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The grad student are not competing against the undergrads, you're competing
against your own group. I separated the grade up top. Whereas in the histogram I lumped
them all together. For the undergrads the high was 86. The low was 35. The average wads
65 and a half. For the grads, the high was 82, notice that the undergrads beat you there.
The low was 46. And the average was not so different, considering the statistics on both
were not so great. The distributions are not so different. Remember the class is graded
on a curve. For the undergraduates I don't know how hard an exam it until you tell me.
So the grade will be a b, 3.0 for the undergrads. The thing for you to think about is
where you sit relative to that average. If you're to the right of that affirm on this
plot, you've been doing fine. If you want to do better keep going if that direction
if
you're to the lower side of that it's not too late to do something about that. The final
will count for 40%. Home of, there's still lots of opportunity to move yourself up in
this.
STUDENT: Do you have any values for the standard deviation.
PROFESSOR: I can't go into that. This is about as much as I can do. End points
for the graduate students the numbers are actually small. There are about 10 graduate
students who are officially registered, the, go into that level of detail especially for
the first midterm, I don't think it's warranted.
We do our best if term of grading but we're human. We do make mistakes. If you feel
you've been mistreated in term a grade I'd like to you write a little note inside the
booklet where you think you made a mistake. Give it toe me and we'll look over T and we
may or might have do anything about it. But at least you have a opportunity to complain
about it. Any questions about that?
Then what I'd like to spend most of the rest of class period doing is going over the
exam.
Season let's go through it problem by problem. So the first one, dealt with an
electrical generator based on radioactive source. And the problem was that if we had
a
gram of polonium 210 which is an alpha emitter with an energy of five (on screen). The
first part of question was what would be the power that that thing could produce in Watts.
So the first thing I want you to think about was the activity, number of decays per second
which is responsible for producing this power. So what you had to remember was activity
is n lambda where n is the number of radioactive pieces present. And lambda is the
decay constant. Appropriate for that particular isotope. So in the way the problem was
stated it said there was 1 gram of polonium two ten. in order to calculate n you take
1
gram and divide it by the atomic mass of polonium 210. which is 210. if you didn't know
that, from looking at the notation for the isotope, there was a periodic table included
in
that so-called useful information. And it says the mass of polonium was 209. They took
a
different isotope. Won't have taken off any points for that. That's how many grams you
had. That's the number of grams in a mol. The, you multiply all those together and find
the number of polonium atomic in that was 2.87 times 10 to the 21. Activities of that
is
n time lambda, lambda is decay constants, log two divided by half-life. In order to
get
the units decay per second, 138 days and momentum by number of seconds in a day, we get
lambda is one point sex seven times 10 to the four is second. Each decay produces 5.3
MeV
worth of ax energy. Total amount of power produced by decays is the number of decays
per
second time the energy produced per decay, times the conversion factor to go from MeV
to
joule which was also provided. And that works out to 141.5 jewels per second. A joule
per second is a watt. So it's 141.5 Watts is amount of power variable from that radio
isotope generator. Any questions about that.
STUDENT: Are we going to have access to these.
PROFESSOR: Absolutely. I didn't post them ahead of time. I wanted to go through
this. This will all be on bSpace this afternoon.
Okay. Part b asked you for the same quaint. Namely the power level but not a year
later. This was your opportunity to show you understood the exponential decay law,,. The
idea was that 1 gram was there sometime we call zero those atomics were decaying. At
later time there will be fewer of them present. The number present at given time t is
power level will fall off exponentially, just because the activity will fall off
exponentially. So the power a year later is good to be the initial power that you
calculated in Part A time the e to the minus lambda t. Now because of units are
simpler this way I do it in days. Half-life is 138 days, time is one year, which is 365
days, power level a year later is 22.6 Watts. Most people got this more or less. Any
questions about that?
Okay. Number two, supposed to be more fun. This had to do with a hypothetical 70
elements and isotopes. An almost called bedlam. One hundred twenty-five protons in its
nucleus. And what it says was that this is the isotope that we're talk about 303
bedlam. In Part A is decays by alpha mission, by another element call tan interim
supposed to write a balanced equation that describes this alpha decay. The point was
to
understand in all the nuclear reactions we don't create and destroy neutrons and protons,
we just rearrange them. In alpha decay the alpha particle is a helium four nucleus. Some
people just wrote alpha and I took off a few points because I want you to explain
explicitly an alpha particle is a balanced state. We have 125 protons here. Emit two
protons in the alpha particle and therefore the tantrum nucleus must there have been have
123 protons in it. We can conserve neutrons as well. And although the neutrons number
isn't written here you could certainly write it if you wanted to. The point is the
helium four which is the two protons and two neutrons as total atomic mass of four since
we started with 303 and emitted something before you have to have 299 as the tantrum
isotope that gets produced.
The next part asks to explain why alpha emission is more likely than any other kind
of decay here. And I guess I was a little disappointed. People had more trouble with
than that I thought they would. We haven't gone through the full quantum mechanical yes.
but we discuss north side class the idea as you're building up heavier nuke the coulomb
repulsion between these grows. Some points it becomes enough for the nucleus to decay
by
emitting particles that carry away positively charged parliament because the alpha
particle is doubly magic state of two protons and two neutrons bound together because of
that Kra binding energy the most energy effective way for the nucleus to decay is to emit
this balanced state of two protons and two neutrons rather than emitting a proton or
a
deuteron or something like that. So it's the coulomb repulsion class driving T then it's
the extra binding energy of alpha particle in particular which favors article particle
emission over other thing some people were arguing if you look at the nuclei, the bedlam
and tantrum, they're extremely neutron rich which is true. People were arguing that
emitting an alpha particle you're lowering the neutron to proton ratio. It's not true
if
you do the imagination. By emitting helium four nucleus you increase the neutron to
proton ratio. That doesn't help. It's not excess neutron, it's the excess of protons
something called pre-***. The question was to write a balanced equation for that.
In
beta minus decay you're emitting an electron it's important to identify the electron as
the beta particle. Since it has negative electric charge had a you're really doing is
turning a neutron notice nucleus no eye proton. So the proton number has it increase by
one unit. For number of people who said it decreased, that can't be true for beta minus
decay because we have to conserve electric charge. So we have 123 positive charges over
here. Need to have 123 positive charges here. And by having 124 minus one, that
balances.
And in order to conserve angular momentum and linear momentum and all those other
good things this other particle that we haven't talked very much about but was illustrated
in the beta decaying thing we talked about back on day one there's a electron ain't neuron
emitted. And the final part of this problem had toot with estimate is the radius of this
particular isotope. The radius goes as some constant times a to the one-third, because
the density remains a constant. So you basically building up a sphere, at ever larger
radius but with constant density as you add more and more nucleons to the system. So the
constant turns out to be 1.2 tent meters. Some people had that, we took a few pion off
for that. The major part was get the formula, plug it in and get that it's about eight
fentometers.
STUDENT: One question about part c, Krane a discussion in chapter six of beta
decay, inaccurately a neutrino will bee mighted but was not -- because it doesn't effect
the identity of other particles.
PROFESSOR: You simply write down neutrino, we probably took off one point. The
reason I feel justified in doing that, go back to the chart of nuclides it shows you
the
correct equation. I think that's right.
We'll go back through my notes.
STUDENT: Part c, is the electron that's produced part of the atom.
PROFESSOR: We haven't gone into the details therefore yet. But that's a good
question. That electron is going to come whizing out and will not remain bound to the
atom.
STUDENT: That atom ionized them.
PROFESSOR: What you're pointing out is I have 123 protons here and 124 and this
electron disappears. In fact you have a positively charged ion sitting there right after
the decay much very good point. If you wrote that down we should have given you extra
credit because we haven't gone into that level of detail in discussing beta decay yet. In
two week is that exactly what we're going to do. It will turn out that keeping track
of
these electrons both this one which gets emitted and flies out and the fact that you end
up with positively charged ion afterwards is very important to figure out the total
decay
energy or the q value for these decays because the way we do that is to imagine we
started out with a neutral atom here. When we go to calculate q values we want to end
up with neutral x. This thing is not going to be neutral initially as you just pointed
out. Other questions about that?
Okay. Problem three was the dreaded quantums problem the idea is was you had an
infinite square well that extended from x equals minus a to plus a. And was
infinite everywhere outside this region and potential was zero inside there. in the first
part of problem it said show the following set wave function satisfied the Schrˆdinger
equation and the boundary conditions. A lot of people jumped to solving, showing they
satisfied the boundary conditions without showing they satisfied the Schrˆdinger equation.
Notice I didn't give I the Schrˆdinger equation, that was part of thought process I want
you you to go through, was to know what the Schrˆdinger equation was. But to actually
write it down, and say that I've got a problem here whether the potential is exactly zero
between these limits, minus a to a, continue it's infinite everywhere else. Since the
potential is infinite everywhere outside that region the wave function has to be zero
outside. I can't have a particle outside that region because it's absolutely forbidden.
Height is infinitely high and it goes on forever. So the particle is really constrained
to be between these limiting values of X-inside that region, the potential is exactly
zero. So inside this region, the particles is a free particle. Except that it's just
bangs into these walls at the extremes. And so inside the Schrˆdinger equation simplifies
to be just minus h bar squared over 2M. . Equals a constant we call the energy times
the wave function. And from that you could say by inspection things you know of that
satisfy that equation with sines and cosines being because when you take the second
derivative of a sine you get a sine back, and when you take the second derivative of
a
cosine you get a cosine back. A number of you didn't do that and we took off four points
for that.
The next part had to do with showing these wave function satisfied the boundary
conditions. And that's the idea that the wave function has to match the boundaries of
this region. Since we know the wave function has to go to zero at x equals minus a
and x equal plus a the wave function inside better go to zero as it approaches those
limits. That's what this is saying, if you plug in the number sine of x equals zero
for x equals an integer time pi in radians. And the cosine is equals zero for a half
integer multiple of pi. So if n is an integer, and even integer, this will work for the
sines. And it's it's an odd integer it will work for the cost of goods sold. That's what
we wanted you to show. In fact, those cases go to zero appropriately at x equals minus
a and x equals plus a. Part b had to do with drawing the probability decisions for
the first four wave functions. You have the wave functions here. The idea was plug in
one, two, three, four and remember the probability for finding the particle as a function
of position is the square of with. It's not just the wave function. Those are the
amplitude, you have to square it to get the probability distribution. When you do that
this is what you get. For n equals one, which is that sign term it starts at zero, goes
up, reaches a max and go to zero again. n equals two, two lobes to the distribution,
n equals three, there are three. Notice the maximum value is S1 over a, that's from
the normalization condition if the wave function. When you integrate the, you get one.
In other words the probability to find the particle somewhere is exactly one. And all
these wave functions and probabilities distributions have that property.
The next part was to draw as best you could what the probability distribution would
look like for the thousandth wave function. For n equals a thousand. The idea was
that by going through these you could see the number of lobes is going to increase as
n,
now there are going to be a thousand of them. I don't expect you to draw a thousand but
just show it's a very fast owes lating high frequency oscillation here that goes from
zero
up to one over a. Just like before. That's because of normalization condition. In
fact, the average of all these is going to be half a. Which shouldn't be a surprise.
Then finally the question was what does the classical distribution look like. I the
classically the particle could be anywhere in the box. There's no special place. It's
just a box and if I put a particle in it could be anywhere. So the probability, constant
inside the box. The probability finds it somewhere must be one. And so therefore the
value of that probiility must be one over two a. Multiply by the length of box which
is
two a, I get I find it somewhere. Multi-by the lengths of box. It's all easy when you
see it.
Now I have to turn my head a little bit here. Somehow my scanner didn't work right.
lithium six ground state in terms of hydrogen three, tritium and helium three. And what
I
said was that both hydrogen three and tritium three have a spin of a half and positive
parity in their ground states. And then use that go information to try to figure out what
lithium six would be. And so the idea was that in both of these nuclei, you've got an
even number of one kind of particle and odds number of other. In the hydrogen three you
have an odd proton. In the helium three you have odd neutron. We're assuming rightly or
wrongly, assuming the total spin of each of these nuclei can be determine by the spin
of
the odd particle, the odd neutron or odd proton. When we go to determine the spin of
lithium six, just going to add those spins vectorially. In lithium six I've got a spin
half coming from the Triton. Spin a half coming from helium three and I have to add them
vectorially. So they're not necessarily pointing in the same direction. That means there
are two possible outcomes for spin after half I have either have the two particle spins
in
the same direction in which case I ends up with spin one or the opposite direction in
which case I come up with spin zero. And the parents will be the products of parents and
since they're both positive it will be positive. If I simply said zero plus or one plus
as possible outcomes we would give full credit for that. If you looked more carefully at
other odd-odd nuclei, it's more often the case that the ground state will be the situation
where the odd neutron and odd proton spins are actually in the same direction. So based
on this you could say it's nor likely the spin will be one plus and that turns out to
be
true.
The next part had to do with knew on 21. Here the idea was you start with neon '20s
and add a neutron to it. This wasn't meant to be a shell model problem. In fact, I think
that came up during the examine, I said it wasn't a shell model problem. The idea is
you
think of neon '20s and add a neutron to it what you want to do is imagine the total spin
of that resulting neon 21 is coming all from that extra neutron. The issue is that the
neutron has intrinsic angular momentum of a half unit but also have orbital angular
momentum associated with it depending on where we put it. The idea was figure out all the
possible total j values that would be associated with that neutron. So the j of the
neutron is the vector sum orbital angular momentum plus intrinsic spin. Intrinsic spin
is
a half. Lample which is quantized into, is zero, one, two, three, and so add that
vectorially to a half and you end up with a half, three halves, five halves and so on.
To
get the parity you have to remember the parity goes as minus one to the lth power. If
l equals zero interior extra neutron minus one to the zero is plus. It's a half plus.
If l equals one, then I'm going to have negative parity. Because I have minus one to
the one. I can get two possible j values from that l value. I've got one plus a half
vectorially, and again, they can be pointed in the same direction or opposite direction.
And so I can get one half oar three Harvard from this value of l. And both will have
negative parity. I can get the same j value being spin three halves but with positive
parity if l equals two, with l equal two, two possible outcomes, either l and a are
parallel or anti-parallel.
I think going up to l equals three was plenty good.
And the final part had to do with calculating q value for a reaction. You had
homework about q values way back when. The reaction was the fusion of two deuterons to
make helium three plus a neutron. The important point was to remember how you define q
values, take the mass of what you start with, and subtract the mass of what you end up
with and momentum by C square. we started with two deuterons. Ep ended up with you
helium threw and a neutron. Masses given in the useful information sheets. You don't
have to worry about the electron masses if you always use atomic masses in the
calculations. And the masses given were actually atomic masses so you deny even have to
think about that. When you do the math the difference between to deuteron masses and
sum
of helium three plus mass is .00351 atomic mass units. And the conversion factoring to,
turns out to be 3.270 MeV. Any questions about that?
Yes these will all be posted this afternoon. Like I said, this is 20% of your grade.
You should take it seriously but don't get too discouraged or cocky about it. We're a
third of whitespace through the class. There's lots of room to go in either direction.
STUDENT: One comment about the mechanics of the class. I felt like I spent a
significant amount of time juggleing between papers and claritys because of the very small
desk.
PROFESSOR: I'm really sensitive to that issue. In fact, the reason we moved across
the way was as you can see you're really crammed if tight here. Woe tried to provide more
space. That isn't a great room either. I will do my best to find a better place for the
next exam.
STUDENT: Maybe one option is half-and-half.
PROFESSOR: I thought about that too. There are issues with that. We had Beth
rooms and I could have done that. Maybe I should have but I didn't. I'm sensitive to
that issue. If it's any consolation to you I know when it comes time for the final
they're going to give us an enormous room. In fact, sometimes we do it in one of gyms,
it's actually a basketball court. But that the not for a couple of months. Any comments
or questions about that?
So time does March on. We keep going. I'm not going to lecture a lot today and not
going to start the next chapter. I'm going to just finish up our final topic having to
do
with nuclear structure. And just to show you where we are going.
So the next major topic is going to be the details of alpha decay. And that's
chapter eight of the Krane. So I'm going to give a homework assignment today which will
be due Monday of next week. I won't actually start lecturing on this until Wednesday burr
you're welcome to start right now. I have office hours if you want to come see me about
that or anything else. This is posted already.
Okay. So just to finish up the discussion of the shell model, pointed out a couple
of different times that when dealing with odd a nuclei in this class at least and asked
to come up with guesses for spin and parities and of the ground state and first few
excited states the only real tool you have is the seismological. In fact, it's it this
extreme single particle shell model where we explain the spin and parity and nuke -- I
showed you in fact that is that isn't always a good approximation to what really happened.
Part of reasons for that were discussed earlier. Is that one of basic assumptions of
shell model is the nuclei are spherical. And you already nuclei that that isn't always
true. And the second effect which we talked about when we start putting particles into
very simple states the assumption is as they go in, they pair up to give spin zero and
it's always that last part we attribute spin to. But that isn't necessarily true either.
Because of these so-called valence particles. The ones just outside the closed-shell are
interacting with one another. It's not necessarily true that they will pair up -- if you
go back to the shell model around the D3 halves, half seven halves shell where the magic
numbers are 20 and 28 as we add particles I don't understand n equals 10 and tens, you
expect the ground state spins to have spin and parity seven halves minus and that the
first excited state is likely to be when I take that odd particle and push it up to the
next shell model orbital which is the three halves and because it's a p orbital it will
also have negative parity. Oy showed this before, in fact, the ground state spins work
out fine. As we do that, go to calcium 41 and 43 and so on, that odd proton and and
neutron goes in the, as soon as you get into the excited states you have trouble. When
I
So I mentioned this just for completeness but this is not going to be on the exam,
there's a way to accommodate the fact that many nuclei are deformed in the context of
the
shell model.ing it, take two, just to show you so you have seen it before if you ever
encounter it again, this idea of orbital angular momentum is really only a good quantum
number if I have a spherical system. If it's deform that isn't a good couple number.
What's used instead to define the quantum numbers associated with a given state in a
deform nucleus is the total angular momentum j which is the vector sum of l and a. And
the components of j along what we call the symmetry axis. Remember I showed you two
limiting cases of the shapes of the deform nuclei, one was prolate, look looked like
a
football. The other was oblate, like looked like a squashed sphere. Interesting talking
about the spin -- this is out of Krane. Just to show you what's going on. The idea here
is that in the shell model, we assume that all the states of a given spin regardless
of
what their z projections of that spin are have the same energy. When we talk about,
f7 halves particle for p three halves particles the assumption all the particles in
the f so much halves orbital have the same total energy because attributed to l and a
coupling business. And we aren't saying to make any difference what the projection of
that spin is on the afltion of nucleus because in this case of nuclei are all spherical.
So it doesn't matter. Once I have a deform nucleus that isn't true. It's a little
complicated but what he's trying to show is that if we look at f7 halves orbital, in the
shell model all the z projections have the same energy. In these deform nuclei they
won't. If I look over here first of all in the upper figure, that's the prolate form.
And so it's ellipsoid with so many major axis in this direction. So the long axis is in
this direction. Imagine the thing is coming out the board like that. So that's the
football. He's showing you different z projections of total angular momentum along that
sum industry Ali Mughayat Syah. And if you look at the he one label J4, the one that's
pointing most closely in the same direction as symmetry axis, that's the one that has
the
largest z projection. In fact, it will be the seven halves projection. Now what he's
saying if you think about where the particle would have to be moving in order to produce
a z projection of total angular momentum pointed over there it's going to move in this
orbital labeled four. So it's moving like this. And it's spending most of its time far
away from the bulk of nucleus. The bulk of the nucleus is inside the solid ellipsoid
in
there. Because it's far away and because the nuclear force is attractive that is less
tightly bound than one spending more of its time close to the nucleus. That would turn
out to be orbital one which is close to the, that's the one-half. What he's trying to say
is this state of spin seven halves which in the shell model and the spherical nucleus
has
one energy, splits into four different energies depending upon the z projection of
angular momentum along that symmetry axis. And similarly if you go to the oblate shape,
this is the one that looks like the flattened earth, it's exactly the same drawing except
the orientation of nucleus is different now. The symmetry axis is here. And so now these
same vectors give you the same projections of angular momentum but the one which is
pointed here going like this, is spending more time near the bulk of nucleus and the
one
going like this and pointing here spends less time. So the relative energy of these
states are completely opposite. Now the state with the highest spin ends up with the
lower energy and the state with the lowest spin ends up with high energy. This is
complicated way of saying shell model split now in the deform nuclei depending upon the
orientation of the angular momentum relatively to the symmetry axis. You get this very
complicated drawing which you don't have to worry about at all. It's called the Nilsson
diagram because Mr. Nilsson is the guy who figured this out initially. What they're
trying to show is here are shell model states and as you deform the nucleus away from the
spherical shape they do split. Some go up in energy. Some come down in energy. If you
have a heavy deform nucleus and wanted to know what the spin and parity actually was
you
have to know something about what the shape of in nucleus was before lands it figure out
where it's most likely to put that last neutron or proton. Alternative if you could
somehow determine the quantum numbers associated with given particle in the nucleus you do
read off the deformation it's beyond the scope of this class but just so you know you
heard of Nilsson diagram here are some example showing how it manifests itself in heavy
odd a nucleus. This is hafnium 177. It's an odd neutron nucleotides. And if we go
and look at ground state, it has a spin of seven halves and negative parity. Remember
this is in the region of nuclei between mass 150 and 190. Where what kinds of excitation
mechanism is responsible for most of the excited state.
STUDENT: Rotation.
PROFESSOR: Rotation, exactly. They're deformed. So we can make them spin. We
talked about in the context of even-even nucleus where the ground state has had spin zero.
Here we're starting out with nucleus that has odd neutron number. And therefore the
spin
and parent of ground state is not zero. But nevertheless I can make that whole nucleus
rotate. And if I don't disturb in any other way I buildup this band. This is rotational
band that's built on this particular ground state. And remember that when we had
even-even nucleus, the sequence of spins went zero plus, two plus, four plus and so on.
We increased spin by two units each time. For various reasons turns out when you have
odd a nucleus the spin increases just by one unit. But no parity change. So we have
seven halves, nine halves, leave hatches and so on. And what you would do to figure out
the moment of inertia of this particular band, is just like we did before, in terms of
even-even nuclei, we looked at the difference in energy between the ground state and the
first excited state. And said that's this quantity h bar squared over two i much we
assumed the moment of inertia didn't change as we increased angular momentum. We could
predict energy then of other states. you can do the same thing here. Starting with the
ground state band everything is identical is to what we said before. Notice I have
another band over here, it starts with spin of nine halves. It's not parts of this band.
It's got the wrong parity for one thing what this is trying to show is it have a different
projects, in one way you can think of this particular bands as having a shape. A certain
shape this particular band has a different shape. It does the same nucleus, same number
of neutrons and protons but they're argued differently will I somehow. Now I take this
different shape and start it spinning. So there's a different moment of inertia
associated with this band. And this band. And the way you can tell that ask by looking
at the energy difference between these two levels and comparing to the energy different
between these two. Turns out they're different. If you say each of theme is that h
bar≤ over two i you imply there's a different moment of inertia for this band and this
band. When you calculate the energies of members of this bands you have to start at, with
it state as the lowest member of that band. So essentially call this zero. And then
determine the other energies relative to that.
Okay. There's a homework problem that you did this work that's sort of related to
that. And at that point I'm going to quit.
STUDENT: You mentioned how you can predict the next MeV levels of h bar≤ over two
i, can you reiterate the reason why that prediction is a little off. Is it because of
beta.
PROFESSOR: What we're assuming it this nuclei are behaving like a rigid body. So if
I have a sold only. As I started spinning, it's a reasonable assumption the shape doesn't
change. I make it spin faster and faster, what that means is the moment of inertia which
forts energy spacing. What we're saying is real nuclei aren't really solid object.
They're closer to a liquid. What happens with a liquid, a drop the water or oil and spin,
it will elongate a little bit. That's why our estimates are a bit off. It's actually
surprising they're as good as they are. Because we totally neglected that. When you go
to higher and higher angular momentum, meaning the nuclides is spinning faster and faster
you see larger and larger diversions between the prediction of how to measure. Questions?
All right. So this is a good place to stop for today. On Wednesday we'll start the
discussion on alpha decay. And that's the homework due Monday. If you haven't picked
up
your exam, they're up here. And see you Wednesday.