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visit MIT OpenCourseWare at ocw.mit.edu. 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. And in fact, where we left
off last time is in order to get positron decay, the
Q value of the reaction has to be at least
1.022 MeV, better known as at least two times the
rest mass of the electron, because in this case, to
conserve charge and energy, you shoot out a
positron, and you also have to eject an
electron in order to conserve all the
charge going on here. So there you have it. Now let's look at the lower
energy decay of technetium to molybdenum,
which had something like 3 MeV associated with it. So we'll click on technetium,
and its energy is 3.2 MeV. Let's take a look at
its electron capture. Significantly simpler. Already what do you notice
about these positron to electron capture ratios? Anyone call it out. AUDIENCE: Electron capture
is much more likely. PROFESSOR: Indeed. When the energy of
the decay goes down-- notice that only these
decays are allowed-- The electron capture suddenly
becomes much more likely. But notice that it does
not let you go directly from 3.2 MeV to 0. There is no
allowable decay here. So this is probably a
change of-- that's 3.2. That's 1.3. A little less than 2 MeV. All of a sudden,
electron capture becomes much more likely,
but positron decay is not disallowed yet. So we can say electron capture
or positron decay right there. Everyone with me so far? So let's go to the
really low energy one. We'll click on
molybdenum-93 and see how it decays with an energy
of 0.405 MeV to niobium. Anyone want to guess
what's allowed? AUDIENCE: Electron capture only. PROFESSOR: Electron
capture only. There's not enough energy
for positron decay. And, indeed, it draws
funny, because there's a metastable state. But if you scroll
down here, there are two pathways allowed, both
of which by electron capture. Decay diagram's
quite a bit simpler. So we leave this one here
by saying it can only decay by electron capture. Any questions on
the odd A before we move on to the even, which
is a little more interesting? Cool. OK. Let's move on to the even case. So for here, I'm
going to go back to the overall picture
of the table of nuclides. Click on around where I
think potassium-40 is. Looks like I got there. And I want to point out
one of these features. If you wanted to
undergo decay change and maintain the
same mass number, that's diagonally from
upper left to lower right. See how all the isotopes here
have a 40 in front of them. The really interesting part
is as you cross this line, you go from stable to unstable
to stable to unstable again. The colors here is dark
blue represents stable, and dark gray represents
long lifetimes of over 100,000 years. So this is one of the
reasons you find potassium-40 in the environment. In fact, 0.011% of all
potassium in you and everything is potassium-40. It's what's known as
a primordial nuclide. It's not stable,
but its half-life is so long that
there's still some left since the universe
began or whatever supernova that formed Earth
got accumulated into the earth. But notice it can come from-- it can decay by a couple
of different methods. So let's pick one of those
stable isotopes, calcium-40, and put that as the bottom of
the parabola on this diagram. So we'll put calcium here. And in a relative sense, we'll
put a calcium point right there for its total mass. And it could have
come from beta decay from potassium-40 or electron
capture from scandium-40 40. So let's look at potassium-40. It can beta decay for
about, oh, 1.3 MeV. So potassium is right here. Let's say it could beta
decay with about 1.3 MeV. And we'll trace potassium
back a little bit, figure out where would
it have come from. Interesting. Doesn't tell us. OK, forget that. Let's trace calcium back
and say there's scandium-40. And scandium-40 can
decay with-- wow-- an enormous 14.32 MeV. Let's put that like here. Anyone want to guess which mode,
electron capture or positrons, much more likely? AUDIENCE: I think positrons. PROFESSOR: Probably positron. Let's take a look. Oh boy. Another complicated one. But the whole way down,
positron, positron, positron for all the
most likely decays. You won't find a drawing
to every single line. I believe that they
know that at some point, drawing extra lines is futile,
and they just all overlap each other. So I don't know exactly
how the algorithm works, but it does draw up to some
number of possible decay chains. If you want to see
every single one, they are tabulated in a very,
very long list down below. I'm never going to ask you to
do something with all of these, because that would
be insane unless it's a relatively simple
decay, like that has two or three possibilities. And let's see. This could have come
from electron capture from titanium-40 with 11.68 MeV. Wow. OK. Up here. And there's titanium. And let's go in the
other direction. So I do know that potassium-40
can decay into argon-40. So let's jump there. Argon is a stable isotope too. So potassium-40 can decay into
argon-40 by electron capture. OK, good. A more respectable 1.505 MeV. Is positron decay allowed? AUDIENCE: Yes. PROFESSOR: Yes. Why is that? AUDIENCE: [INAUDIBLE] PROFESSOR: Over 1.022 MeV. Yeah. Anyone have a question? No? OK. So we've got kind of a
kink in our mass parabola. Yeah? AUDIENCE: Actually,
yeah, so it's possible if it's over 1.022,
but it's still very unlikely. PROFESSOR: That's correct. AUDIENCE: Until we get to
these higher orders, like 10. PROFESSOR: Yep, so once
the Q value's satisfied, it is technically possible. But if you had
something with the decay energy of, like, 1.023 MeV, it
would be exceedingly unlikely. So in fact, we can
take a look at this. This, I would say,
is also going to be on the exceedingly unlikely
level, and we can take a look. So if we look at
the decay diagram, we know it makes positrons. They're not even really listed. Interesting. So that process
would not be allowed, but this one, because
that's about 1.5 MeV, should be allowed. But since that branch
ratio or the probability of that happening is already so
low, I wonder if it even says. Yep, beta ray with a max or
average energy of 482.8 MeV. We're going to go over why
that energy is so low when we talk about decay next week. With the relative intensity of
something with a lot of zeros before the decimal place. So there you go. Like you said, energies near
1.022 MeV, slightly above it, are extremely unlikely but
possible and measurable. Cool. And then let's see what
could have made argon 40. Could have been beta
decay from chlorine-40. So chlorine maybe was here. And I don't think I
have to draw any more. So we've got a funny-looking
parabola with a kink in it, because really, you have two
mass parabolas overlapping. I'm going to go
back to the screen so that the diagram
from the notes makes a little more sense. What we've kind
of traced out here is that there's two
overlapping parabolas here. There's the one
with the-- what is it-- the odd Z and the even Z. So there you go. Just like the one on
here, which I think is for a different mass number. Yep. 102. We get the same kind of behavior
where things will mostly follow the lower mass parabola,
but sometimes if something gets stuck here, it can go
either way to get more stable. So I want to stop
here for new stuff, because this is
precisely where I thought we'd be at the end of the week. And in the next
half an hour, I'd like to open it up to
questions or working things out together on the
board or anything else you might have had. Yeah? AUDIENCE: I have a question
about the parabola things. PROFESSOR: Sure. AUDIENCE: There's-- you
said multiple paths, so it doesn't have to do the
little peak in the middle? Like, could it follow the
lower parabola or the upper, or does it have to jump over? PROFESSOR: It's going
to go in whatever way makes it more stable. So you're never going
to have a nucleus that's going to spontaneously
gain mass in order to get to a different path. You can only go down
on the mass axis. But let's say you happen to be
starting here at potassium-40. You can go down via either
mechanism to the next mass parabola down. AUDIENCE: But if you were
argon, you would go up. That's what you would do. PROFESSOR: That's right. If you're at argon,
you're stuck. And in fact, if you
want to take a look, what do I mean
scientifically by "stuck"? I mean stable. Argon-40 is a
stable nucleus that comprises 99.6% of the argon. So that's what I mean
by "stuck" is stable. And if we look at the rest
of the table of nuclides for similar-looking places-- so let's hunt near potassium-40. So notice potassium-40
right here has got stable isotopes to the
upper left and the lower right. If we look back over
here, manganese-54. Same deal. It's got a stable isotope to
the upper left and a stable one to the lower right. How much you want to bet that
when we click on manganese-54, it's got two possible
parent nuclides-- or I'm sorry, two
possible decay methods. So let's take a look. Manganese-54 can either electron
capture and positron decay to chromium-54 or
beta decay to iron-54. Let's take a look at one more
to hammer the point home, and I think that'll
probably be enough. Cobalt-58 right near
nickel-58 and iron-58. Interesting. That one's not allowed unless
there's more down here. So you can electron
capture to iron-58, but there's no
allowed decay to-- what was it? Nickel-58. OK. Let's look for more. Chlorine-36 has argon and
sulfur on either side. There it is. Beta decay and electron capture. And how much you want to
bet there's basically never a positron here? But basically, not
actually never. So you get a positron 0.01%
of the time and electron capture 1.89% of the time. Where is the other
98-and-change percent? Right here in the beta decay. So in this case, chlorine-36
will preferentially beta decay. If you also notice, it's a-- let's see. I don't know if that
actually matters. But I am going to say it's
more likely to beta decay. So when you sum
these up, you get 100% of the possible decays. Let's see how many energy
levels there are there too. Hopefully not too many. That qualifies as not too many. Yeah? Sean? AUDIENCE: Are the changes
in mass always going to be attributed to
beta decays or electron captures or positron
[INAUDIBLE]?? PROFESSOR: They'll be
due to those as well as some other processes, which
we're going to cover on decay. But if you notice,
I've been giving you a lot of flash-forwards
in this class. We've introduced
cross-sections as a thing, the proportionality
constant between interaction probabilities. We're going to hit
them hard later. I've also been kind of
introducing or flash-forwarding different methods of decay. So there's also alpha decay. There's also isomeric
transition or gamma emission. There's also
spontaneous fission. This is the whole
basis behind how fission can get working
without some sort of kick-starting element. So maybe now's a good
time to show you. Let's go to uranium-235
and see how it decays. It goes alpha decay to
thorium-231 most of the time. And if you look how,
it's not terrible. We can make sense of this. It also undergoes SF, which
stands for spontaneous fission. So one out of every seven-- what is it? Seven out of every
billion times, it will just spontaneously
fizz into two fission products. And this is why if you put
enough uranium-235 together in one place, you can
make a critical reactor. In reality, you
don't tend to want to put enough U-235 together to
just spontaneously go critical. We use other isotopes
as kickstarters. For example, californium,
I think it's 252. Let's take a quick look. There we go. Californium-252 undergoes
spontaneous fission 3% of the time. It's even heavier,
even more unstable. So there is a
reactor called HFIR, or the high flux isotope reactor
at Oak Ridge National Lab. One of its main outputs is
californium kickstarters for reactors. So to get things going, you
put a little bit of californium in as a gigantic neutron source,
and then you don't really need it anymore
once it gets going. So it's one of the safer
ways of starting up a reactor is put in a
crazy neutron source, and then once it gets going,
take it out or leave it in and burn it. I'm not actually sure
which one they do. Yep? AUDIENCE: Is the
name californium based on California? PROFESSOR: It is. When we get to-- now is a good time to introduce
the super heavy elements since you asked. So a lot of these older
elements were named after-- actually this is kind
of a hobby of mine. So I don't know if you guys
saw the periodic table outside. I collect elements,
because if you're going to collect
something, you might as well collect everything that
everything else is made of. It's the same reason I
went into nuclear energy. I started off course 6, or
6.1, specifically electrical. And I was like, well, I
could be designing, like, the next screen
for a cell phone, or we could solve
the energy problem, which is the problem all
others are based off of. So my whole life theme
has been go to the source. That's why I came here in
high school and never left. That's why I declared course 22. That's why I collect
elements and probably is the reason for many other things
which only a psychiatrist could diagnose. But let's look at some
of the other elements. For example, yttrium. I think it has a isotope 40. Anyone know-- no, it
doesn't have a 40. What about a 50? 60? 100? Whatever. At least it knew
that Y was yttrium. Anyone know-- AUDIENCE: 89. PROFESSOR: --where
this is coming from? 89? Seems high. Oh my god. You're right. AUDIENCE: I work with it. PROFESSOR: OK. Gotcha. You work with it. Awesome. Anyone know what this
is all about, yttrium? There's a town called
Ytterby in Sweden where large deposits of
yttrium and ytterbium, or Yb, tend to be found or
Db, named for dubnium. So let's say the really
basic elements tend to come from Latin. Fe stands for iron, which
actually stands for ferrum. Lead is plumbum. Gold is aurum. Silver, Ag, is argentium. I don't know if I'm
saying that right. I never took Latin,
and I've never heard it spoken, of course. And then a lot of the heavier
and heavier elements as we go are being named for
more and more famous scientists or places where
they tend to be made like Db. I'm going to guess
260 for a mass there. Oh, nice. For Dubna in Russia
that has got one of the few gigantic super heavy
element colliders where they're constantly synthesizing
and characterizing these super heavy elements. So finally they
said, you know what? They've made enough
of these in Dubna. Let's name one of the
elements after them. Or Sg, seaborgium,
for Glenn Seaborg. Or No, nobelium,
for Alfred Nobel. Yep? AUDIENCE: I just have a question
for the actual mass parabola. PROFESSOR: Uh-huh. AUDIENCE: Like, do the parabolas
ever, like, reach each other? PROFESSOR: Do they intersect? AUDIENCE: Yeah. PROFESSOR: I've never seen
a case where they intersect. That would make for a
crazy situation indeed. However, part of what the
homework assignment's about is to derive an analytical
form for a mass parabola and then check the data
to see how well it works. So for any cases where you
have an even mass number, and you have either odd-odd
or even-even nuclei, you can check those
equations analytically to see if they'll intersect. AUDIENCE: And for the
case of A equals 40, I'm not really sure what
the top parabola is. PROFESSOR: So the top
parabola for potassium-40-- let's take a quick look at how
many protons and neutrons it has. Potassium has a proton
number of 19, which means it has a neutron number of 21. So the top parabola
is odd N and odd Z, where the bottom one
is even N and even Z. Whereas for odd
mass number nuclei, it has to be either
odd-even or even-odd, else it would be even, which
is a funny sentence when you say it all out loud. Yeah. So that's the idea
here is that notice that the even-even parabola
tends to be further down. All those nuclear magic
numbers, 2, 8, 20, 28-- I'm not going to quote the rest. Those the little ones I know. All even numbers. So any other questions
on these mass parabolas before we launch
into super heavy elements? Yeah? AUDIENCE: [INAUDIBLE]
the bump on the right? PROFESSOR: Uh-huh. AUDIENCE: How do you know
that that's where [INAUDIBLE]?? PROFESSOR: Analytically
or experimentally? Which question? AUDIENCE: Analytically. PROFESSOR: So analytically. Analytically there
should be some isotope of-- well, not potassium. That wouldn't be allowed. So in this case, the
stable element positions have got to kind of switch
off, shouldn't they? So if that's potassium-40,
that would still have to be potassium. You don't really
have another choice. There isn't really a position
there, is there, analytically? That's the interesting
thing is that you can either be odd-odd or even-even
for an even mass number. But you can't just take off
one neutron from potassium-40, and then you've
got potassium-39. Then you're on a
different mass number. Or if you exchange a
proton and a neutron, which you pretty much do in
either of these directions. There's no way to get
straight down here. AUDIENCE: Right. PROFESSOR: Yeah? AUDIENCE: For odd-odd,
delta is negative, right? PROFESSOR: For odd what? For-- sorry? AUDIENCE: For odd-odd,
delta is negative? PROFESSOR: Yeah, let's
go back to that slide just to make sure. You mean the pairing term in
the semi-empirical mass formula? AUDIENCE: Yeah. PROFESSOR: Yeah, so
for odd-odd nuclei, indeed, delta's negative, which
means lower binding energy, which means higher mass. And that's why we see it bump
up on the mass right here. Yeah? And do you have a second
part of the question? AUDIENCE: It was
more so how to relate [INAUDIBLE] like the binding
energy to that mass parabola. PROFESSOR: We can
actually relate-- so we can relate
the binding energy to the mass and the mass
parabola analytically, because the binding energy is
equal to Z times protons plus N times mass of neutron
minus the actual mass of that same nucleus, A comma
Z. So they're actually directly related, just negatively. So something with
a higher mass is going to have a low binding
energy, which means it's less bound and less stable. And indeed, the further
up the mass scale we go, the higher those
beta or electron capture or positron
energies are. And there's another
thing you can check too, which is the half-life. Half-life is what we'll be
talking about on Tuesday. It's how long before an
average amount of a substance has undergone radioactive decay. So let's look at some
of these isotopes and start looking
at half-life trends as another measure of stability. So potassium-40 has an
exceptionally long half-life. So it's relatively stable. Let's take a look not at
either the stable isotopes, but let's go up
the mass parabola chain in one direction. Calcium-40, scandium-40. So let's take a
look at scandium-40. Scandium-40 has a half-life
less than a second. And it's got quite a high
decay energy by whatever method you want to use. Let's go up to titanium-40. Anyone want to guess? Do you think the half-life
is going to go up or down? Let's see if the
half-life goes down. We know the decay
energy goes up. Indeed. Half-life goes down from
182 milliseconds to 50 milliseconds. And let's say titanium-40
could have come from-- wow. Two proton decay from Cr-42 with
a half-life of 350 nanoseconds. So as we go up the mass ladder
and down the stability ladder, the half-life decreases, which
kind of follows intuitively. Something that's
exceptionally stable should have a
half-life of infinity, and something that's
exceptionally unstable should just blow
apart instantly. Like, remember the
first week of class, we talked about helium-4
grabbing a neutron, becoming helium-5, and
instantaneously going back to helium-4. If you look at
helium-5, its half-life is measured in MeV, or 7 times
10 to the minus 7 femtoseconds. So if helium-4
absorbs a neutron, it simply doesn't want
it and gets rid of it in 10 to the minus 7
femtoseconds, which would tell us that it's
exceptionally unstable. So I hope that's a
long-winded answer to that question about what
does it mean to be going up in the mass levels. Any other questions on mass
parabolas or the liquid drop model or stability in general? Yes. AUDIENCE: For something that
goes upwards [INAUDIBLE] just because the
mass [INAUDIBLE].. PROFESSOR: So if you're
changing one neutron to a proton in each case,
you're switching back and forth from the odd-odd to the
even-even mass parabolas. So if I were to redraw
these dots more to scale, this would have to
be on the odd-odd. And, well, let me draw
them a little better. Yep. AUDIENCE: OK. PROFESSOR: That's
on the odd-odd, and that's on the even-even. AUDIENCE: OK. PROFESSOR: Yeah. So excuse my poor
drawing skills. But if you're switching
one proton to a neutron or vice-versa, by
definition, you're jumping back and forth
between these parabolas. AUDIENCE: OK, thank you. PROFESSOR: That's a good
question for clarification. You had a question too? AUDIENCE: Yeah, about the
semi-empirical mass formula. When do you use that to find
binding energy as opposed to, like, any of the other ways? PROFESSOR: I'm sorry. The semi-empirical mass
formula is a good way to get an analytical
guess at most of them. If you want an
exact answer, always use the actual binding energy. AUDIENCE: So, like, how
often is it used now? PROFESSOR: I would not say
it's used much now except-- well, that's going to be one
of your homework questions is this formula predicts that
as you get heavier and heavier and heavier, nuclei should just
continuously get less stable. And that was, as of
when this was derived, let's say decades ago, we
now know something different is happening. So if you look at the
table of nuclides, you can sort of see some swells
in the number of black pixels until it cuts off. And this region actually where
we think super heavy elements happen, I want to jump to the
actual table of nuclides, which I'll say is our snapshot
of knowledge today, and go all the way to the top. And our knowledge kind of cuts
off at these elements, which are, for now, temporarily
named in a very uncreative way. We don't even know
anything about them. Uun is probably going to
have a proton number of what? 110. I don't know what
the prefixes are, but UUU would be un-un-un 111. Probably has 111 protons. Beyond here, off the screen or
probably up into the next room, it's predicted that once you
approach the next magic number in nuclei, there should
be an island of stability where it may not necessarily
be totally stable, but the half-lives
should go up again. And we should be able to
synthesize super heavy matter. And if you actually
graph neutron number versus half-life-- so notice
how we were looking at half-life as a measure of stability. It starts to go up, then
comes down, and then, to the extent of
our knowledge, it is going back up again to the
next predicted magic number. So what we think
should be happening is half-lives should be
continuously going up. And yeah? You had a question? AUDIENCE: Like, what do we
do with these weird things? PROFESSOR: Well,
whatever you want. It's going to be-- it should be dense as all heck,
because nuclear matter is quite a bit denser than
ordinary matter, and quite a bit is
quite an understatement. So what would you do
with super heavy matter? A lot of it could be used to
probe the structure of matter. There's a lot about
how the nucleus is constructed that we don't know. And beyond the
scope of this course would also be an understatement. There's folks that are
making their careers now on figuring out what are
the forces between nucleons? Why do things spontaneously
fizz at the rates that they do? You'll even hit a little
bit of this in 22.02 when you can calculate the rough
half-life for alpha decay using quantum tunneling through the
potential barrier in a nucleus. And so the more nuclei we
have to mess around with, the more data and real
examples we have to study. But practical applications,
well, I could imagine, we might find something
denser than osmium. Osmium right now has a
density of about 22 grams per cubic centimeter. This stuff, zirconium,
is about 6.9 or so. Steel is like 8. Lead's like 11. Mercury is like 19. Have any of you ever played
with liquid mercury before? This is a "don't try this at
home, kids" kind of moment. My grandfather happened
to be a dentist, so we happened to have a lot
of mercury to mess around with. And it's, like,
unintuitively heavy. It's unbelievable. A 1-pound jar is about that big. I think it would
be cool if we could find something even denser. And then really,
really dense matter happens to make really, really
good photon shields and gamma-- not the Star Trek thing. I mean this in the actual
nuclear physics sense. The best way to stop gamma
rays for gamma shielding is just put more
matter in front of it. And if we find a denser state
of matter that's earth stable, you then have a
smaller gamma shield. So there are practical
applications too in radiation shielding. They also might make
awesome nuclear fuel, because you better
believe they're going to fizz like crazy. So who knows? Maybe we can-- I don't think
that would be cost-effective, but it would probably work. So the way they're
doing this is actually slamming calcium 48 nuclei
into other super heavy elements that have exceptionally
long half-lives. So if you can't read
what the screen says, this here is berkelium,
for Berkeley, with proton number
97, mass number 249. Let's take a look at the
Bk-249, which happens to be way beyond uranium. So it's definitely
not a stable isotope, but it has a
half-life of 320 days. That means you can
make a bunch of it, chemically separate it,
make it into a target, and fire calcium-48
nuclei into it. Anyone want to guess,
why do we use calcium-48? And I'll give you a hint
and write the proton number for calcium. The isotope that we
use is calcium-28-- or calcium-48. Anyone want to take a guess? Why start here? Why not just smash two
berkeliums into each other? Calcium 48 happens to
be exceptionally stable, because it's got
two magic numbers. Its proton number is 20, one
of those peaks of stability. And its neutron number is 28. So start with
something super stable, something with a lot more
binding energy to begin with, and you maximize your
chance of making something with more binding energy
that won't just spontaneously disappear. So there are reasons
calcium 48 was chosen and not something
heavier or lighter. If we go back to
that article, you can see what happens here is
you make some element 117, which has yet to be made, and it
undergoes alpha decay until it reaches some rather stable-- you know, 17 seconds. That's pretty exceptional. And if you notice the
trends here, as you decay, the alpha energy
steadily goes down, and the half-life
steadily goes up. And so what you do is you make
a super, super heavy element, hoping that it
will decay and rest in one of these
islands of stability beyond the magic numbers
that we know right now. Which I thought
this was super cool, because this is
actually happening now. Like, new elements are made. I think we've been
seeing one a year or so for the past
few years on average. There might have been a year
when there was more than one announced recently. There's only a few places
in the world doing them, but you can start to-- already
with two weeks of 22.01, you can start to
get a handle for why do they use the nuclei they do,
and then what sort of things are you looking for? Decays with lower
and lower energy mean you're already starting
to get less steep on whatever imaginary mass parabola. Don't quite know how
to draw this one, because it's beyond
anything we know. And as the half-lives
keep going up, you can tell that it's reaching
a measure of stability. However, to get you
started on the homework, for the open-ended
ended problem-- I think I'll bring
it up right now so you guys can take a look. So let's go to the Stellar site. Hopefully it doesn't call me. Good. And two problem set two. This is the way I
know that everyone's seen the P set seven days before
it's due, because I'm going to put it up on the
screen so you can see it all the way at the end. Predicting the
island of stability. Does the semi-empirical
mass formula predict the island of stability? Well, let's start you
off with the easier part of the question,
which is yes or no. And I'm going to leave you
to the why and the how. If we graph binding energy per
nucleon versus mass number, the semi-empirical mass formula
predicts something like this. What happens as we go beyond
the realm of known mass numbers? Anyone? How should I extend this curve? What did you say, Alex? AUDIENCE: I don't know. PROFESSOR: Just keep going. Yeah. Does this predict an
island of stability? I don't think so. So that's one of the
few questions I'm asking you in this homework. And it's up to you guys. Use your creativity. Again, this is an
open-ended problem. I'm not looking for
a specific answer. I want to see how
you think and how you would change this
formula to account and actually predict the
island of stability while still satisfying the mostly correct
predictions from the elements we know. So-- sorry, go ahead. AUDIENCE: So should it,
like, converge a little bit? PROFESSOR: Well, you're
on the right track. If you want to show
stability, you'd want it to maybe have a
higher value right here. Higher binding
energy per nucleon would correspond to a
lower mass, which would correspond to higher stability. So how would you predict
this island of stability? And then more
specifically, how would you reconcile the inaccuracies
in the semi-empirical mass formula? Because we know it doesn't
work very well for all cases. There are some cases like right
around here where it works great, and there's some like
right here and right here where it really doesn't. You can get things
wrong by like 10 MeV, which is pretty significant. You know, that's like four
digits on the mass scale, like, the fourth decimal place. That's huge to a
nuclear engineer. So that's something
to get thinking about. And remember I did tell
you that there will be some open-ended problems. I'm going to mark
them as open-ended so you actually know. We're not looking for a
right or wrong answer. This is one of those
kinds of things where we want to
see how you think and what do you
think is missing. There's other hard problems
where we give you the answer, because I'm not interested
in you deriving some insane expression and getting it right. I'm interested in the
derivation process. What are the steps you choose? What sort of
assumptions do you make? What sort of terms can
you neglect and say, that's in the ninth
decimal place. I'm going to forget it. So in this case, we
give you the answer, because we're going to
grade you on the process. And you can use the answer
to check your process and see if you're on
the right track or not. For the skill-building
questions, we actually do
want you to come up with some sort of an
answer like explaining the terms in the
semi-empirical mass formula or modifying an equation to
calculate something else. We will be looking for
a right answer there. But those are questions
to make sure that you get the basics of the material. If you can answer all of the
questions in the first half of these P sets fairly quickly,
let's say in three or four hours, you're totally
on the right track. The hard ones is
because this is MIT. And we want you to
think beyond just knowing what's in the
Turner book or the Yit book. Like I said, you guys are
the leaders of this field. So any other questions
on stability in general? Yes? AUDIENCE: Just a
real quick reminder. When you say, like,
even-even, are you talking protons, neutrons? PROFESSOR: Correct. So that would be even
N and even Z or odd N and odd Z like in
the reading and like on these mass parabolas. Yep. Any other questions? Yes. AUDIENCE: Is the
only proof or reason that we say that there's
an island of stability because the mass increases
up to the point of unknown? PROFESSOR: There's a few-- so the question was,
is the only reason people think there will be super
heavy elements because the mass increases, right? AUDIENCE: Yeah. PROFESSOR: So in this case,
the mass will always-- are you talking about
now the total mass or-- AUDIENCE: Why is
the idea that there is this island of stability? PROFESSOR: Ah, OK. AUDIENCE: If this
doesn't prove it, do we have other
reasons [INAUDIBLE]?? PROFESSOR: We have a
few things to go on. There are a number of different
aspects of nuclear stability that are all pointing
to the same conclusion. One of them, you can
see on this graph here. If you look at the
alpha decay half-life as a function of
neutron number, it doesn't just increase or
decrease monotonically. It swells up and down. And it reaches a
relative maximum near certain magic numbers. We can confirm that with
the lower mass nuclei. It doesn't work for
really low mass, because tiny things don't
tend to undergo alpha decay. But there are patterns
that we're simply recognizing and saying, well, if
this is the next magic number, it should continue to increase. And I should mention too,
this scale is logarithmic. So the top right here is
like 10 to the 4 seconds. Just so you know, there
are 86,400 seconds in-- what is it-- a day. And 3 times 10 to the
7 seconds in a year. So if this graph-- let's say for Z 111-- were to continue
on its track, it should reach like
10 to the 9 or 10 to the 10, which could be
like 100-year lifetimes or 100-year half-lives,
which means definitely you can chemically separate them
and do things with them. I don't know if they would
be safe enough to deal with. But we also don't really
know what's going to happen. You can see that there
is some uncertainty, and things don't always
follow the trend. Even the error bars are
outside the dashed lines. But so we have this to go on. We have the alpha
decay half-life. We also have the
alpha decay energy. As you approach an
island of stability, something that's more
stable won't give off as much kinetic energy
to its alpha particle. There is also-- for the
ones that you can actually measure that live
long enough, you can measure their
mass to charge ratio and actually get a good
picture of their actual mass. So we would expect
the mass defect to follow a certain
trend as we go up. The mass is always
going to increase. If you add more nucleons,
it's going to increase. But the mass defect,
which is the real mass minus the atomic number mass-- if stability were
to increase, do you think the mass defect
would increase or decrease with more stability? Let's take a quick look at this. If A were to stay the same,
a shrinking real mass-- and remember, lower mass
means more stability-- would mean a higher
or a low mass defect? It would mean a lower mass
defect or a lower excess mass as you'd call it. So in this case, you would
expect the mass of the nucleus to be smaller than its A if
it was going more stable. And all of these trends
work in the same direction, which is saying, OK, so
we have the alpha energy. We have the mass defect. We have the half-life all
pointing to the same thing that something should
be more stable. And we have some
patterns to go on, but our understanding
is kind of incomplete. So-- yeah? AUDIENCE: So if there's
super heavy elements, do they exist
somewhere in space, or do stars make them, possibly? PROFESSOR: Ooh. AUDIENCE: Or are they-- PROFESSOR: Good question. AUDIENCE: --currently made? PROFESSOR: So the question is,
if super heavy elements exist, do they exist out
there in space? I think there would be a
couple places they would exist. The source of most of
the elements beyond iron is supernovas, where regular old
fusion doesn't cut it anymore. When you hit the maximum of
this binding energy per nucleon curve, you're at about iron 56. That's why stars tend
to form a core of iron before it goes really
bad in whatever way it does for a star. There are multiple ways. When you get a supernova,
you have an insane explosion, and the core gets
compressed from the outside, forcing fusion
of heavy elements to happen. That's because you're putting
in extra kinetic energy. So it's like you have an
endothermic reaction where if Q is less than zero, how do
you make that reaction happen? Add kinetic energy,
which can come from a tremendous explosion
outside of the outer regions of the star. So who's to say that some of
these super heavy elements aren't formed in supernovas? I think they would be. But would they actually make
it out to be part of Earth and then, let's say, live the
5 billion years that Earth's been around? We don't know if their
half-lives are long enough. There very well may have
been some 5 billion years ago or when the supernova was made. But we haven't detected
any here on Earth. So we know that they're
not 5 billion years stable. Rather, I wouldn't
even say we know that, but we have a pretty good idea. That's a great question is
like, are they naturally made? Probably. Yeah. Any other questions? I like these outside
the material ones. We can take things beyond
our known universe, start to explain them.
