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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,