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This lecture deals with atomic and nuclear structure. 1 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. 2 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. 3 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. 4 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. 5 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. 6 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. 7 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. 8 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. 9 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. 10 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. 11 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. 12 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. 13 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. 14 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. 15 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. 16 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. 17 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. 18 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. 19 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. 20 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. 21 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. 22 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. 23 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. 24 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. 25 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. 26 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. 27 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. 28 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. 29 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. 30 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. 31 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. 32 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. 33 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. 34 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. 35 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. 36 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. 37 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. 38