Download This lecture deals with atomic and nuclear structure.

This lecture deals with atomic and nuclear structure.
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We will begin the lecture by describing the atom using the Bohr model.
Next, we will look at electronic shells and their quantum numbers.
We will, in particular, be concerned with binding energies of the various electronic
shells. A large part of this course addresses energy transfer, and binding energies
are very important when looking at energy transfer.
We will then look at the production of characteristic x-rays and Auger electrons.
The study of characteristic x-rays and Auger electrons will also give us a better
understanding of how energy can be transferred in interactions.
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We will then turn to the nucleus, describe its structure, and calculate nuclear binding
energies.
We will finish off the lecture by looking at factors that affect the stability of nuclei.
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We will start by looking at a model of the atom. I think that for most of you this is a review. Most, if not all, of
you have had a serious course in quantum mechanics. You are probably wondering why do we take a couple of
steps back and look at a model of the atom that dates from 1913 rather than using either the Schrödinger
equation, Dirac equation, or quantum electrodynamics.
The reason for using the Bohr model is that we are going to try to use as simple a model as possible that still
explains the phenomena that we observe. Certain phenomena will require us to use some more involved
quantum mechanics in order to explain. However, we are not going to go through heavy quantum mechanical
calculations; you’ve outgrown that stage in your lives, I hope. Maybe you might revisit it if you go into some
really detailed radiation interactions. What I really want to do is give you an intuitive feel for what goes on in an
atom when interactions occur. And the simplest way to do that is with a very simple model and that is the Bohr
model.
As all of you remember, the Bohr model is a planetary model consisting of a nucleus and electrons. The nucleus
contains a positive charge and most of the mass of the atom. Its diameter is about 10-12 centimeters.
The nucleus is surrounded by the electrons that revolve around the nucleus in well-defined orbits. That’s why
we call it a planetary model.
The number of protons and electrons in an atom is equal, consequently, the net charge of an atom is going to be
zero.
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Here we have diagrams of a carbon atom and a silicon atom based on the Bohr
model. We see that the carbon atom has 6 positive charges in its nucleus surrounded
by 6 electrons. Two of these electrons go into one shell called the K shell and four
of them go into the outer shell called the L shell. Silicon has 14 positive charges in
the nucleus. Electrons occupy the K, L, and M shells, three different shells.
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One of the observations made when the Bohr model was first developed was that it
violated classical physics.
The model has electrons in these orbits that are repelling each other. Consequently,
an atom ought to be unstable. Moreover, an electron in a circular orbit is constantly
changing its velocity. Whenever a charge is changing its velocity it gives off
radiation. Consequently, an electron should be losing energy while it revolves
around the nucleus and it eventually should spiral in to the nucleus and the atom
therefore is not stable.
I remember very distinctly in my sophomore college physics course, a course in
electricity and magnetism, the last problem set of the course was to try to calculate
the lifetime of a typical atom, assuming classical physics.
Did any of you have to do that in your undergraduate physics course? Did you
remember what the lifetime of an atom is in classical physics? I think it is
something like 10-4 seconds.
The whole idea of this problem set was to prepare us for the quantum mechanics
which we were going to get that fall, our junior year.
Anyway, we know that atoms last a lot longer than 10-4 sec.
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So, in order to explain why atoms have lifetimes greater than fractions of a second,
Niels Bohr came up with two postulates.
First, he said that electrons revolve in specified orbits with fixed radii. Second, he
said that electrons gain or lose energy when they jump from one orbit to the other.
Basically Bohr had the gall to say that classical physics was wrong and he was right
and I guess it turned out that he was right, because his model of the atom really
explains a lot of the phenomena we see in the atom.
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When Bohr started doing calculations using his theory, he labelled these orbits with
quantum numbers. He then calculated that the gain or loss of energy as an electron
went from one orbit to another was given by this equation. Sure enough, this
equation really modeled spectroscopic observations. This equation did a pretty
good job of quantitatively predicting how atoms gained or lost energy. That was a
good sign as far as explaining atomic behavior. We are going to say that the Bohr
atom is an adequate model for describing atomic phenomena we see in radiological
physics.
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So let’s look at the components of the Bohr atom.
First, we have electrons. These electrons have charges. Each electron has a charge
of -1.6  10-19 Coulomb. The electron has a mass, which is very small, 9.1  10-31
kilograms. We will use the charge of the electron as a practical unit of charge and
worry about this 1.6  10-19 factor a little bit later, so we will say the electron charge
is -1.
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Now when we look at Bohr’s theory, we find the theory gives us specified orbits for
the electrons. Bohr used the letter n to identify the orbit. n is called the principal
quantum number and we know that n can have integral values, 1, 2, 3, 4, etc. Once
we have the value of n we can calculate the radius of the orbit and the energy of the
orbit. So n can be used to identify the shell of an atom. We say that the electrons
are in shells indicated by the quantum number n.
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The way we fill electron shells is to put electrons to go into the lowest total energy
state. A general principle of physics states that any system tries to achieve the
lowest total energy state. You are well aware of this. When your alarm clocks goes
off in the morning you would rather be in a horizontal position, which is a lower
energy state than getting up out of bed. You need to expend some energy to make
an 8 am class. Glad you could make it to your class.
So what is the lowest energy state in an atom? Zero energy is the electron infinitely
far from the nucleus. As the electron approaches the nucleus of the atom, it sees an
attractive force resulting from the positive charge of the nucleus. Remember that
unlike charges attract. Because the force attracts the electron, as it goes into an
orbit it has negative energy.
The lowest energy is the innermost orbit. One would think, then, that initially all
the electrons would fill that innermost orbit. We find that’s not the case, as we are
going to see a little bit later.
It turns out that if we are given the index n of a shell, the maximum number of
electrons in a shell is given by 2n2.
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That means that the innermost shell, at n = 1, which is called the K shell, has 2
times 1 squared or 2 electrons. The next shell n = 2, which is called the L shell, has
2 squared or 4 times 2 or 8 electrons. For n = 3, 2 times 3 squared is 18 electrons,
and so forth.
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There are also energy sub-levels in each shell.
We use 3 additional quantum numbers to describe the electrons.
We have an azimuthal quantum number, which is a measure of the eccentricity of
the orbit, that is, how elongated the orbit is. This value, denoted as ℓ, can be any
value from 0, 1, 2, 3, up to n-1. The value of ℓ = 0 implies a circular orbit. A value
of ℓ equal to 1 indicates a slightly stretched out orbit. For ℓ equal to 2, the orbit is
even more elongated.
Then there is a magnetic quantum number m that tells us what orientation the
electron of the orbit takes when a magnetic field is applied. The quantum number m
can have integral values between –ℓ and +ℓ.
Finally, the spin quantum number s represents an intrinsic electronic property. Spin
can have a value of ±½.
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So electrons can have all these different quantum numbers. The basic idea of the
Pauli Exclusion Principle is that no two electrons can have the same set of quantum
numbers. If you accept the Pauli Exclusion Principle and start assigning quantum
numbers to electrons, you wind up with the result that for a given value of n you can
have no more than 2n2 electrons in a shell. You can work out the math if you have
time.
So because of the existence of quantum numbers and the Pauli Exclusion Principle
we have certain electronic shells that the electrons go into.
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An important quantity is the binding energy, which is the energy required to remove
the electron from whatever shell it’s in completely out of the atom to bring it out
infinitely far from the nucleus. We normally identify the binding energy to be a
negative number, which means that energy must be supplied to the electron to
overcome the attractive force of the nucleus and get it out to infinite distance from
the nucleus.
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Let’s look at some binding energies. Here are some values. For the different shells, we show binding energies
for both hydrogen and tungsten.
There are two very important trends you need to look at. One is that as you get further and further away from
the nucleus, as you go from an inner shell to an outer shell, the magnitude of the binding energies decreases.
That means that it takes less energy to remove an electron from an outer shell than it does from an inner shell.
That makes sense.
In an outer shell, you’re further away from the nucleus. Remember that this attractive force is a 1 over R2 force,
so if you’re further away from the nucleus the attractive force is less, so less energy is required to overcome this
force. A second point to note is that if you’re further away from the nucleus, the inner shell electrons can
partially shield the nucleus. Consequently, the effective charge that an outer shell electron sees is less than what
an inner shell electron sees. So again less energy is required to pull an outer shell electron from the nucleus.
Notice also the difference in binding energies for a small nucleus such as hydrogen, with one proton, versus a
big nucleus such as tungsten, with 74 protons. It takes a lot more energy to pull an electron from a tungsten
atom than it does from a hydrogen atom.
Notice that typical binding energies for inner shell tungsten electrons are on the order of kiloelectron volts. You
don’t have to memorize these numbers, but I think it is important that you get a feeling for the order of
magnitude of these numbers. Recognize that for hydrogen, binding energies are on the order of electron volts;
for tungsten inner shell electrons, binding energies are on the order of kiloelectron volts.
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Let’s summarize what we know about binding energies.
The binding energy is a measure of the attractive force between the nucleus and the
orbital electron. It is a measure of the energy required to remove an electron from
an atom. The smaller the value of n, that is, the closer the electron is to the nucleus,
the stronger the attractive force from the nucleus. The closer the electron is to the
nucleus, the less the shielding of the nuclear charge by other electrons. As a
consequence the magnitude of the binding energy is greater for smaller values of n.
It therefore requires more energy to remove an electron from the n = 1 shell than it
does from the n = 2 shell.
The greater the atomic number of the nucleus, the greater the number of protons in
the nucleus; the stronger the attractive force from the nucleus, hence, the greater the
magnitude of the binding energy. It requires more energy to remove an electron
from the n = 1 shell of tungsten than it does from the n = 1 shell of hydrogen.
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Let’s say we have some kind of interaction. Radiation interacts with an electron, in
particular, with an inner shell electron, and the incident radiation has enough energy
to overcome the binding energy of that inner shell electron. If the radiation can
transfer enough energy to the electron to overcome the binding energy of that
electron, it’s going to pull that electron out and remove it from the atom.
For example, suppose we have an incident photon with 80 kiloelectron volts of
energy incident on tungsten. Remember that it only takes 69.5 kiloelectron volts to
overcome the binding energy of that electron. The photon can then transfer its
energy to the inner-shell electron, overcoming the binding energy, ejecting the
electron.
What’s left behind is a vacancy, an empty space, a place where an electron could
otherwise be. We call this vacancy a hole.
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Now because there is a vacancy in the inner shell, an outer shell electron can fall into the hole and fill the vacancy without
violating the Pauli Exclusion Principle.
The outer shell electron goes from a higher energy state to a lower energy state. In going from a higher energy state to a lower
energy state it gives off energy. The energy given off by this electron going from a higher energy state, an outer shell, into an
inner shell vacancy, is called characteristic radiation. This energy is characteristic of the atom and the shells. So we can talk
about L to K characteristic radiation in tungsten. The energy of this characteristic radiation is simply the difference in binding
energies between the L shell and the K shell. Any of you remember what the L shell binding energy is, -11 kiloelectron volts.
The K shell binding energy is -69.5 kiloelectron volts. So the difference in energies is roughly about 58 kiloelectron volts, and
that is the energy of the characteristic radiation coming from a tungsten atom.
So you come into a tungsten atom with a high-energy photon or an electron, that inner shell electron is going to be knocked
out. An outer shell electron, say an L shell electron, can fall into the K shell giving off a characteristic X-ray of energy about
58 kiloelectron volts.
In addition, an M shell electron can fall into a K shell vacancy. No reason why it can’t. The M shell electron has a binding
energy of about 3 kiloelectron volts so that the characteristic radiation is 69 minus 3 or 66 kiloelectron volts of energy.
We can have an M shell electron fall into the L shell vacancy. We previously had an L shell electron fall into the K shell
vacancy and we had a hole cascading so we can have about another 9 kiloelectron volt characteristic X-ray.
A lot of things can happen once you’ve created a hole.
The other thing that can happen with the 58 kiloelectron volts of energy is that because the energy is greater than the binding
energies of an L shell electron or an M shell electron, there’s a finite probability that the energy can be used to ionize an L shell
electron or an M shell electron, and so forth. Those electrons are called Auger electrons and the process is called the Auger
effect. What we observe is a single event in which a photon comes in and ionizes an inner shell electron resulting in a whole
cascade of events. We can have characteristic X-rays from various transitions taking place. We can have Auger electrons also
being ejected.
I think one of the problems on the problem set asks you to calculate what happens when an event occurs. If it is not in this
problem set it is in a subsequent problem set. But you actually get an opportunity to track all these events.
Remember that one of the mantras of this course is going to be follow the energy. And when an interaction takes place, we
will have to track the energy that goes on. We have to identify what the events are that can take place and how energy gets
transferred from incident photon to electron to ejected characteristic X-ray to ejected Auger electron.
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One more word about electrons and that has to do with valence electrons, which are
the electrons in the outermost shell. One of the rules for building up electron shells
is that we allow no more than 8 electrons in the outermost shell.
The valence electrons determine the chemical properties of elements, and the
presence of valence electrons explains why we have the periodic table. There is a
periodicity of similar chemical behavior—this material should all be like what you
learned in freshman chemistry. You have long since forgotten it but I am hoping to
jog your memory a little bit.
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Let’s move now from electrons to the nucleus, and focus now on what’s going on in
the nucleus.
To a first approximation, the nucleus is composed of two kinds of particles, protons
and neutrons. The generic term for protons and neutrons is nucleons.
The mass of both protons and neutrons is very similar, about 1.6 10-27 kilogram;
that’s about 2,000 times the mass of an electron. The charge in the proton is +1.6 
10-19 Coulomb. That’s the same charge as that of an electron, but positive rather
than negative. Thus the proton charge is the same as the electron charge but the
sign is opposite. The neutron has no charge.
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Several numbers are used to characterize the nucleus.
We’ll start with the mass number, abbreviated by the letter A. The mass number is
the number of nucleons in the nucleus; that is, the total number of protons plus the
total number of neutrons. The mass number gives some indication of the mass of
the nucleus.
The atomic number, denoted by the letter Z, is the number of protons in the nucleus
and the number of electrons in the neutral atom.
The neutron number, N, is the number of neutrons in the nucleus.
The mass number is equal to the atomic number plus the neutron number.
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If we want to express the mass more precisely, we talk about the actual mass of a
nucleus and use a unit called the atomic mass unit. The atomic mass unit is defined
as the mass of the carbon nucleus that has 6 protons and 6 neutrons.
I should say that some sources say mass of the carbon atom that has 6 protons and 6
neutrons and 6 electrons. I was looking at a few of the textbooks. Johns and
Cunningham say nucleus. Khan says atom. Hendee, Ibbott and Hendee say
nucleus. Various websites say nucleus while others say atom. So we can collect
different interpretations.
The fact of the matter is that the difference between the two definitions is the
difference of mass of electrons, which is on the order of 1/2000 the mass of the
atom. I think we will be concerned about that difference in only one situation later
on in the course.
So an atomic mass unit is 1.6605  10-27 kilograms. The mass of a proton is roughly
1 amu; the mass of a neutron is roughly 1 amu; and the mass of an electron is
roughly 0 amu.
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Let’s be more precise now. The mass of an electron is 0.00055 amu; the mass of the
proton is 1.00727 amu; and the mass of the neutron is 1.00866 amu.
Notice that the mass of the neutron is slightly greater than the mass of the proton.
In fact, when we look at some nuclear interactions, we are going to roughly say the
mass of a neutron is approximately the mass of a proton plus the mass of an
electron. Keep that in mind. We are going to be talking about that a later on in a
couple of lectures from now.
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What does a nucleus look like?
There are a couple of models of the nucleus. Niels Bohr had such a good model of the atom
and decided to design a model for the nucleus. His model was called the liquid drop model
and in his model of the nucleus was composed of closely-packed nucleons.
Maria Mayer came up with a little more sophisticated shell-type model more analogous to
Bohr’s model of the atom. She said there are discrete energy levels in the nucleus analogous
to the discrete energy levels in an atom. There is some experimental evidence of why there
might be energy levels in the nucleus. One of Mayer’s observations that validates a shell
model is that nuclei with atomic numbers of 2, 8, 20, 82, and 126 were exceptionally stable.
These numbers correspond to filled nuclear shells analogous to atoms for which certain
numbers of electrons corresponded to filled shells. Atoms with filled electronic shells are
non-reactive. In the shell model of the nucleus these particular values of the atomic number
suggest filled nuclear shells.
The other thing that we are going to see is that nuclei can give off energy because there are
energy levels. For a given nucleus, we have a ground state, the lowest energy state, as well
as excited states. This is analogous to ground states and excited states of atoms. We are
going to see nuclei going from excited states to lower energy states giving off radiation.
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Now, one of the measures of nuclear stability that we use is very important. That is
the n over p ratio, which is the ratio of neutrons to protons.
If we plot stable isotopes, we find they all lie along a line of stability. For low
atomic numbers, the number of protons and the number of neutrons for the most
stable nuclei are equal. So an oxygen nucleus with 8 protons and 8 neutrons is very
stable. If you either have too many protons or too many neutrons, you tend to have
a nucleus that is unstable.
Why this is important is that depending on whether we have an excess of neutrons
or an excess of protons will tell us what kind of process a radioactive nucleus will
undergo in trying to achieve stability. If we have too many neutrons, then we are
going to look at a process or processes that tend to decrease the number of neutrons
and/or increase the number of protons. Conversely, if we have too many protons,
we are going to be looking at processes that either increase the number of protons or
decrease the number of neutrons or both. So this line of stability is going to be very
important in telling us what kind of radioactive process will take place when a
radioactive nucleus decays. Keep this in mind for our radioactivity lectures down
the road.
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The other thing is that nuclei with even numbers of protons or neutrons tend to be
more stable that those with odd numbers. So the most stable nuclei have even
numbers of protons and even numbers of neutrons. If we have an odd number of
protons or an even number of neutrons we will have some stability, but not quite as
much stability. Notice how few stable nuclei have an odd number of protons and an
odd number of neutrons.
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So, let’s summarize the indicators of stability.
Indicators of stability include magic numbers that correspond to filled nuclear
shells, a line of stability for N=Z, for low Z, and somewhat greater than Z for higher
Z, and nuclei with even numbers of protons and/or even numbers of neutrons, being
more stable than those with odd numbers.
So by looking at a nucleus, we should be able to get some indication of how stable
this nucleus is.
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One problem that we have not addressed is how the nucleus stays together.
Protons are positively charged. Positive charges repel; therefore, protons should repel each other and nuclei
should not exist. Why, then, do stable nuclei exist?
It turns out there is another force called the “strong nuclear force,” which is approximately 100 times stronger
than the electrostatic force, but it only operates over short ranges. Call it “nuclear contact cement.”
Suppose you have two protons. To bring them together, you will experience a large repulsive force; but once
you overcome that repulsive force and let the two protons come into contact with each other, or come at very
close proximity to each other, the strong nuclear force will overcome the electrostatic force to hold the protons
together and cause the nucleus to become stable.
The particles of nuclear contact cement are called mesons. If you can blow apart a nucleus, you can also knock
some of these mesons out and you can produce negative pi mesons, or negative pions.
There was actually some interest about thirty years ago in using negative pions for radiation therapy. We will
understand why there was some interest in that in a subsequent lecture. The program never did work out, but
there are some very interesting stories about using them.
We have this short-range force. We can also calculate what kind of energy is incorporated in this short range
force to hold a nucleus together. Calculations of nuclear binding energies are very important.
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Let’s do such a calculation.
If we look at the mass of a nucleus and compare it to the mass of the individual
particles that make up the nucleus, we find the mass of the nucleus is less than the
sum of the masses of the individual particles. In other words, the whole is less than
the sum of its parts. This difference in mass is called the mass defect, and it
represents the energy that is used to hold the nucleus together.
We can take the mass defect and convert it to energy using Einstein’s formula, E =
mc2 – that’s one of the formulas I want you to remember in this course.
What we’ll do is convert the mass in atomic mass units to energy equivalents. This
is actually a number you ought to remember. One atomic mass unit is 931 million
electron volts of energy. You will be doing a lot of problems involving that so by
the end of this course you should be comfortable with that conversion. The other
conversion that you really ought to know is 1 electron mass. The mass of an
electron is 0.511 MeV. You will see that even more than 931 MeV equals an atomic
mass unit.
So, if we look at the mass of a nucleus and subtract it from the mass of the
component nucleons, we will find the mass defect. Convert that mass defect into
energy, and you will find the energy that is used to hold the nucleus together.
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Let’s look at some numbers.
Let’s take 6 protons and combine them with 6 neutrons to form a carbon nucleus.
The mass of 6 protons is 6 times 1.00727 atomic mass units. Multiply those two
numbers together and we get 6.04362 amu. We need to keep a lot of significant
figures because we are going to be doing subtraction.
Add to the protons six neutrons. The mass of 6 neutrons is 6 times 1.00866 amu or
6.05196 amu.
The total of the masses of these 12 nucleons turns out to be 12.09558 amu.
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In this calculation, amu are relative to the mass of a carbon-12 nucleus. So the mass
of a carbon-12 nucleus is 12.00000 amu. The difference in mass of 0.09558 amu is
called the mass defect and is the energy that is used to hold this carbon-12 nucleus
together.
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Let’s convert this mass into energy units. Multiply it by 931 MeV per amu and we
find that the binding energy of the carbon-12 nucleus is 89.0 million electron volts.
More important than that is the binding energy per nucleon. If we have 89 million
electron volts holding 12 nucleons in place, that’s about 7½ MeV per nucleon. I am
not interested if you memorize 7½ MeV per nucleon, but you should at least have a
feeling for the order of magnitude of this number because we will see this later. If
we can pump in about 7½ MeV into a carbon-12 nucleus we will have enough
energy to overcome the binding energy of that carbon-12 nucleus, and we can eject
a nucleon. So any interaction involving incident radiation with energy greater than
7½ MeV will be sufficient to cause photonuclear disintegration and knock out a
nucleon from a carbon-12 nucleus.
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Let me digress slightly by pointing out that saying that the binding energy per
nucleon is the energy required to remove a nucleon from the nucleus is somewhat of
an approximation.
If we want to be more precise, we can define a quantity, the neutron separation
energy, as the energy required to remove one neutron from the nucleus with atomic
number Z and neutron number N. The neutron separation energy is the difference
between the binding energy of a nucleus with atomic number Z, neutron number N,
and the binding energy of a nucleus with atomic number Z, neutron number N-1.
In an analogous manner, we can define the proton separation energy as the energy
required to remove one proton from the nucleus (Z,N). The proton separation energy
is the difference between the binding energy of a nucleus with atomic number Z,
neutron number N, and the binding energy of a nucleus with atomic number Z-1,
neutron number N.
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If we plot binding energy per nucleon versus mass number, we see that for very
small nucleons, the binding energy per nucleon is very small, say 1, 2, 3 MeV. The
binding energy per nucleon reaches a maximum value of around 9 MeV in the range
of mass numbers of 60 – 80 and then it gradually decreases.
What does this mean? This means that nucleons with large mass numbers tend to
undergo interactions that will increase the binding energy per nucleon or
interactions that will cause smaller nucleons to take place. In other words, we are
likely to observe nuclear fission. So nuclear fission is an interaction that will occur
for large mass number nucleons.
Nuclei with small mass numbers will undergo interactions that will increase the
binding energy per nucleon, or interactions that will cause larger nucleons to take
place, that is, nuclear fusion. Again, we want to try to get to a mass number of 60
thru 80, which has the largest binding energy per nucleon.
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So what is nuclear fission?
If we take a nucleon with high mass number, such as uranium-235, and have it
absorb a slow neutron, it will split to give rise to smaller nucleons, for example,
krypton plus barium plus three new neutrons plus energy. The amount of energy
released is approximately 200 million electron volts per fission. Very important is
this energy release because we can use this energy for various applications.
Also notice that one neutron goes in and three neutrons come out.
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These neutrons that are produced can cause more fission to take place. In fact we
can have what’s called a chain reaction. If we have an uncontrolled chain reaction,
that is a euphemism for an atomic bomb. If we control the chain reaction by
absorbing some of these neutrons and not allowing them to cause other fissions, we
can use the energy in a nuclear reactor.
This chain is controlled by absorbing some of the neutrons produced in fission. If
we have a small amount of the nuclear fissionable material, most of the neutrons
will come out of the surface and not be involved in additional fissions. However, if
we look at the surface to volume ratio, the number of neutrons produced is going to
be related to the volume. If the surface to volume ratio gets too small, we will
produce more neutrons than escape, and we will have what’s called a critical mass
of the fissionable material.
When I took my graduate nuclear physics course many, many years ago, when we
learned about nuclear fission, one of the problems was to calculate the critical mass
of uranium-235. So once you know how many neutrons are coming out, what the
density is, presumably you can calculate the critical mass. I remember on the
problem set solutions the instructor basically set up the problem and said “I can’t
solve the problem for you, because that’s still classified information. But all of you
are capable of solving that problem.” So that’s how you determine a critical mass.
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I want to conclude today’s lecture with just some nomenclature. This is some
nomenclature we use in various amounts in this course. We talk a lot about
isotopes. Isotopes have the same atomic number but different mass numbers -- same
number of protons, same number of electrons, but different numbers of neutrons.
For example, let’s take carbon. One isotope of carbon has 6 protons and 6 neutrons,
while another isotope has 6 protons and 8 neutrons. Thus we have two isotopes,
carbon-12 and carbon-14. They are both isotopes of carbon.
You could have the same number of neutrons but different numbers of protons, and
they are called isotones. An example of two isotones is hydrogen-3 and helium-4.
Hydrogen-3 has 1 proton and 2 neutrons whereas helium-4 has 2 protons and 2
neutrons -- so they are isotones.
You have isobars, which have the same mass number but different numbers of
protons and neutrons. For example, oxygen-18 has 8 protons, 10 neutrons, 18
nucleons; whereas fluorine-18 has 10 protons, 8 neutrons, 18 nucleons. You get to
see fluorine-18 haunting you when we talk about nuclear imaging because that’s the
isotope we use for a lot of our PET imaging.
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