Because of the strong nuclear force, the nucleons in a stable nucleus are held tightly together. Therefore, energy is required to separate a stable nucleus into its constituent protons and neutrons, as Figure 31-3 illustrates. The more stable the nucleus is, the greater is the amount of energy needed to break it apart. The required energy is called the binding energy of the nucleus.

Figure 31-3   Energy, called the binding energy, must be supplied to break the nucleus apart into its constituent protons and neutrons. Each of the separated nucleons is at rest and out of the range of the forces of the other nucleons.

Two ideas that we have studied previously come into play as we discuss the binding energy of a nucleus. These are the rest energy of an object (Section 28.6) and mass (Section 4.2). In Einstein’s theory of special relativity, energy and mass are equivalent; in fact, the rest energy and the mass m are related via (Equation 28.5), where c is the speed of light in a vacuum. Therefore, a change in the rest energy of the system is equivalent to a change in the mass of the system, according to . We see, then, that the binding energy used in Figure 31-3 to disassemble the nucleus appears as extra mass of the separated and stationary nucleons. In other words, the sum of the individual masses of the separated protons and neutrons is greater by an amount than the mass of the stable nucleus. The difference in mass is known as the mass defect of the nucleus.

As Example 2 shows, the binding energy of a nucleus can be determined from the mass defect according to Equation 31.3:

	(31.3)

Example 2   |   The Binding Energy of the Helium Nucleus

The most abundant isotope of helium has a nucleus whose mass is . For this nucleus, find (a) the mass defect and (b) the binding energy. Reasoning The symbol indicates that the helium nucleus contains protons and neutrons. To obtain the mass defect , we first determine the sum of the individual masses of the separated protons and neutrons. Then we subtract from this sum the mass of the nucleus. Finally, we use Equation 31.3 to calculate the binding energy from the value for . Solution a.   Using data from Table 31-1, we find that the sum of the individual masses of the nucleons is This value is greater than the mass of the intact nucleus, and the mass defect is b.   According to Equation 31.3, the binding energy is Usually, binding energies are expressed in energy units of electron volts instead of joules : In this result, one million electron volts is denoted by the unit MeV. The value of 28.3 MeV is more than two million times greater than the energy required to remove an orbital electron from an atom.

In calculations such as that in Example 2, it is customary to use the atomic mass unit (u) instead of the kilogram. As introduced in Section 14.1, the atomic mass unit is one-twelfth of the mass of a atom of carbon. In terms of this unit, the mass of a atom is exactly 12 u. Table 31-1 also gives the masses of the electron, the proton, and the neutron in atomic mass units. For future calculations, the energy equivalent of one atomic mass unit can be determined by observing that the mass of a proton is or 1.0073 u, so that and In electron volts, therefore, one atomic mass unit is equivalent to

Data tables for isotopes, such as that in Appendix F, give masses in atomic mass units. Typically, however, the given masses are not nuclear masses. They are atomic masses—that is, the masses of neutral atoms, including the mass of the orbital electrons. Example 3 deals again with th nucleus and shows how to take into account the effect of the orbital electrons when using such data to determine binding energies.

Example 3   |   The Binding Energy of the Helium Nucleus, Revisited

The atomic mass of is 4.0026 u, and the atomic mass of is 1.0078 u. Using atomic mass units instead of kilograms, obtain the binding energy of the nucleus. Reasoning To determine the binding energy, we calculate the mass defect in atomic mass units and then use the fact that one atomic mass unit is equivalent to 931.5 MeV of energy. The mass of 4.0026 u for includes the mass of the two electrons in the neutral helium atom. To calculate the mass defect, we must subtract 4.0026 u from the sum of the individual masses of the nucleons, including the mass of the electrons. As Figure 31-4 illustrates, the electron mass will be included if the masses of two hydrogen atoms are used in the calculation instead of the masses of two protons. The mass of a hydrogen atom is given in Table 31-1 as 1.0078 u, and the mass of a neutron as 1.0087 u. Figure 31-4   Data tables usually give the mass of the neutral atom (including the orbital electrons) rather than the mass of the nucleus. When data from such tables are used to determine the mass defect of a nucleus, the mass of the orbital electrons must be taken into account, as this drawing illustrates for the isotope of helium. See Example 3. Solution The sum of the individual masses is The mass defect is . Since 1 u is equivalent to 931.5 MeV, the binding energy is , which matches that obtained in Example 2.

To see how the nuclear binding energy varies from nucleus to nucleus, it is necessary to compare the binding energy for each nucleus on a per-nucleon basis. The graph in Figure 31-5 shows a plot in which the binding energy divided by the nucleon number A is plotted against the nucleon number itself. In the graph, the peak for the isotope of helium indicates that the nucleus is particularly stable. The binding energy per nucleon increases rapidly for nuclei with small masses and reaches a maximum of approximately 8.7 MeV/nucleon for a nucleon number of about . For greater nucleon numbers, the binding energy per nucleon decreases gradually. Eventually, the binding energy per nucleon decreases enough so there is insufficient binding energy to hold the nucleus together. Nuclei more massive than the nucleus of bismuth are unstable and hence radioactive.

Figure 31-5   A plot of binding energy per nucleon versus the nucleon number A.

Check Your Understanding 2

The following table gives values for the mass defect for four hypothetical nuclei: A, B, C, and D. Which statement is true regarding the stability of these nuclei? (a) Nucleus D is the most stable, and A is the least stable. (b) Nucleus C is stable, whereas A, B, and D are not stable. (c) Nucleus A is the most stable, and D is not stable. (d) Nuclei A and B are stable, but B is more stable than A. (The answer is given at the end of the book.)   A B C D Mass defect, 0 kg Background: The mass defect of a nucleus is the sum of the individual masses of the separated nucleons minus the mass of the intact nucleus. The binding energy of a nucleus is the energy required to separate it into its constituent protons and neutrons. The greater the binding energy, the more stable is the nucleus. The binding energy and the mass defect are related. For similar questions (including calculational counterparts), consult Self-Assessment Test 31.1, which is described at the end of Section 31.5.