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PROFESSOR: All right let's get start.
I just want to go back over a couple of point we were discussing, a couple of points
we were discussing on Monday. The first has to do with looking at the decay energies for
nuclei that can undergo electron expenditure and/or positron, if it's a positron decay
the
only way that can happen is if the difference between the initial and final atomic mass
is
greater than twice the rest mass energy of electron. If that isn't the case, then
positron emission is not energetically possible but electron capture will be. And just to
show you a couple of examples, you have a homework problem related to one isotope which
is
beryllium seven. It undergoes electron capture into lithium seven because the Q-value, I
think in the homework problem you have to calculate it. But I don't know the number
the top of my head but it's something like 0.85 MeV. And that's less than twice the rest
force, what's going on here is this has four protons, and three neutrons. And this has
three protons and four neutrons. So you are converting a proton into a neutron. In this
case the only way it can happen is via capture of a electron. On the other hand in the
case talking about the other day which has sodium 22, it's going to be neon 22, and this
is 11, 11. And 10 and 12. Turns out this q value is more like two and a half MeV. Or
something like that. And so in this case, and for reasons that we'll see in a little
bit,
the decay actually goes to an excited state preferentially. Which then decays by emiting
a gamma ray. The point is this energy is greater, then positron emission will happen.
But in addition electron capture decay will also happen. If you have a situation where
positron emission is energetically allowed it will happen but ec decay will also
happen. If you have less than 22 MeV. And there are ways to calculate relative
probability of positron emission versus ec. We'll talk about that in a little bit.
STUDENT: For calculating the Q-value for electron capture, how do we find the
bn, binding energy of the electron.
PROFESSOR: That's not something you're expected to be able to calculate. You want
to look that up. In the back of the table of isotopes, there are binding energies for
electron shell, that's something you look up online.
At the end of class on Monday we were discussing this problem about why is it that we
have a free neutron, in fact, radioactive, and undergoes beta minus decay. I said it
had
a half-life of 10 minutes. Why when we stick neutrons inside a complicated nuclei they
don't all decay. There was skepticism first of all of whether you could actually observe
a neuron undergoing beta decay. I claim it has been done. It's an old paper, it
illustrates the technique. The concerns that was raised when neutrons will strongly
interact with materials they encounter. And more likely be captured before they decay.
That's true. So if you want to study the beta decay of a neutron you have to make sure
you keep it away from interacting with other matter so it doesn't capture and has the
opportunity to decay.
And because neutrons are in fact radioactive, we don't just dig them out of the
ground as free particles. We have to make them. If you don't know already you'll see
in
a couple of weeks that nuclear reactors are sources of free neutrons. A lot of
experiments that make use of neutrons for various purposes use as their source a nuclear
reactor. And that's the way this experiment was done. There was a nuclear reactor and
neutrons were brought out of the core of the reactor and allowed to travel down this tube.
And these were thermal neutrons, meaning they scattered a lot and lost energy and come
into thermal equilibrium with the environment in which located. And we'll talk more about
that later too. The idea was traveling down this pipe. This region over here is what's
called the decay region. Surrounding that region there's a magnetic field, vectors
pointing down in the vertically, and there are what are called plastic scintillators.
These are detectors sensitive to charged particles passing through them. The idea is most
neutrons will go whizzing by and nothing have anything happen. Occasionally some decay
while in the decay volume. The electrons will spiral around the magnetic field lines and
run into a scintillator which is either above or below the decay region. When they go
into the scintillator they deposit the energy. You can measure how many you see and also
the energy spectrum. What's shown here is the spectrum of electrons that are detected
in
plastic scintillators from the decay of the free neutron. And what was determined from
this experiment was the half-life was about 10.8 minutes.
STUDENT: What's the single crystal used for.
PROFESSOR: This is used to monitor how many neutrons are passing through because,
it's not just enough to see the beta spectrum you have to know how many neutrons you're
sensitive to. Measure the number of neutrons here and measure again there will infer how
many decayed along the way.
STUDENT: So the neutrons spiral around the magnetic field because of their magnetic
moment.
PROFESSOR: Not the neutrons. The neutrons are traveling through this pi. the
magnetic field is there to trap the electrons. So when the neutrons beta decays, what's
happening is the neutron is turning into a proton plus a electron, plus an electron
anti-neutrino. It turns out for the, well, because the proton is, essentially has massive
as neutron, the electron and neutrino take away almost all the energy. The proton will
pretty much keep going in the same direction the neutron has been transformed. The
electron and neutrino come in all directions. Idea of magnetic field is to direct the
electrons toward where the plastic scintillator is. Because of charge of electron and its
momentum that it will spiral around the magnetic field line. And tend to go that way or
that way. Is there another question.
STUDENT: How fast going.
PROFESSOR: How fast are the electrons going? The neutrons? These easy said are
thermal neutrons. That means their kinetic energy is about one-fortieth of a electron
volt. If you plug in the mass of neutron and say that this is equal to one-half mv≤
you can work out the velocity and not to spoil the answer because we'll talk about in a
few weeks, it's 2200 meters per second. That's about how fast the neutrons are traveling.
So on velocity scales we're used to that's pretty quick. But it's very small compared
the speed of light.
STUDENT: Do you expect the rate of neutrons to be constant.
PROFESSOR: Across this? Not necessarily. You're right some of the neutrons will
run into something and undergo a reaction. That's why you want to monitor number of
neutrons here and downstream to get a feel for how many are in the volume at any one
time.
It's a tricky experiment. There are lots of other losses for neutrons. So you have to
know how many actually were sensitive to that decay solve for a half-life. If you look at
the history of measurements there are, it still isn't settled what the value S is. This
is a rather old paper. From '67. More modern experiments don't do it this way. They
actually use what's called a magnetic bottle to confine neutrons and then study their
decay while isolated inside this bottle.
STUDENT: How's the magnetic field keep the neutron.
PROFESSOR: It's related to, what Sam was talking about. Because the neutron has a
magnetic moment that we can use the magnetic field to confine it.
Okay. So that's just to free free neutrons really do decay. Now the question is why
only way but here's one way. Let's look at the example I was discussing which is carbon
12. Let's think about carbon 12 in the context of a shell model. Carbon telephone 12 has
six protons and six neutrons. If I look at the shell model from chapter five and where
where the six protons and neutrons going to be located, it's as Sloan in the upper part
of
this figure here. So I've got six neutrons, six protons. The lowest state you can put
them into according to the shell model is one minus half orbital. I stick two neutrons
in
there. and I can stick two protons there because they are not identical particles. Notice
that the proton levels I shifted up a little bit relative to the neutron levels. That's
because of coulomb interaction which the protons feel butts neutron don't. I have two
separate wells. The proton well was shifted up a little bit. The amount by which is it
cyst shifted up dens on how many protons in the nucleus. I have two protons and two
neutrons in the one s half orbital. The next orbital available is the one p3 halves
which can accommodate four particles. Stick four neutrons there, four protons in the
corresponding one p3 halves orbital for the protons. Now I've beinged for the ground
state of carbon 12. Now I'm to see what would be required if one of those neutrons was to
decay and into a proton. You can see the problem. Suppose I take the neutrons which are
in the one p3 halves orbital. In order for them to decay they have to go into some
other shell model orbital which is not already filled. And the only one that would be
available is one P1 half which is higher in another than the one p3 half-lives, it's not
energetically possible for it to undergo beta decay. Why don't go in the one s half
orbital? And the problem is it's already filled too. Even though this is energetically
possible because not allowed because of Pauli exclusion principle. It's said the reason
these beta decays in complicated nuclei don't happen is because they're Pauli blocked.
That means the Pauli exclusion principle, if that state is already full, the beta decay
is
not allowed.
Just to show you how it works where it is allowed. Here is boron 12. Still 12
neurons but now I have five protons and seven neutrons. I do the same drawling. The
seventh neutron goes into the one p half orbital. Since I volume five protons to start
with there's an opening, a vacancy in the one p3 half-lives s orbital this is
energetically allowed and not Pauli blocked. In a that case beta decay is allowed. Okay.
So it's interesting the role the Pauli exclusion principle plays, it's preventing
thing from hang. Preventing decays from a happening. That's why some people talk about
the Pauli principle as being jet another force and it's manifesting it self in that way
here.
Okay. We talked early Oron about this idea, in certain kind of radioactive decay,
particles already exist and are moving around. So in the discussion of alpha decay we
imagination that we have an alpha particle running inside a heavy nucleus and tunneling
through a barrier to get out. That's not the situation in beta decay. I think when
talked about the Heisenberg uncertainty principle I went through a calculation that showed
that if you actually confine a electron to the volume of an atomic nucleus, by the
uncertainly principle it would have 10 or 20 MeV kinetic energy. And yet the beta
particles that generally come out through beta decay have one MeV. That can't be the
case, can't be that I have a electron running around inside the nucleus because it would
have to have much higher energy and then the only way we think of both beta decay and when
we talk about gamma decay, is that the radiations which emerge from the nucleus are
created in the decay process. In the case of beta decay the way we think of it is I
have
a neutron. It doesn't go directly in a proton, a electron and neutrino, it goes through
an intermediate step. Where the neutron turns into a proton. And what is produced is the
carrier of weak interaction. In this case it's a W minus boson. In the case of positron
emission it's a w plus boson. For each force there's a particle which is being
exchanged between the particles participating in the force in the case of gravity it's the
graviton, in the case of electric magnetic timple it's the photon. In the case of beta
decay which is a weak interaction, it's the ws and the related particle called the z
which were being exchange. If you go and look in particle data group booklets or look
online for what the mass of the w is, it turns out it's about 80 GeV. Eighty billion
electron volt. We're talking about beta decays that have energy on the order of one MeV.
So this is not energetically allowed. And the way it happens is through one, a virtual
processes I was talking about earlier, the neutrons steals energy from the universe for
a
little bit of time. And produces a w. But it has to pay the bill back quickly
according to the uncertainty principle. The way it does it is w decay in a electron and
anti-neutrino. When all is said and done energy momentum and everything are conserved and
you end up with electrons or positron that can have a few MeV but not 80 GeV. Because
he
don't have that much energy available in the decay.
MeV.
What does beta decay do? We looked at mass parabolas seven week ago when discussion
the semi-empirical mass formula. What I said was if you have an odd amount of you'll have
a bunch of different isotopes that have different protons numbers and hence, different
neutron numbers but all sum up in this case to a equal 125. And nuclei which are
located over on the left-hand side of the parabola have more neutrons than the ones
over
on the right-hand side. And nature always trys to move to the lowest energy state. The
way it can move in mass parabolas is through beta decays. So what's happening here on
this side is neutrons are turning into protons. Emitting electrons, and so I'm increasing
the number of protons every time I come down here. And the neutrons number is decreasing.
And through a sequence of beta decays I can ends up at the bottom of this mass parabola
with the one stable isotope, one isotope which is stable for that given a, which turns
out to be tellurium 125 much similarly if offer on the right-hand side these nuclei
have
too many protons for the number of neutrons. They can increase their proton number, sore
are decrease their proton number by positron emission and/or electron capture. And so
along here decreasing the z, increasing n and lowering total energy of system. When
you've got an even-A situation, remember there are two different mass parabolas because
of that pairing term in the nuclear binding energy formula. There's one parabola, lower
energy parabola which is associated with those nuclei that have both even number of
protons and even number of neutrons. And then there's an upper parabola which is for the
odd-odds. I have odd number of protons plus odd number of neutrons that still sum
newspaper this case to A equals 130. In that case thing get more interesting because if
I start over here, with this guy, he can beta decay down to here. And he can beta decay
down to here. And he can beta decay down to here. These are albeit minus decays. This
point is not the true momentum for this particular mass value. This a. The true
minimum is here. But this nucleus can't undergo beta decay minus decay because the scalar
product heavier in mass. So this isotope occurs in nature, as well as this one. If you
look on the chart of nuclides they're both listed as stable. In fact, this one isn't.
But it can't undergo beat minus decay because that would require energy. What can happen
is somehow undergo two beta decays simultaneously it can get down here and lower the total
energy. And that's a process known as double beta decay. And it's a very rare and very
slow process. A couple dozen nuclei for which this is possible. This is a little more
detail on mass 130. Here's the tellurium 130. There's the xenon 130 that would be the
lowest mass state for A equals 130. But the single beta decay would lead to iodine 130
and that's energetically not allowed. So the two beta decay would get you down there in
two neutrons inside the nucleus basically simultaneously decay into two protons, two
electrons, two proton anti-neutrino. And the proton stay inside the nucleus. So the
tellurium is transformed into xenon. This process has been observed in the laboratory.
It's half-life is about 10 to the 21st years. Remember the age of universe is 10 to the
tenth years so this is enormously long compared to that. It's been seen a couple of
different ways. The first way it was inferred to have happened was using mass
spectrometry and geo chemistry. I talked about this back in the first week of class. And
the idea here is you get an ore who's age you somehow know. Use radioactive dating,
rock you know is about a billion years old. Use potassium argon dating or uranium lead.
And you know you have a very old rock. Hope my it's a rock that in case has some
tellurium in it. You heat the rock up and evolve any gases which are trapped in it.
Take
the gases and run through a mass spectrometer. This is mass number of the xenon isotopes
which come out of that rock. These little horizontal lines are what the normal
atmospheric abundance of xenon is. If you took the air which is in this room, turns
out
there's a little bit of xenon there. if you ran it through mass spectrometer you get
these values. That's, but the one which is a high, is xenon 130. And so the
interpretation of this data is that element all the xenon 130 which is trapped if the
rock
was produced by the double beta decay of tellurium 130. But knowing how many tellurium in
the rock and how much xenon came out and knowing the age of rock the half-life is inferred
to be two times 10 to the 21st years. There are a lot of uncertainties associated with
that. How well do I know that the rock held onto the xenon produced there? How do I
really know it was produced by tellurium 130 and the honest answer is you don't. You'd
like to do a direct measurement in the laboratory. That was first done by Michael Moe at
us U.C. Irvine, these are some data. This is a cloud chamber. So what he had were a
bunch of strips of material. These horizontal lines are little strips. In this case it
was not tellurium, but it was selenium 82. Another one of these nuclei that is
energetically allowed to undergo double beta decay but not single beta decay. And again
it's a cloud chamber. He's got a magnetic field. In the cloud chamber so the charge
particles will produce tracks and tracks will be curved. Depending upon the direction of
curvature you can determine whether it's positively or negatively charged particle. What
he saw, enough to be able to measure the half-life is an event like this. You have one
point here where you see a particle coming out like that. and a particle coming out like
that. Two particles that have the same sense of curvature and hence, both being
identified as negatively charged electrons. Emerging from a common point inside the
chamber. That isn't necessarily proof you've seen double beta decay because there are
other ways you can get two electrons coming out more or less simultaneously. There's
another image where one of background events is shown. You have an electron coming out
here and another electron coming out there. you've also go heavy straight line coming
out.
Turns out that's a decay where there was a single beta decay followed by an alpha. And
this very strong track here is alpha particle. Those are backgrounds you have to
distinguished. Moe and his collaborators your able to study two or three different
nuclei. They were the people who get the credit for having observed double beta decay in
the laboratory for the first time. It's an interesting curiosity about the way mass
parabolas work.
Going to spend a little bit of time going through more formal discussion of beta
decay. And the reason we do this is to show you that in fact, we can do it. It has been
done. And the only really important point we're going to get out of this is to try to
understand why beta spectra have the shapes they do. Remember if I had alpha decay, what
I end up with is energy spectra that look like this. (on board). Whereas in beta decay
if I make the same kind of plot, q alpha, (on board), and what I try to argue the other
day was it has to do with energy sharing. In the case of the alpha decay, you've got
some
parent z and a and it's zeeking into D minus two and A minus four plus the helium
four. It's a two body decay. These two particles have to share the energy and that means
the alpha particle comes out with a discrete energy. So this is alpha. Whereas in beta
decay, let's say it's beat minus, so this goes to z plus one and A plus an electron
plus an electron anti-neutrino. And now they're actually three particles in the final
state. Not just two. So these three particles can share the energy. Not in a totally
arbitrary way which is had a I want to show today. Because I've got three particles, I
can ends with with a situation where occasionally the electron ends up with essentially in
energy or it ends up with whole energy of decay or anything in between. And our friend
Fermi figured this out back in the 1930s. So he was a very bright guy. And we called
this the term a theory the beta decay in his honor because he was the first person to
figure out. It's complicated and I'm nothing going to require you to understand the
details of this. I'm going to spend the revert of class going through T because it makes
use of some of the ideas of quantum mechanics we've already talked about. Non-the details
of this are going to be on any homework or exam.
Okay. So what Fermi did was go back it his famous golden rule which allows you to
cleat the rate for any process, whether it be radioactive decay or nuclear reaction.
I
wrote this did you know several week AGS in the quantum mechanics discussion at that says
the rate for a particular transition is something constants, two pi or h bar time
something we call the me. ≤ times something even more measures at the yours, the
densities of states. vif we can calculate in principle if we know the interaction
responsible for the particular transition we're talking about. In this case going from
some initial state to some final state. I'd like to be able to evaluate at that integral.
This is the operator that actually transforms the initial state to the final state. There
are several pieces to that. One is the intrinsic strength of interaction. There's some
somebody which is a measure of how strong the particular interaction is. Strong
interaction is the strongest, oftentimes the strength of that interaction is characterized
as being one in some dimensionless units. The electric magnetic interaction is next
strongest and in the same dimensionless units it would be approximately one over 137, the
fine structure constant. g is measure the weak interaction. And there would be a
similar constant for the gravitational interaction and that coupling constant would be
about 10 to the minus 40 the that's the relative scales of those interactions. What's
written here is the initial W that would be the wave function. Initial nucleus
starting W ends up with three particles. Beta decay. I've got the final nucleus I've
got an electron and a neutrino. This rho v is measure of how many distinct quantum
mechanical states are available as a function of energy of the decay. Turns out the
spectrum of beta decay is largely determined by this density of states. How many final
state are actually available depending pong how the energy the decay gets shared between
the electron, neutrino and resulting nucleus. If I think about it, I've argued that
electron and neutrino are not in the nucleus to start with. They got produced by beta
decay and come flying out. Once they're out of the nucleus to first order their free
particles at least we're going to treat them that way initially. The wave function of a
free particle can be written in this form. This solve the Schrˆdinger equation if free
particle. This v is meant to allow you to normalize the wave function. When I do psi
star psi, and integrate over the appropriate region of spacey get one. The way this is
stated is you imagine the beta decay occurs inside a box of volume V. When all is said
and done the volume of box doesn't matter. It goes away. This is win that way it
constant.
Okay. So now let's think about the kinematics a little more. Tip DALY beta decay
has we said has about an MeV of energy. What that means is the kinetic energy of electron
on average is greater than its rest mass energy. Therefore, we have to use relativistic
kinematics to describe the electrons which means I have to write down the energy this
way.
As the square of p≤ V≤ (refer to Powerpoint 11). I can calculate the kinetic energy
using at that formula. If I take the kinetic energy to MeV then the total energy of that
electron is the sum of its kinetic plus rest mass energy. Using this formula I can
calculate the momentum of that electron is. And it turns out to be 1.4 MeV over c.
Then I want to calculate what is the argument of that exponential. Remember I've try it
figure out what these things are. I want to make life as simple as I can. So I want to
calculate p over h bar, and I play usual tricks because I don't remember h bar.
Multi-numerator and denominator by c because pc just going to be 1.4 MeV. h bar
c is one of those numbers either, 197 MeV fentometers. p over h bar works out to
be.does he 07 inverse fentometers, a small number which is good for our purpose. What
I
want to do now is to expands this exponential. In the tailor series, e to the
ip.r over h bar is one this, (refer to Powerpoint 11). Plus a bunch of other terms
which is this squared or cubedded or. What I tried to argue is that if I have some normal
nucleus, who's radius is a few tent fentomenters p times r is going to be a few time
this number, it's a very small number. There have been I can basically neglect T and say
these wave functions are just one. That make life easier when I keep going on. That's
because the nucleus is small. and these energies we're talk are quite small.
Okay. So now consider the electron and neutrino produced by the beta decay confined
in this box of volume V. I have to calculate the number of final state available this
seems mysterious. But the way I can think of it if you good back to the uncertainty
principle, remember, we said delta p delta x is on the order of h bar. You can
think of that meaning that had I have a quantum mechanical state, it takes up a certain
volume if a space which is made up of both momentum and position vectors. It's call
configuration space. So a given quantum mechanics at state will take up a volume of
roughly h bar cubed given bit uncertainty principle. How many final states there are
depends on how big a volume in this configuration space I've got. This is the way you
calculate it. If you have a particle who's momentum is p, and you ask how many states
there are within a little region dp an that. This is the number, four pi p≤, that's
the first is area of sphere who's radius is P time the synthesisness of shell dp
times the volume, that's the position information divided by h cubed. Do I this for the
electron and separately for the neutrino because they both have their own phase spaces
available. I go back and rewrite a differential rate for decay as a function of final
energy in this complicate complicated way. Another constant related to the strength of
reaction. This element squared times this space face factor. This me. Is some number
which will depend on the particular decay we're talking about. And we're not going to
worry about the details of that. It will be large for some decays, small for other. But
to first order it will be independent of energy. So it's outside the integral. (refer to
Powerpoint 12).
There are some approximations we can make because public health information we
already next the neutrino while not being exactly massless has a type pass compared
to
that the electron. I can neglect that and say the neutrino mass is in these colleaguess
can be taken to be zero. That means the relativistic another momentum relationship take
on the extreme value that the momentum of the neutrinos times c is what we call the
energy the neutrino. The final energy of the decay which is the sum of energies of the
electron and neutrino are just the energy of the electron plus q time cq being the
moment of neutron. Because of way the formula is written as a dependents, I have to
evaluate the derivative of the neutron momentum with respect to the total final synergy
energy. It's just one over the spied of light. You put it all together. You find that
that mass we had on the previous page n of pdp can be written as some constant
which depends on the decay we're talking about times the product of the momentum of the
electron≤ times the momentum neutrino≤ dp. And the momentum of neutrino is Q-value
for the decay minus whatever energy the electron happens it carry away divided by the
speed of light. And so we rewriting that qb minus the kinetic energy of electron
divided by c. And so the number of electrons function of momentum looks like this.
It's a mess. (refer to Powerpoint 13). We're going to rewrite in tether of energy. Now
rewrite the number of electrons as a function of kinetic energy. And you finally get
this. So it's some constant divided by the speed much light to the fifth power time the
kinetic energy of the electron≤ plus two kinetic energy times the rest mass energy
all of
that to the half power. Times q beta minus the kinetic energy of electron≤ times eight
kinetic energy of (refer to Powerpoint 14). And you say so what? The reason it's
important it note the limiting values this takes on. This thing goes to zero for two
extreme cases. If the kinetic energy of the electron is zero, this first term is DOE.
And the whole thing is then zero. What that means is there there be no electrons emitted
with zero kinetic energy. That's a good thing. On the other hand if the kinetic energy
of the electron happens to be equal to Q-value the middle term zero and then the whole
thing is zero. It goes to zero here. In between it look like that. Qualitatively this
thing has the right shape it explain had a we see experimentally. Namely, that you get
a
continuous distribution that goes from kinetic energy being zero to the kinetic energy
being the full decay available in the decay. That was Fermi's contribution. It's really
a question of how energy gets shared will you have three particles like this in the
final
state. What's very important is the fact one of these particles namely the residual
nucleus that's produced is way more massive than the other two. It's interesting if you
look at a case where that isn't true, where the masses are all similar. And the decay
spectrum looks very different in this case. But for nuclei, this is the way beta decays
go. Okay.
Now, in that derivation which I waved my handy realize, I sewed that the electron and
positron and neutrino that come out are free particles. That's not really true if you
think about it. Suppose I'm a electron, a beta minus coming out of the nucleus, the
nucleus has a positive charge and the electron has a negative charge. Just because it's a
beta decay doesn't mean there's no electric magnetic interaction between the particles.
What we've done is to neglect the coulomb interaction between the beta particle and
the
nucleus that's left behind. You can think of it as the electrons are attracted to the
nucleus. Whereas the positrons were expelled away. You can correct for this because we
know how to calculate the interaction. So the spectrum we just draw over there get
modified by something called the Fermi function. And this depends on z of the daughter
nucleus because that's going to influence the strength of interaction and also it depends
on the kinetic energy of electron that emitted. The other things was it say the energy
spectra doesn't depends on nuclear matrix element. That's not exactly true either. One
can accommodate that. The way it's done is modify the formula for the shape of beta
spectrum. So there's this phase space factor which I mentioned earlier. We can include
the coulomb interactions that's called the Fermi function. Then also include the energy
dependence of matrix element. Energy dependence of matrix element. The way this gets
written to write it down quantity q beta minus te and say it's proportional to the
means later. if you take this funny quantity here,≤ of mpf squared over f and plot
that, you measure how many electrons you see as a function of their momentum or energy,
and plot that versus the kinetic energy, turns out what you see is the data will fall on
a
straight line if it's an allowed decay. Okay. In other words if this term over here is
one, I can neglect it. If it's in the ray straight line we call it ray for by endecay.
By measuring the deviations of this shape from a straight line you can figure out how
forbidden it is.
The total decay rate you can get from that differential decay rate will integrate
over the entire spectrum and because of this Fermi function now which depends on the
energy and hence, the momentum you have to integrate that thing over all possible momenta.
You end up with the term a integral. These are not thing you're going to calculate, if
you needed to use them, you would like them up. There are tables of these. But I don't
think you'll need this. For a given daughter nucleus z and given energy willing decay,
people have done these integrals and you can just look them up and read off the value.
So this tells you the decay rate, remember the relationship between the decay rate
and half-life, the decay rate lambda is log of two over the half-life. And so you can
calculate what is called the comparative half-life, the half-life times this f value,
can be written in this very simple way. At least it hides a lot. It looks simple, it
isn't. So this half-life times this integrated Fermi function is log of two times two pi
cubed h bar to the seven. (refer to Powerpoint 17). All the dirt is hidden in here.
This is the matrix element squared.
The bigger the mai terrorism almost the small the half-life will be and vice versa.
And turns out there are literally thousands of atomic nuclei that have been studied
in terms of beta decay. There's a lot of data on what ft values are. Because of that
enormous range of half-lives we talked about the other day from milliseconds to 10 to the
20th years, it's typically written down as the log of the ft values. Take the log 10
so the range doesn't like so big. And for all the decays that have been studied, the
ones
which have the small. log ft values are those that occur between states of zero plus
initial and zero plus final spins.
And experimentally turns out the range of log ft values for these decays is very
narrow. All in the range of three to four. And those are what are called super allowed
decays. The faster beta decays we know of. An example of that is shown here. (refer to
Powerpoint 18). It's the positron decay of carbon 10 into boron 10. The ground state
of
carbon 10 like all even-even nuclei is zero plus. Turns out the ground state of boron
10
is not zero plus. It's three plus. And so the decay does not go to this level. It
actually goes to another zero plus excited state. It's one of these super allowed states.
Now, this seem a little funny to talk about, allowed and forbidden decays, if you
think about these words allowed is something that can happen and forbidden is something
that's not allowed to happen in which beta decay, speak if you like, it suspect exactly
that way. What we mean is for an allowed decay it will happen at a fast rate. A forbid
and decay will still happen but all the much slower rate. Beta decays what we mean by an
allowed decay will turn out to represent a decay where the electron and the neutrino
which
are emitted carry away no orbital angular momentum. Remember L is orbital angular
momentum. The electron and neutrino are each fermions. They each have intrinsic angular
momentum of half h bar. In addition to that in principle they can also have orbital
angular momentum. For allowed decays they don't have. What that means is, if I have
a
transition that is known to be an allowed decay, whatever change in total spin between
the
initial and finite nucleus must come from the intrinsic spins of the leptons I'med.
The
electron and neutrino. And there are two possible relatively options of the spins of
electron and neutrino that are emitted in beta decay. They can be in the opposite
direction. One is spinning up and one is spin down. That's known as Fermi decay. And so
the intrinsic spins add up to zero. Where we're assuming there's no orbital angular
momentum involved in this decay. Therefore, there can't be any change P spin between and
initial and final nucleus states because the leptons that can away, get emitted carry away
no intrinsic in that spin and no orbital angular momentum. So if you have a transition
where the spin change is zero and the change in parity is no, because the parity always
going as minus one to the L. That's what we call an allowed Fermi transition. So the
transition I was mentioning a second ago, this one, is an allowed Fermi transition.
On the other hand it is possible that the spins of electron and neutrino happen to
point in the same direction in which case those add to up to one. This is northern as
Gamow-Teller decay. If I look at the initial and final state of nucleus, I start out with
nucleus of a given angular momentum momentum and parity. Ends up with a final nucleus of
spin and parity. But I've emitted a particle, actually a pair of particulate that carry
away one unit of angular momentum remember this is a vector. So I have to add one to
whatever the initial spin of nucleus was vectorially, and what that means is I can ends up
with a spin of the finite nucleus which is either the same as the spin of the nucleus
or
it can different by one unit. Again there won't be any parity change because there's
no
orbital angular momentum carried away. This is known as a Gamow-Teller decay. So I have
a decay zero plus to one plus, something like that (on board), this is an allowed
Gamow-Teller decay because I have one change of spin, no change of parity. So delta I
is one. Delta pi is no. So this is allowed but it has to be the Gamow-Teller variable.
It has to be this change of spin is the result of the lepton spins happening to point in
the same direction.
Forbidden decays. Like I said, aren't really forbidden, what makes them different is
the leptons carry away nonzero amounts of angular momentum. Nonzero amounts of orbital
angular momentum. The reason they're less likely to occur has to do with difficulty
of
radiation caring away orbital angular momentum given what the energies we have have
evolved and the size of region from which they're being emitted. So remember when talking
about alpha decay I made the same argument. Doesn't go through the math. Here I'm doing
to show you how difficult it is for beta or neutrino to carry away much in the order of
orbital angular momentum. Again taking a typical beta decay energy of a MeV, the momentum
associated with one MeV electron, has I showed a minute ago is 1.4 MeV over C. Let's
assume this electron is being emitted from a medium weight nucleus. So it's radius given
the r is zero a to the one their formula is something like six fentometers. What is
L? Again I do this semi-classically because I don't know how to do it any other way, I
imagine I've got this nucleus the of radius R, the six fentometers. Imagine emiting
electron from the surface, it goes away with some momentum PE. And L maximum it could
be is r cross p which is the way I've drawn it would be r time p. Son numerically
for this dais L would be 8.4 MeV fentometers divided by c. If you convert into units
of h bar it's .0 4-H bathroom that's small compared to one. If you think about it how
in the world could this happen? One way to picture it is the electron isn't really
emitted from here. It's emitted from way out here. Why is this a lot less likely. The
probability of finds I a neutron out there is not exactly zero but it ain't very high
either. It's difficult for radiation to carry much angular momentum. There are the
probable for it to happen is going to be lower than if the L value required happens to
be zero, namely the value for the allowed decay.
Okay. So selection rules, just to reiterate, for allowed decays the orbital angular
momentum carried away by the front and neutrino is zero whatever spin difference therein
anti-parallel which leads to no change in nuclear spin and no change many parity. The
other variety for allowed decays is where the neutrino and electron spins are parallel.
I
can get spin change of zero or within and no parity change. What's going on here is
in
the Fermi decay turn a neuron in a proton, because there's no spin change, whatever
nuclear state the neutron was in, the proton ends up being in the same nuclear state
without any possibility of flipping spin. In this case I can take a neutron in a given
state and turn into a proton who's spin is in the opposite direction of where the -- we'll
stop for today and on Friday going into more detail about forbidden decays and what
sequences they have. Okay. (refer to Powerpoint) 21.