182 Binding Energy

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The more tightly bound a system is, the stronger the forces that hold it together and the greater the energy required to pull it apart. We can therefore learn about nuclear forces by examining how tightly bound the nuclei are. We define the binding energy (BE) of a nucleus to be the energy required to completely disassemble it into separate protons and neutrons. We can determine the BE of a nucleus from its rest mass. The two are connected through Einstein’s famous relationship \(E=\left(\Delta m\right){c}^{2}\). A bound system has a smaller mass than its separate constituents; the more tightly the nucleons are bound together, the smaller the mass of the nucleus.

Imagine pulling a nuclide apart as illustrated in (Figure). Work done to overcome the nuclear forces holding the nucleus together puts energy into the system. By definition, the energy input equals the binding energy BE. The pieces are at rest when separated, and so the energy put into them increases their total rest mass compared with what it was when they were glued together as a nucleus. That mass increase is thus \(\text{Δ}m=\text{BE}/{c}^{2}\). This difference in mass is known as mass defect. It implies that the mass of the nucleus is less than the sum of the masses of its constituent protons and neutrons. A nuclide \({}^{A}\text{X}\) has \(Z\) protons and \(N\) neutrons, so that the difference in mass is

Thus,

where \({m}_{\text{tot}}\) is the mass of the nuclide \({}^{A}\text{X}\), \({m}_{p}\) is the mass of a proton, and \({m}_{n}\) is the mass of a neutron. Traditionally, we deal with the masses of neutral atoms. To get atomic masses into the last equation, we first add \(Z\) electrons to \({m}_{\text{tot}}\), which gives \(m\left({}^{A}\text{X}\right)\), the atomic mass of the nuclide. We then add \(Z\) electrons to the \(Z\) protons, which gives \(\text{Zm}\left({}^{1}\text{H}\right)\), or \(Z\) times the mass of a hydrogen atom. Thus the binding energy of a nuclide \({}^{A}\text{X}\) is

The atomic masses can be found in Appendix A, most conveniently expressed in unified atomic mass units u (\(1\phantom{\rule{0.25em}{0ex}}\text{u}=\text{931}\text{.}5\phantom{\rule{0.25em}{0ex}}\text{MeV}/{c}^{2}\)). BE is thus calculated from known atomic masses.

What patterns and insights are gained from an examination of the binding energy of various nuclides? First, we find that BE is approximately proportional to the number of nucleons \(A\) in any nucleus. About twice as much energy is needed to pull apart a nucleus like \({}^{\text{24}}\text{Mg}\) compared with pulling apart \({}^{\text{12}}\text{C}\), for example. To help us look at other effects, we divide BE by \(A\) and consider the binding energy per nucleon, \(\text{BE}/A\). The graph of \(\text{BE}/A\) in (Figure) reveals some very interesting aspects of nuclei. We see that the binding energy per nucleon averages about 8 MeV, but is lower for both the lightest and heaviest nuclei. This overall trend, in which nuclei with \(A\) equal to about 60 have the greatest \(\text{BE}/A\) and are thus the most tightly bound, is due to the combined characteristics of the attractive nuclear forces and the repulsive Coulomb force. It is especially important to note two things—the strong nuclear force is about 100 times stronger than the Coulomb force, and the nuclear forces are shorter in range compared to the Coulomb force. So, for low-mass nuclei, the nuclear attraction dominates and each added nucleon forms bonds with all others, causing progressively heavier nuclei to have progressively greater values of \(\text{BE}/A\). This continues up to \(A\approx \text{60}\), roughly corresponding to the mass number of iron. Beyond that, new nucleons added to a nucleus will be too far from some others to feel their nuclear attraction. Added protons, however, feel the repulsion of all other protons, since the Coulomb force is longer in range. Coulomb repulsion grows for progressively heavier nuclei, but nuclear attraction remains about the same, and so \(\text{BE}/A\) becomes smaller. This is why stable nuclei heavier than \(A\approx \text{40}\) have more neutrons than protons. Coulomb repulsion is reduced by having more neutrons to keep the protons farther apart (see (Figure)).

There are some noticeable spikes on the \(\text{BE}/A\) graph, which represent particularly tightly bound nuclei. These spikes reveal further details of nuclear forces, such as confirming that closed-shell nuclei (those with magic numbers of protons or neutrons or both) are more tightly bound. The spikes also indicate that some nuclei with even numbers for \(Z\) and \(N\), and with \(Z=N\), are exceptionally tightly bound. This finding can be correlated with some of the cosmic abundances of the elements. The most common elements in the universe, as determined by observations of atomic spectra from outer space, are hydrogen, followed by \({}^{4}\text{He}\), with much smaller amounts of \({}^{\text{12}}\text{C}\) and other elements. It should be noted that the heavier elements are created in supernova explosions, while the lighter ones are produced by nuclear fusion during the normal life cycles of stars, as will be discussed in subsequent chapters. The most common elements have the most tightly bound nuclei. It is also no accident that one of the most tightly bound light nuclei is \({}^{4}\text{He}\), emitted in \(\alpha \) decay.

There is more to be learned from nuclear binding energies. The general trend in \(\text{BE}/A\) is fundamental to energy production in stars, and to fusion and fission energy sources on Earth, for example. This is one of the applications of nuclear physics covered in Medical Applications of Nuclear Physics. The abundance of elements on Earth, in stars, and in the universe as a whole is related to the binding energy of nuclei and has implications for the continued expansion of the universe.