Electrostatic energy in nuclei
We will now take up another example of electrostatic energy in atomic physics, the electrical energy of atomic nuclei. Before we do this we will have to discuss some properties of the main forces (called nuclear forces) that hold the protons and neutrons together in a nucleus. In the early days of the discovery of nuclei—and of the neutrons and protons that make them up—it was hoped that the law of the strong, nonelectrical part of the force between, say, a proton and another proton would have some simple law, like the inverse square law of electricity. For once one had determined this law of force, and the corresponding ones between a proton and a neutron, and a neutron and a neutron, it would be possible to describe theoretically the complete behavior of these particles in nuclei. Therefore a big program was started for the study of the scattering of protons, in the hope of finding the law of force between them; but after thirty years of effort, nothing simple has emerged. A considerable knowledge of the force between proton and proton has been accumulated, but we find that the force is as complicated as it can possibly be. What we mean by “as complicated as it can be” is that the force depends on as many things as it possibly can. First, the force is not a simple function of the distance between the two protons.  At large distances there is an attraction, but at closer distances there is a repulsion. The distance dependence is a complicated function, still imperfectly known. Second, the force depends on the orientation of the protons’ spin. The protons have a spin, and any two interacting protons may be spinning with their angular momenta in the same direction or in opposite directions. And the force is different when the spins are parallel from what it is when they are antiparallel, as in (a) and (b) of Fig. 8–6. The difference is quite large; it is not a small effect.  Third, the force is considerably different when the separation of the two protons is in the direction parallel to their spins, as in (c) and (d) of Fig. 8–6, than it is when the separation is in a direction perpendicular to the spins, as in (a) and (b). Fourth, the force depends, as it does in magnetism, on the velocity of the protons, only much more strongly than in magnetism. And this velocity-dependent force is not a relativistic effect; it is strong even at speeds much less than the speed of light. Furthermore, this part of the force depends on other things besides the magnitude of the velocity. For instance, when a proton is moving near another proton, the force is different when the orbital motion has the same direction of rotation as the spin, as in (e) of Fig. 8–6, than when it has the opposite direction of rotation, as in (f). This is called the “spin orbit” part of the force.  The force between a proton and a neutron and between a neutron and a neutron are also equally complicated. To this day we do not know the machinery behind these forces—that is to say, any simple way of understanding them. There is, however, one important way in which the nucleon forces are simpler than they could be. That is that the nuclear force between two neutrons is the same as the force between a proton and a neutron, which is the same as the force between two protons! If, in any nuclear situation, we replace a proton by a neutron (or vice versa), the nuclear interactions are not changed. The “fundamental reason” for this equality is not known, but it is an example of an important principle that can be extended also to the interaction laws of other strongly interacting particles—such as the $\pi$-mesons and the “strange” particles.  This fact is nicely illustrated by the locations of the energy levels in similar nuclei. Consider a nucleus like B$^{11}$ (boron-eleven), which is composed of five protons and six neutrons. In the nucleus the eleven particles interact with one another in a most complicated dance. Now, there is one configuration of all the possible interactions which has the lowest possible energy; this is the normal state of the nucleus, and is called the ground state. If the nucleus is disturbed (for example, by being struck by a high-energy proton or other particle) it can be put into any number of other configurations, called excited states, each of which will have a characteristic energy that is higher than that of the ground state. In nuclear physics research, such as is carried on with Van de Graaff generators (for example, in Caltech’s Kellogg and Sloan Laboratories), the energies and other properties of these excited states are determined by experiment. The energies of the fifteen lowest known excited states of B$^{11}$ are shown in a one-dimensional graph on the left half of Fig. 8–7. The lowest horizontal line represents the ground state.  The first excited state has an energy $2.14$ MeV higher than the ground state, the next an energy $4.46$ MeV higher than the ground state, and so on. The study of nuclear physics attempts to find an explanation for this rather complicated pattern of energies; there is as yet, however, no complete general theory of such nuclear energy levels.  If we replace one of the neutrons in B$^{11}$ with a proton, we have the nucleus of an isotope of carbon, C$^{11}$. The energies of the lowest sixteen excited states of C$^{11}$ have also been measured; they are shown in the right half of Fig. 8–7. (The broken lines indicate levels for which the experimental information is questionable.) Looking at Fig. 8–7, we see a striking similarity between the pattern of the energy levels in the two nuclei. The first excited states are about $2$ MeV above the ground states. There is a large gap of about $2.3$ MeV to the second excited state, then a small jump of only $0.5$ MeV to the third level. Again, between the fourth and fifth levels, a big jump; but between the fifth and sixth a tiny separation of the order of $0.1$ MeV. And so on. After about the tenth level, the correspondence seems to become lost, but can still be seen if the levels are labeled with their other defining characteristics—for instance, their angular momentum and what they do to lose their extra energy. The striking similarity of the pattern of the energy levels of B$^{11}$ and C$^{11}$ is surely not just a coincidence. It must reveal some physical law. It shows, in fact, that even in the complicated situation in a nucleus, replacing a neutron by a proton makes very little change. This can mean only that the neutron-neutron and proton-proton forces must be nearly identical. Only then would we expect the nuclear configurations with five protons and six neutrons to be the same as with six protons and five neutrons. Notice that the properties of these two nuclei tell us nothing about the neutron-proton force; there are the same number of neutron-proton combinations in both nuclei. But if we compare two other nuclei, such as C$^{14}$, which has six protons and eight neutrons, with N$^{14}$, which has seven of each, we find a similar correspondence of energy levels. So we can conclude that the p-p, n-n, and p-n forces are identical in all their complexities. There is an unexpected principle in the laws of nuclear forces. Even though the force between each pair of nuclear particles is very complicated, the force between the three possible different pairs is the same. But there are some small differences. The levels do not correspond exactly; also, the ground state of C$^{11}$ has an absolute energy (its mass) which is higher than the ground state of B$^{11}$ by $1.982$ MeV. All the other levels are also higher in absolute energy by this same amount. So the forces are not exactly equal. But we know very well that the complete forces are not exactly equal; there is an electrical force between two protons because each has a positive charge, while between two neutrons there is no such electrical force. Can we perhaps explain the differences between B$^{11}$ and C$^{11}$ by the fact that the electrical interaction of the protons is different in the two cases?  Perhaps even the remaining minor differences in the levels are caused by electrical effects? Since the nuclear forces are so much stronger than the electrical force, electrical effects would have only a small perturbing effect on the energies of the levels. In order to check this idea, or rather to find out what the consequences of this idea are, we first consider the difference in the ground-state energies of the two nuclei. To take a very simple model, we suppose that the nuclei are spheres of radius $r$ (to be determined), containing $Z$ protons. If we consider that a nucleus is like a sphere with uniform charge density, we would expect the electrostatic energy (from Eq. 8.7) to be \begin{equation} \label{Eq:II:8:22} U=\frac{3}{5}\,\frac{(Zq_e)^2}{4\pi\epsO r}. \end{equation}  where $q_e$ is the elementary charge of the proton. Since $Z$ is five for B$^{11}$ and six for C$^{11}$, their electrostatic energies would be different. With such a small number of protons, however, Eq. (8.22) is not quite correct. If we compute the electrical energy between all pairs of protons, considered as points which we assume to be nearly uniformly distributed throughout the sphere, we find that in Eq. (8.22) the quantity $Z^2$ should be replaced by $Z(Z-1)$, so the energy is \begin{equation} \label{Eq:II:8:23} U=\frac{3}{5}\,\frac{Z(Z-1)q_e^2}{4\pi\epsO r}= \frac{3}{5}\,\frac{Z(Z-1)e^2}{r}. \end{equation}  If we knew the nuclear radius $r$, we could use (8.23) to find the electrostatic energy difference between B$^{11}$ and C$^{11}$. But let’s do the opposite; let’s instead use the observed energy difference to compute the radius, assuming that the energy difference is all electrostatic in origin. That is, however, not quite right. The energy difference of $1.982$ MeV between the ground states of B$^{11}$ and C$^{11}$ includes the rest energies—that is, the energy $mc^2$—of all the particles. In going from B$^{11}$ to C$^{11}$, we replace a neutron by a proton and an electron, which have less mass. So part of the energy difference is the difference in the rest energies of a neutron and that of a proton plus an electron, which is $0.784$ MeV. The difference, to be accounted for by electrostatic energy, is thus more than $1.982$ MeV; it is \begin{equation*} 1.982\text{ MeV}+0.784\text{ MeV}=2.766\text{ MeV}. \end{equation*}  Using this energy in Eq. (8.23), for the radius of either B$^{11}$ or C$^{11}$ we find \begin{equation} \label{Eq:II:8:24} r=3.12\times10^{-13}\text{ cm}. \end{equation}  Does this number have any meaning? To see whether it does, we should compare it with some other determination of the radius of these nuclei. For example, we can make another measurement of the radius of a nucleus by seeing how it scatters fast particles. From such measurements it has been found, in fact, that the density of matter in all nuclei is nearly the same, i.e., their volumes are proportional to the number of particles they contain. If we let $A$ be the number of protons and neutrons in a nucleus (a number very nearly proportional to its mass), it is found that its radius is given by \begin{equation} \label{Eq:II:8:25} r=A^{1/3}r_0, \end{equation}  where \begin{equation} \label{Eq:II:8:26} r_0=1.2\times10^{-13}\text{ cm}. \end{equation}  From these measurements we find that the radius of a B$^{11}$ (or a C$^{11}$) nucleus is expected to be \begin{equation*} r=(11)^{1/3}(1.2\times10^{-13})\text{ cm}= 2.7\times10^{-13}\text{ cm}. \end{equation*}  Comparing this result with (8.24), we see that our assumptions that the energy difference between B$^{11}$ and C$^{11}$ is electrostatic is fairly good; the discrepancy is only about $15\%$ (not bad for our first nuclear computation!). The reason for the discrepancy is probably the following. According to the current understanding of nuclei, an even number of nuclear particles—in the case of B$^{11}$, five neutrons together with five protons—makes a kind of core; when one more particle is added to this core, it revolves around on the outside to make a new spherical nucleus, rather than being absorbed. If this is so, we should have taken a different electrostatic energy for the additional proton. We should have taken the excess energy of C$^{11}$ over B$^{11}$ to be just \begin{equation*} \frac{Z_Bq_e^2}{4\pi\epsO r} \end{equation*}  which is the energy needed to add one more proton to the outside of the core. This number is just $5/6$ of what Eq. (8.23) predicts, so the new prediction for the radius is $5/6$ of (8.24), which is in much closer agreement with what is directly measured. We can draw two conclusions from this agreement. One is that the electrical laws appear to be working at dimensions as small as $10^{-13}$ cm. The other is that we have verified the remarkable coincidence that the nonelectrical part of the forces between proton and proton, neutron and neutron, and proton and neutron are all equal.