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MICHAEL SHORT: Want to give a quick review
so that we can launch into some more technical stuff today.
We started talking about this reaction for boron neutron
capture therapy was the focus of today's lecture.
And I want to get this one up on the board
since I'm going to move to different slides later.
We had boron 10 captures a neutron
and becomes lithium 7, helium 4.
There's a gamma ray.
And there's going to be some q value, or energy
either released or consumed, in this case, released
in the form of the kinetic energy of the recoil products
as well as the gamma ray energy.
So let's also give a quick review of the Q equation
since I think we covered that last week.
If you remember, if you have a general system of, let's say,
a small initial nucleus i firing into a larger nucleus capital
I, after the reaction, off comes some small final nucleus
and some large final nucleus.
If I were to draw these arrows to scale,
the little one would probably be moving faster.
Then we can figure out how much mass and kinetic energy
each of these nuclei are by just conserving everything.
So we can write, let's say, the mass of nucleus i,
c squared plus the kinetic energy of i
plus the mass of big I c squared.
Plus the kinetic energy of big I has to equal the mass of little
f c squared plus the kinetic energy of little f
plus the mass of big F c squared plus its kinetic energy.
And what this tells us is if the total amount of mass,
or the total kinetic energy change,
they've got to exchange energy kind of equally.
So we can write the difference in mass or energy by just,
let's say, taking all of the mass terms
and putting them on one side.
So we can say that--
let's just say everything here is multiplied by c squared.
We have M i plus M I minus M little f minus M big F
has got to equal the sum of the final kinetic energies
minus the initial ones, which we also call this Q value.
And so by getting the difference in the masses
or the kinetic energies at the end,
you can figure out whether this reaction
is exothermic or endothermic.
If you remember, we said if Q is greater than 0,
it's exothermic.
If Q is less than zero, it's endothermic.
In a nuclear reaction like a chemical reaction,
you've got to input extra energy into the system
beyond just the rest masses of the particles
to make an endothermic reaction happen.
For example, since we found out that this BNCT reaction is
endothermic, you can make it go the other way,
but you have to impart kinetic energy to one
or both of those nuclei to overcome the Q value that--
let's see-- yeah, to overcome the Q value that you'd get.
And I want to put up a couple of other terms just
to quickly review.
We started looking at the table of nuclides.
We learned how to read it specifically to find things
like excess mass and binding energy,
the definitions for which I want to leave up here on the board.
So we had the excess mass, which, again, doesn't
have a physical significance.
It's the difference between the actual mass of a nucleotide
and the integer approximation of its mass
just from the number of protons and neutrons,
and the binding energy.
Since some of you asked for a nucleus uniquely
defined by its total mass number a and number of protons z.
So let's say this is a functional quantity
because you can have any nucleus with a certain number
of protons and a certain number of protons plus neutrons
would be the sum--
let's see-- the sum of its individual nucleons
minus its actual mass, which also
is a function of z and a, or its proton
number and its total atomic mass number.
So let's leave these things up on the board
while we move into some new stuff.
So back to where we were on Tuesday.
We wanted to figure out, well, how do we
calculate this Q value?
You can do it any of three ways.
You can use the masses, like we have over here.
So the masses in amu, or atomic mass units, times c squared
gives you the Q value in MeV.
Or the difference in kinetic energies
gives you the Q value in MeV, although you don't usually
just know these off the bat, especially
for the final products.
Or you can use the binding energies.
Because the binding energy is directly related
to the mass additively, you can substitute in binding energies
here and you'll get the same thing.
And so if we have the table of nuclides,
you can either use the atomic mass
of any nucleus or its binding energy
and just look it up directly.
The thing with the fewest steps is just
to use the binding energies because those are already
given in MeV or keV.
And then it's just an addition subtraction problem.
If you use the masses, don't forget
to multiply by c squared.
And remember our conversion formula,
which should not be rounded, which is this right here.
Again, 931.49, not 931, not even 931.5, or else
you're off by 10 kiloelectron volts.
And so then the question is, all right,
let's try to calculate the Q value from this reaction.
I have the individual energies and kinetic energies up here.
But just so we have a worked out example,
let's actually do this.
So the Q value of this reaction should
be the binding energy of lithium 7 plus the binding energy of--
what do we have-- helium 4 minus the binding energy of boron 10
minus the binding energy of a neutron.
Now first of all, the easy one.
What's the binding energy of a low neutron?
Let's call it out.
0, yeah.
A lone nucleon is not bound to anything, so that's easy.
That's 0.
And we can just use the table of nuclides
to look up the other three.
Luckily I've got it live over here.
So let's just punch these in.
We'll have boron 10.
And our binding energy right here is 64.75 MeV.
Let's look up helium 4.
Good.
That's showing up on the screen.
I think I'll make it bigger so it's easier for everyone
to see.
The binding energy is 28.296 MeV.
And notice everything's already in MeV,
so this is nice and easy to deal with.
And lithium 7.
39.245 plus minus minus 0 MeV.
Let's see.
Actually, I think my sines for exo and endo are backwards.
Let me fix that right now because the idea is
if you release energy--
there we go.
Sorry about that.
No, wait a minute.
Let's calculate this out, figure what we get.
And again, I can't do six digits in my head,
but I will do it as fast as I can here.
245 plus 28.296 minus 0.75.
Ah, indeed.
2.79 MeV.
I had it right the first time.
So good.
Shouldn't second guess myself.
Cool.
So now we know the total Q value of this reaction.
And I'll bring the reaction back up here.
We also know, which you can find from measurements,
that the gamma ray comes off with an energy of 0.48 MeV,
leaving--
let's see how much--
leaving 2.13 MeV for the sum of the kinetic energies
of the lithium and the helium nucleus.
So let's say Tli 7 plus t for helium.
Now the question is how did I get
to those numbers for the split between the kinetic energies
of those two?
Anyone have any idea?
Yeah.
AUDIENCE: Related to their mass?
MICHAEL SHORT: Yeah.
So it's definitely related to their relative masses.
More specifically, we have this conservation
of energy equation, but we've still
got two variables and one equation.
We need a second equation.
That certainly does relate the mass.
Yep.
AUDIENCE: Conservation momentum.
MICHAEL SHORT: You can relate their momentum.
So we can say that if this initial kinetic energy of boron
and of the neutron was approximately 0, then
it's like these two nuclei, lithium and helium,
were kind of standing still and all of a sudden
moved off in opposite directions, which
means they've got to have equal and opposite momenta.
So let's say the absolute value of the momentum of lithium
has got to equal the absolute value
of the momentum of helium.
This is our second equation, where
we'll use the first one to figure out what are
the relative kinetic energies.
And so we're going to use a quick trick to say if momentum
equals mass times velocity, we can also
say it equals the kinetic energy, which
is 1/2 mv squared, and then multiply by things in order
to make it mv.
So we can multiply by 2.
Let's make a little bit of space here.
If we take the kinetic energy, multiply by 2, multiply by m,
and take the square root, we have t.
Let's see, that would give us 2 m
squared v squared inside the square root gives us mv.
So now we can take these expressions
and we can say that the root 2 mass lithium, t lithium,
equals root 2 mass helium t helium.
And they both have a square root of 2.
We can square both sides of both equations.
And we end up with mass of lithium t lithium
equals mass of helium t helium.
Now we take this equation, rearrange it a little bit.
Let's just call that Q to keep things in variable space.
And we can say that t helium equals Q minus t lithium.
We can take this t helium, stick it in here.
We end up with m lithium t lithium
equals the mass of helium times Q minus t lithium.
There's a missing h right there.
There we go.
And then from here--
oh good.
We've got some blank space right here.
Let's see.
So we'd have-- I'll do out all the steps--
equals mass of helium times Q minus mass of helium times
t lithium.
And we're solving for t lithium, so we can put this term over
on the other side.
So let's say t lithium times the mass of lithium
plus the mass of helium equals Q times the mass of helium.
Then we can just divide both sides
by the sum of those masses
Those two terms cancel, and we're
left with the expression for the kinetic energy of lithium,
which is the mass of helium over the sum of the masses
times the total Q value.
Looks like the two of those might actually
be backwards, huh?
Because the lithium one should be smaller.
I'll correct that for the notes when I put them up online
because this ratio should be smaller than 0.5 Q.
But at any rate, this is how you actually get them.
And I want to point out something
a little flash forward to some of the decay
that you're going to be looking at in terms of nuclear decays.
Let's talk for a second about alpha decay.
In alpha decay, you have a low nucleus sitting around.
And then all of a sudden it emits a helium nucleus as well
as a recoil nucleus.
And one of the questions that came up a lot last year
is why isn't the Q value of an alpha decay
reaction-- let's just write one up on the board, possibly one
that you'll be dealing with very soon, hands on.
Uranium 235 can spontaneously go to a helium nucleus.
That's 92 plus looks like 90 to 31.
What comes before uranium?
I think that is thorium, although I wouldn't-- yeah,
I think that's thorium.
And the idea here is that there is
some Q value associated with the kinetic energy of both
of these.
But it's not the alpha kinetic energy
because the thorium nucleus would have to take away
some of that kinetic energy.
And I want to show you this on the diagrams.
So let's look at U 235.
And we can see that it has an alpha decay to thorium 231.
Awesome.
Got the symbol right.
And it has a decay energy of 4.679 MeV.
Let's take a look at the diagram,
which actually lists all of the possible alpha decay energies,
of which there are many.
Had to zoom out a little bit there.
Yeah?
Anyone have a question?
OK.
So notice that the difference in energy levels is 4.676 MeV.
And if we look at the highest energy alpha ray,
it's less than that.
And that's because just like we showed up here,
the thorium nucleus has to take away
some of that kinetic energy to conserve both energy
and momentum.
So this is a question that came up quite a few times last year,
and I want to make sure you guys don't get tripped up like this.
So again, I think I've said it every single day,
and I'll say it again because it's
the next day, is make sure to conserve mass energy
and momentum.
That's the whole theme of this class.
Yes.
AUDIENCE: How do you get that value
for [INAUDIBLE] mega electron volts?
MICHAEL SHORT: Should have been 2.79 minus-- oh,
did I do a little mental math mistake?
That should be-- oh.
Yeah, no.
A little dyslexia thing.
3, 1.
Yeah.
There we go.
Thank you.
Cool.
OK.
Is everyone clear on how to calculate Q values
from nuclear reactions using either kinetic energies, which
you won't typically know, or masses or binding energies,
which you can look up directly from the table of nuclides?
Yeah.
AUDIENCE: So what did you do down there in the bottom right
corner of the chalkboard?
MICHAEL SHORT: Of this one?
AUDIENCE: Yeah.
MICHAEL SHORT: Yep.
So I took this expression right here, which
is to say the Q value has got to be
the sum of the kinetic energies of the lithium
and helium nuclei, rearranged it thusly to isolate--
ah, there we go--
yep, to isolate the helium kinetic energy,
and then substituted that expression in here
to get this one right there.
AUDIENCE: OK.
MICHAEL SHORT: Yeah.
So this way, we have--
in this case, we had, let's say, two variables
and three unknowns.
But because we have this equation relating them,
we're left down with two variables, two unknowns,
and we can actually solve this thing.
Yep.
Yeah, good question.
Yes.
AUDIENCE: The energy of the gamma, is that just a known?
Like the .48 MeV?
MICHAEL SHORT: That's something either
I tell you or you would measure, let's say.
That's just for completeness to say all right,
this reaction actually gives off a gamma,
and I want to give the right value for the kinetic energies.
And we'll get into what gamma transitions are allowed
and then how you measure them in the next couple of weeks,
actually.
Yes.
AUDIENCE: So do we refer to Q as the Q you calculated up there,
or that 2.31?
MICHAEL SHORT: They're the same one, actually.
AUDIENCE: The 2.7
MICHAEL SHORT: Oh, I see.
That's a good point.
So this wouldn't really be Q, would it?
But it is the sum of the kinetic energies.
This is like Q minus the gamma ray energy.
Let me stick that in there.
L, i.
Yep.
Yeah.
Good point.
OK.
Any other questions before I move on?
We're going to get into a universal formula
to predict in a so-so way what the binding energy of any given
nucleus will be and start looking at stability trends
so you can predict, just from the number of protons
and the number of neutrons, how stable a nucleus will
be with a few exceptions, which we will go over.
And this is what's referred to as the semi-empirical mass
formula.
So I'm going to erase some stuff.
Has everyone got the notes on this bottom board right here?
OK.
Let me know when you're ready, and let's see.
Yeah.
I want to make sure I move at your guys' pace.
Let's say going to have a graph of binding energy per nucleon
on versus nucleons.
Well, anyway, I'll leave that up there
and I'll do the work on this board right here.
So let's say we wanted to figure out a weighted graph
or to predict the binding energy per nucleon.
So I have this binding energy term
over a, where a is the total number of nucleons,
as a function of the number of nucleons--
in a generalized way.
Not accounting for magic numbers or anything else that we'll
get into pretty soon.
And I don't like the term magic numbers,
but that is the parlance that's used in this field
so I'm going to stick with it.
Let's try and think about if you imagine the nucleus
as a kind of drop of liquid-- and one
of the other words for the semi-empirical mass formula
is called the liquid drop formula or the liquid drop
model.
It assumes that the nucleus takes the shape
roughly of a liquid drop, and you can kind of
treat some of the energy terms accordingly.
This is why I have all the different colors of chalk out
for this.
Makes it a little visually easier to see.
So let's start writing a general expression
for the binding energy as a function of a and z,
and start thinking about what sort of terms
would add to or decrease the stability
of a given liquid drop nucleus, where all the nucleons are just
kind of there in some sort of floating, crazy, coulombic,
strong nuclear force soup.
First of all, as you add nucleons
to a given nucleus, what tends to happen
to the binding energy, in general?
Without knowing anything else.
Assemble more nucleons, you convert more mass to energy.
And you end up increasing the binding energy.
So let's call this the volume term.
As you increase the volume of this liquid drop,
its total binding energies start to increase.
So let's say there's some term that's
going to be proportional to A, the number of nucleons that's
in this liquid drop.
And we'll draw this.
Let's see if I can do the trick right.
Yes.
I love doing that.
If we were to graph binding energy per nucleon
as a function of number of nucleotides for this term,
it would just be a flat line because it's related to A.
And there's going to be some constant, which we're
going to call the volume constant, that says, well,
there's going to be some relation
between the actual amount of stability gained
and the number of nucleons.
We don't know what it is yet, but what
we're really concerned with is the functional
form of this thing.
It's proportional to A.
Next up, what also happens to a liquid drop
as you increase its volume?
What other parameters do you increase?
AUDIENCE: The surface area?
MICHAEL SHORT: Exactly.
The surface area.
The idea here is that if this liquid drop
is made of all sorts of different nucleons--
and let's pretend that they're like atoms in a crystal
and they're all binding to each other--
the ones on the outside aren't bound to as many nucleons
as the ones on the inside.
And so the more nuclei there are near the surface
as opposed to inside the liquid droplets, say,
inside some little radius where all it sees around
it are other nucleons, then they're not quite as bound.
And how does the surface area of a liquid drop
scale with its volume?
To what function or to what exponent?
2/3.
I mean, let's take a quick look at the volume of a sphere
is 4/3 pi r cubed.
And the surface area of a sphere is 4 pi r squared.
So if you want to get some expression for how
does area scale as volume, I said cube,
then I wrote squared.
It's going to end up looking like something times
r to the 2-- let's see.
Oh, yeah.
I'm sorry, that's not the expression I want to write.
But the idea here is it's going to scale with r to the 2/3.
So let's pick a different color and say
we're going to have some surface term times number of nucleons
to the 2/3.
And if we then adjust this formula to also take
into account this surface area term--
which is to say for very small nuclei,
there's a lot of nucleotides near the surface,
and as the nucleus gets bigger and bigger,
more and more of them are in the juicy center and don't know
they're near the surface--
we'd have some modification that looks like that.
Now I'm going to erase the stuff over here because I'm
running out of room.
Now these nucleons aren't just untagged, anonymous nucleons.
They're either protons or neutrons.
And what happens when you try and cram a lot of protons
into one space?
AUDIENCE: They want to repel each other.
MICHAEL SHORT: Yep.
They want to repel each other by coulombic forces.
And so every proton--
let's pick a different color for the coulombic forces--
and that should be a minus if I want
to stick with all the notation.
There's going to be some other term to account for the fact
that the nuclei, specifically the protons,
are trying to repel each other.
So in this case, it's going to be proportional to, let's say,
the number of protons that we have.
And every proton should feel a repulsive force
from every other proton.
So let's say it's times z times z minus 1,
so that every proton feels the force of every other proton
except for itself.
And that's going to be mediated by the total number
of nucleons.
So if there are more neutrons in the way,
it won't be quite as bad.
And there's going to be some other--
we'll call it a C for the coulombic term--
and that will say that as you make a bigger and bigger
nucleus, you start to get more and more coulombic
repulsion trying to rip it apart.
So if we were to then modify the purple curve--
oops.
Trying to get it to go the same as the nucleus gets bigger.
I want to make sure it's really to scale-ish.
As the nuclei get bigger and bigger,
it's going to be a little less stable.
And already we're starting to get
a curve that is getting close to looking like the binding energy
curve from the reading.
But there are a couple more terms to reckon with.
So let's pick a fourth color.
What other sort of trends that you
notice in the reading about the stability of different nuclei?
Let's say you were to take a common nucleus like carbon-12.
It's got six protons and six neutrons,
and it's exceptionally stable.
What about carbon 6?
A nucleus of just 6 protons?
Doesn't exist.
Exceptionally unstable.
What about carbon 24?
18 neutrons, six protons.
Sound stable or not?
Not at all.
So there's some sort of asymmetry term going on.
When the number of protons and neutrons
is roughly in balance, especially for light nuclei,
the nucleus tends to be more stable.
So we can write some sort of term--
let's call it an asymmetry term--
that relates to the number of neutrons
minus the number of protons.
And in this case, for reasons I'm not going to get into,
but are derived in a reference in your reading,
there's a squared on it.
But suffice to say if the number of neutrons and number
of protons are equal, then the nucleus is
predicted to be pretty stable.
And this works out quite well for light nuclei.
It starts to break down a little bit for heavier nuclei.
And then divide by the number of nuclei that there are.
I also see a missing 1/3 because let's say
this nucleus has a volume that scales with roughly
the number of nucleons.
Then the distance of that coulombic force
is going to be like A to the 1/3 or the radius
of this nuclear drop.
So let's take the asymmetry term.
That's going to give us a further modification slightly
downward.
And finally, there's what's called the pairing term.
What's the last color I haven't used?
This pairing term delta.
And this is not a smooth function.
It's a piecewise function that depends on
whether you have an odd or an even number
of each type of nucleon, protons or neutrons.
And so what this means, it's going
to add a little bit of jaggedness
to the beginning of the curve and equal out in the end
because this delta term can be something
like plus, let's call it an A pairing.
And it scales with the square root of A, or minus, or 0,
depending on if the nuclei are odd odd,
like odd number of protons, odd number of neutrons,
even even, or odd even.
Now I know that the derivation is a little hand wavy.
That's why we call it semi-empirical.
We're taking each of these additive terms
and saying it kind of comes from a fairly OK, a little poor
approximation of the nucleus.
But what we end up with is a formula
whose constants are fit, whose terms, the actual variables,
are derived somewhat from physical intuition.
These constants were then fit later by some other folks,
and the references for this are in the reading.
They're all in MeV, and this gives you a binding energy
in MeV for a given nucleus.
Now it works some places and it doesn't work in other places.
Yeah.
AUDIENCE: So the lower case a [INAUDIBLE]
no matter what the nucleus looks like?
MICHAEL SHORT: That's right.
So this is the universal, semi-empirical, usually works
formula for the binding energy of a nucleus.
So the constants don't change because the variables here
are z, [INAUDIBLE] a, and n.
And don't forget-- because you'll
need to remember this on the homework--
that A equals z plus n.
So for example, if you want to express
what is the most stable nucleus, you
could take the derivative of this formula with respect
to A or z.
And don't forget that you can substitute this expression
into there.
That's giving you guys a hint for the homework.
And let's look at what this actually looks
like as far as theory compared to experiment.
So the red points are theoretical predictions.
The black points are experimental predictions.
And all of the different nuclei are shown here.
First of all, the curve looks quite a bit like the one
that we just hand wavy made on the board right here.
And second of all, there aren't too many exceptions.
It's hard to see what the exceptions are,
so it's a little easier to draw them
in terms of relative error.
So you can see where does this formula work
and where does this formula not?
So if you notice, for the small nuclei,
approximating it as a liquid drop
is not a very good approximation because you
can't treat this as a homogenized,
smeared liquid drop.
It's much more-- well, there's either
two or three nucleons, and very few protons and neutrons
in each.
But then as you get to larger and larger nuclei,
it starts to hit very close with a few exceptions
that I want to point out right now.
If you zoom in on that part, you can actually
see that at certain neutron numbers, or certain proton
numbers, there is an exceptionally high stability
of a lot of those nuclei.
And that's as you start to approach
these what's called magic numbers,
or numbers of nucleons which, say, fill all energy
levels at a certain level.
And again, it's not for every nucleus as a function
of neutron number.
But even drawing an envelope around this curve,
you can see that the nuclei around 82,
around 50, around 28, are a whole lot more stable
than the ones in between.
And this pattern kind of repeats with larger and larger
periodicity.
And it kind of looks like right here,
at the edge of our knowledge of nuclei,
we haven't quite gotten to the next peak yet.
This is something we're going to talk about Friday on the quest
for super heavy elements, or SHE's,
as you'll see in the reading.
Yeah.
AUDIENCE: So the most stable nuclei peaks were closest to 0.
MICHAEL SHORT: Closest to 0 is the closest agreement
between experiment and theory.
So the ones that are exceptionally stable,
which are not predicted by this very simple formula,
are up here at the peaks of these magic numbers.
And actually, I want you to take a look right here at some
of these very small nuclei.
Like helium 4 is probably way up here somewhere, all
the way over on the right.
It's an exceptionally stable nucleus
that is not very well approximated by liquid drop
model because it's got four nucleons.
They're all on the surface.
There's nothing on the inside of a helium nucleus, let's say.
And then if you look at stability trends in terms
of are the nuclei more stable if they have odd numbers
or even numbers, you can graph the two separately
and look at the number of stable nuclei
that have an odd total mass number or an even total mass
number.
And there's a few things to note here.
One of them is the even numbers tend to have
a lot more stable nuclei.
This is something I mentioned on the second day of class.
If you look at the [INAUDIBLE] table of nuclides--
let's go to their home page--
and just look at it sort of in a color way.
The blue colors are stable nuclei,
and you notice that every other row of pixels
here has a whole lot more stable ones.
And that's the same thing that we're seeing right here,
is that there's a lot more even nuclei that are more stable.
If I jump back to our semi-empirical maths formula,
notice that this binding energy goes up for even, even nuclei.
So when there's an even number of protons
and an even number of neutrons, the semi-empirical mass formula
does predict an increase in stability,
which you can actually see on the table of nuclides,
and on this sort of stability trend.
And so let's look a little closer
and see how many nuclei for each proton number or each neutron
number are actually stable.
And we graphed the odd and the even ones separately.
And what's important to note here is one,
the odd is way lower than the even.
There's usually either 2 1 or 0 stable nuclei at that number.
And what other sort of features do you guys notice about this?
It's not smooth, first of all.
Where are those peaks?
Where do you tend to find that most stable nuclei?
4?
Where do you tend to find the least?
What about these two right here?
No stable nuclei at these proton numbers.
And remember, proton number uniquely defines an element.
Anyone know what these two might be?
What sort of elements?
Look to the back of the room if you want.
There's a periodic table in the back wall.
And you can see, except for the super heavy things
down at the bottom, there's a couple
of elements that have no stable isotopes.
These are technetium and promethium, which
are relatively light isotopes--
I say relatively light compared to things like uranium--
with no stable isotopes.
They're also fairly far away from these so-called magic
numbers or other regions, where you tend to have a spike
in the number of stable nuclei due to--
well, things that you'll learn in 22.02,
in terms of nuclear shell occupancy and stability.
But you see the same thing when you graph the neutron number.
You can see a couple of sudden spikes
right here at 20, 28, 50, 82, and 126.
When everything gets really stable, all of a sudden you've
got one last gasp of a stable isotope
before you go off into nowhere land.
So let's start looking at relative stabilities of nuclei,
let's say, for a given mass number or a given proton
number.
Anyone mind if I cover this board
because you can't roll it up.
You all got the notes from here?
Cool.
That one has less to erase.
And I want to keep these formulas up for our reference.
Let's say I pose this problem.
I want to find out, make sure I solve the right one-- actually,
I'm going to check my notes real quick--
that what is the-- for a given A, or for a given mass number--
what is the most stable number of protons?
for A, given A.
This is the question that I like to answer here.
How would you approach this question
using the semi-empirical mass formula
that-- well, you can't see here, so I will bring it
up back on the screen.
Here.
How would you find this out?
Well, let's say for a given mass number
A, for a given poor approximation of the total mass
of a nucleus, the more binding energy it has, the more stable
it is.
Therefore, if you want to find the minimum of a mass
for A and Z, given a fixed A, that
will give you the most stable nucleus
because it will tell you which value of Z
gives you the smallest m, or rather,
the most tightly bound nucleus that
has the most binding energy.
So let's start writing this out.
First of all, we can use one of the two equations
we already have up here, a relation between the mass
and the binding energy.
The second one, well, we have it right here.
So let's substitute in our binding energy equation
and express it in terms of mass.
So let's say our mass of a nucleus A and Z is equal to Z
times the mass of hydrogen plus A minus Z times the mass
of a neutron minus the binding energy
because in this case, what I've done right here
is I've added mass to each side of the equation, subtracted
binding energy from each side of the equation,
and we can just take negative that expression
and write it all out together.
So we have minus AVA plus A surface A to the 2/3 plus AC z
times z minus 1 over A to the 1/3,
and plus AA symmetry and minus z squared over A and minus delta.
So we've got one expression for the total mass.
We've fixed the value of A because we're
going to take some fixed--
we're going to choose some fixed value of A.
Let's say A equals 93.
And that's the example that I've kind of worked out in my head.
It so happens that niobium has a stable isotope that a mass of A
equals 93.
And we just found out in some of our research
that niobium doesn't stick to chromium very well.
That's why I've got it on the brain.
So this is what I was thinking about this morning.
So for a fixed A equals 93, we want
to find what is the most stable A. How do we do that?
Anyone have an idea?
Yeah.
AUDIENCE: Differentiate?
MICHAEL SHORT: Differentiate.
Sure.
Let's take, we'll just say the derivative of-- oh, no,
it is a partial derivative because we
have two variables here.
Take the derivative as a function of z,
set it equal to 0.
This will give us the z number that gives us
the minimum mass for a fixed A. So let's actually
do this right now.
So let's see.
This gives us mh.
And this term expanded out is AMN minus ZMN.
So we have a minus MN.
I'm going to make one quick correction right here.
I want to make sure everything's in the same units.
All of these semi-empirical mass terms are in MeV.
These right here are in AMU, atomic mass units.
What do we have to add in order to get these all in MeV?
AUDIENCE: [INAUDIBLE] the conversion factor.
MICHAEL SHORT: Conversion factor, yeah.
Or in this case, we'll just stick a C squared.
I'll make a little bit of room so I can
stick the C squared in there.
C squared.
Now everything's in MeV.
We're all in the same units.
So let's say we have mh minus mn.
These are in C squared.
And minus VC plus 2/3 AS times A to the negative 1/3 plus--
I'm going to expand this out to call it z squared minus z.
I'm also going to stick in n equals A minus Z
so that this expands out to A minus 2Z squared.
Is everyone with me here?
Yeah.
AUDIENCE: Shouldn't the A terms, like the AVA and the AS,
stay at 2/3?
MICHAEL SHORT: Oh.
You're right.
I'm deriving with respect to the wrong variable.
Thank you.
Yep.
So we want to do this as a function of z.
So that term disappears.
That term disappears.
Thank you.
Let's work on these ones right now.
So we have AC time z squared over A to the 1/3.
So that will give us A plus AC over A to the 1/3 times 2z.
And then we have--
let's see- minus AC over A to the 1/3.
That's it, actually.
Oh, times plus 1, OK?
And then we have the A minus 2z.
So let's expand this out just so we can see it all on the board.
So we have AA over A times A minus 4z squared
plus 4 minus 4 AZ plus 4z squared.
So let's take the derivative with respect to z of that.
That term goes away.
That becomes 4A.
So we have plus AA over A times 4A.
And there's a minus sign there.
And that becomes 8z.
So we have plus AA symmetry over A times 8z.
And the delta term goes away because there's
no z dependence.
And what we end up here is the solution for what is the most
stable z as a function of A?
This is a linear equation.
There's only one solution for it.
If we actually want to graph this m as a function of A
and Z, we end up with what's called a mass parabola, which
is to say you can graph the binding energy per nucleon,
or the mass, or pretty much similar things, of a nucleus,
of all nuclei with A given as a function of z.
Think I can do this on the remaining space right here.
So let's say 4A equals 93 if this is Z
and this is m as a function of A and Z. Let's actually
look at a concrete example.
So let's go live to the chart of nuclides
and start looking at things with a mass number of 93.
Looks like I clicked a little too high.
There it is.
Let's see.
Moly 93 I was looking at, and that
becomes niobium 93, which is the stable isotope I was thinking.
So let's put niobium right here.
I haven't given an actual scale to this
because I just want to show you in sort of relative terms.
So let's say niobium is the stable one.
So it's going to have the lowest actual mass, even though it's
got an A number of 93.
If you look here on the chart of nuclides,
you can see it could have come from a couple
of different places, either from zirconium 93
or from molybdenum 93.
And now is a good time to start introducing
these different modes of decay so you can figure out, well,
how would a nucleus decay to get to the most stable place?
Let's say it came from zirconium 93 and--
let's see, miobium has a proton number of 41.
So if we go to zirconium, it beta
decays to niobium 93 with an energy of .091 MeV.
Very, very close.
So we'll draw it slightly higher.
That's about 91 keV.
Zirconium 93 could have come from the beta decay of yttrium
93 you can see right here.
So let's go up the mass parabola and keep exploring.
And now we see that yttrium can decay by beta decay with 3 MeV.
So if we put yttrium on this graph, it would be way higher.
Yttrium itself could have come from--
well, let's see, strontium 93 with the decay
energy of 4.1 MeV.
So let's put strontium here.
And I think 4 MEV would be, like, off the chart.
But whatever.
That's the way we drew it.
Already we've got the makings of a parabola.
And each one of these can decay by beta decay,
or does decay by beta decay, in order
to get to the most stable nucleus.
So let's write the nuclear reaction
for beta decay of one of these, let's say, from zirconium
to niobium.
So we'd have 9340 zirconium spontaneously
goes to 9341 niobium plus a beta and
plus an electron anti-neutrino.
That's the part I don't expect you to know yet.
But that's the whole energy conservation thing.
A little bit of a flash forward.
The beta decay energy is not necessarily
the energy of the electron that you
will measure because some of that energy
is taken away by the anti-neutrino.
But we'll get into how those relate probably next week.
So let's now look on the other side of the parabola
and confirm that the semi-empirical mass
formula, which predicts something
parabolic here and here with respect to z,
actually checks out.
So let's back up to niobium 93 and notice
that it could also have come from electron capture
from molybdenum 93.
So let's put molybdenum right here.
And it decays with an energy of 0.4 MeV
into niobium, which let's say it's around here.
Let's keep going through the chain.
Anyone have any questions so far while we keep going?
Cool.
Let's trace it back up the chain.
Technetium 93 can beget molybdenum 93
by electron capture with a much higher energy, 3.201.
I'm going to extend our graph because we need the space.
Technetium, another 3 MeV.
And let's go one more back.
Technetium can be made by electron capture from rubidium
93 with an even higher energy, which
means a higher difference in mass between these two.
So so far-- let's see, what's 6 MeV for rubidium here?
It would be like there, I guess.
There's our mass parabola, right from the data.
So I like doing this better than just showing you a diagram
because you can actually try it for yourself.
Pick a fixed A, change A, and construct
the mass parabolas yourself.
Now the question is how could these decay into niobium 93,
which is the stable isotope?
I have negative one minutes left,
so I'm very quickly going to tell you for large energy
changes, it can either be positron decay or electron
capture.
And we'll go over what these modes of decay are next week.
This can be, again, positron or electron capture.
And for small amounts of decay energy,
it can only be electron capture because in order for positron
decay to happen, you have to be able to create the positron.
And the positron plus the extra electron
ejected to balance charge has got to be 1.022 MeV,
or same thing as what's known as two times the rest
mass of the electron.
I'm going to stop there, and we'll
pick up with lots of examples and questions tomorrow.
The last thing-- well, I'll go over the next problem
set tomorrow.
I want to make sure everyone's seen it
seven days before it's due.
And the best way to do that is to show it on the board.