra negative charge.

  • You're right.

  • OK.

  • Yep, I was thinking backwards, because we're now

  • climbing up the decay chain in reverse order.

  • So this could have come from yttrium-93 with a much

  • higher energy of three MeV.

  • So let's put yttrium right here.

  • That gives a beta decay.

  • And we'll just go one more back to strontium-93 Has

  • an even higher beta decay energy.

  • So let's put that up here.

  • And let's take a look at its mass real quick.

  • The mass of strontium-93, 92.914 AMU.

  • If we go back to niobium-93, now it's

  • noticeably different to, like, four significant digits

  • instead of six.

  • 92.914 versus 92.906.

  • And so that shows you that a tiny bit of mass and AMU

  • corresponds to a pretty significant change

  • in binding energy by that same conversion factor

  • that we've been using everywhere.

  • 931.49 AMU per big MeV per C squared.

  • Let's see.

  • Yeah.

  • OK.

  • So let's go now in the other direction, in the positron

  • direction.

  • Niobium can also be made by electron capture

  • from molybdenum.

  • So let's put molybdenum right here.

  • Let's say that around half an--

  • what did we have here?

  • It was like half an MeV.

  • Like that.

  • And let's see.

  • Molybdenum-93 could have been made by electron capture

  • from technetium-93 with an energy of 3.201 MeV,

  • even more extreme.

  • We'll go back one more, because there's a trend that I want

  • you guys to be able to see.

  • And this could have come from electron capture

  • from ruthenium.

  • I think I may have said rubidium last time,

  • but Ru is ruthenium-93.

  • And that 6.3 MeV, something like that.

  • And this is where we got to yesterday.

  • Now I'd like us to take a closer look at the decay

  • diagrams, which tells us what possible decay reactions can

  • happen in each of these reactions.

  • Since we're right here on the chart,

  • let's take a look at ruthenium turning into technetium by what

  • it says, electron capture.

  • So note that on the table, you can click on electron capture,

  • and if it's highlighted, then the decay diagrams are known.

  • It's not known for every isotope,

  • but for a lot of the ones you'll be dealing with, it is.

  • And you get something I have to zoom out for--

  • a lot, a lot of different decays.

  • What I want you to look at is this one here on the bottom

  • that I'll zoom in to.

  • That should be a little more visible.

  • So notice that if you want to go down the entire 6.4-something

  • MeV, it usually proceeds by B-plus or positron decay,

  • by either method.

  • And as you go up the chain, as these energy differences get

  • smaller, look what happens to the probability

  • of getting positron decay.

  • It shrinks lower and lower and lower.

  • So there's a trend that the larger the decay

  • energy for this type of reaction, the more likely

  • you're going to get positron decay.