21.1 Nuclear Structure and Stability

Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of ${}_1^1\text{H},$ neutrons. Recall that the number of protons in the nucleus is called the atomic number (Z) of the element, and the sum of the number of protons and the number of neutrons is the mass number (A). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation ${}_{\text{Z}}^{\text{A}}\text{X},$ where X is the symbol for the element, A is the mass number, and Z is the atomic number (for example, ${}_6^{14}\text{C}).$ Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, ${}_6^{14}\text{C}$ is called “carbon-14.”

Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about 10⁻¹⁵ meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10⁻¹⁰ meters. Nuclei are extremely dense compared to bulk matter, averaging 1.8 $\times$ 10¹⁴ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm³. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately 6.4 $\times$ 10⁶ meters, 30,000 times larger). Example 21.1 demonstrates just how great nuclear densities can be in the natural world.

To hold positively charged protons together in the very small volume of a nucleus requires very strong attractive forces because the positively charged protons repel one another strongly at such short distances. The force of attraction that holds the nucleus together is the strong nuclear force. (The strong force is one of the four fundamental forces that are known to exist. The others are the electromagnetic force, the gravitational force, and the nuclear weak force.) This force acts between protons, between neutrons, and between protons and neutrons. It is very different from the electrostatic force that holds negatively charged electrons around a positively charged nucleus (the attraction between opposite charges). Over distances less than 10⁻¹⁵ meters and within the nucleus, the strong nuclear force is much stronger than electrostatic repulsions between protons; over larger distances and outside the nucleus, it is essentially nonexistent.

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Nuclear Stability

A nucleus is stable if it cannot be transformed into another configuration without adding energy from the outside. Of the thousands of nuclides that exist, about 250 are stable. A plot of the number of neutrons versus the number of protons for stable nuclei reveals that the stable isotopes fall into a narrow band. This region is known as the band of stability (also called the belt, zone, or valley of stability). The straight line in Figure 21.2 represents nuclei that have a 1:1 ratio of protons to neutrons (n:p ratio). Note that the lighter stable nuclei, in general, have equal numbers of protons and neutrons. For example, nitrogen-14 has seven protons and seven neutrons. Heavier stable nuclei, however, have increasingly more neutrons than protons. For example: iron-56 has 30 neutrons and 26 protons, an n:p ratio of 1.15, whereas the stable nuclide lead-207 has 125 neutrons and 82 protons, an n:p ratio equal to 1.52. This is because larger nuclei have more proton-proton repulsions, and require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together.

Figure 21.2 This plot shows the nuclides that are known to exist and those that are stable. The stable nuclides are indicated in blue, and the unstable nuclides are indicated in green. Note that all isotopes of elements with atomic numbers greater than 83 are unstable. The solid line is the line where n = Z.

The nuclei that are to the left or to the right of the band of stability are unstable and exhibit radioactivity. They change spontaneously (decay) into other nuclei that are either in, or closer to, the band of stability. These nuclear decay reactions convert one unstable isotope (or radioisotope) into another, more stable, isotope. We will discuss the nature and products of this radioactive decay in subsequent sections of this chapter.

Several observations may be made regarding the relationship between the stability of a nucleus and its structure. Nuclei with even numbers of protons, neutrons, or both are more likely to be stable (see Table 21.1). Nuclei with certain numbers of nucleons, known as magic numbers, are stable against nuclear decay. These numbers of protons or neutrons (2, 8, 20, 28, 50, 82, and 126) make complete shells in the nucleus. These are similar in concept to the stable electron shells observed for the noble gases. Nuclei that have magic numbers of both protons and neutrons, such as ${}_2^4\text{He},$ ${}_8^{16}\text{O},$ ${}_{20}^{40}\text{Ca},$ and ${}_{82}^{208}\text{Pb},$ are called “double magic” and are particularly stable. These trends in nuclear stability may be rationalized by considering a quantum mechanical model of nuclear energy states analogous to that used to describe electronic states earlier in this textbook. The details of this model are beyond the scope of this chapter.

Stable Nuclear Isotopes
Number of Stable Isotopes	Proton Number	Neutron Number
157	even	even
53	even	odd
50	odd	even
5	odd	odd

Table 21.1

The relative stability of a nucleus is correlated with its binding energy per nucleon, the total binding energy for the nucleus divided by the number or nucleons in the nucleus. For instance, we saw in Example 21.2 that the binding energy for a ${}_2^4\text{He}$ nucleus is 28.4 MeV. The binding energy per nucleon for a ${}_2^4\text{He}$ nucleus is therefore:

$$\frac{\text{28.4 MeV}}{\text{4 nucleons}}=\text{7.10 MeV/nucleon}$$

21.9

In Example 21.3, we learn how to calculate the binding energy per nucleon of a nuclide on the curve shown in Figure 21.3.

Figure 21.3 The binding energy per nucleon is largest for nuclides with mass number of approximately 56.

Example 21.3

Calculation of Binding Energy per Nucleon

The iron nuclide ${}_{26}^{56}\text{Fe}$ lies near the top of the binding energy curve (Figure 21.3) and is one of the most stable nuclides. What is the binding energy per nucleon (in MeV) for the nuclide ${}_{26}^{56}\text{Fe}$ (atomic mass of 55.9349 amu)?

Solution

As in Example 21.2, we first determine the mass defect of the nuclide, which is the difference between the mass of 26 protons, 30 neutrons, and 26 electrons, and the observed mass of an ${}_{26}^{56}\text{Fe}$ atom:

$$\begin{array}{l}\text{Mass defect}=\left[\left(26\times \text{1.0073 amu}\right)+\left(30\times \text{1.0087 amu}\right)+\left(26\times \text{0.00055 amu}\right)\right]-\text{55.9349 amu}\\ =\text{56.4651 amu}-\text{55.9349 amu}\\ =\text{0.5302 amu}\end{array}$$

21.10

We next calculate the binding energy for one nucleus from the mass defect using the mass-energy equivalence equation:

$$\begin{array}{}\\ \\ E=m{c}^2=\text{0.5302 amu}\times \frac{1.6605\times {10}^{\mathrm{-27}}\text{kg}}{\text{1 amu}}\times {\left(2.998\times {10}^8\text{m/s}\right)}^2\\ =7.913\times {10}^{\mathrm{-11}}\text{kg}\text{⋅}{\text{m/s}}^2\\ =7.913\times {10}^{\mathrm{-11}}\text{J}\end{array}$$

21.11

We then convert the binding energy in joules per nucleus into units of MeV per nuclide:

$$7.913\times {10}^{\mathrm{-11}}\text{J}\times \frac{\text{1 MeV}}{1.602\times {10}^{\mathrm{-13}}\text{J}}=\text{493.9 MeV}$$

21.12

Finally, we determine the binding energy per nucleon by dividing the total nuclear binding energy by the number of nucleons in the atom:

$$\text{Binding energy per nucleon}=\frac{\text{493.9 MeV}}{56}=\text{8.820 MeV/nucleon}$$

21.13

Note that this is almost 25% larger than the binding energy per nucleon for ${}_2^4\text{He}.$

(Note also that this is the same process as in Example 21.1, but with the additional step of dividing the total nuclear binding energy by the number of nucleons.)

Check Your Learning

What is the binding energy per nucleon in ${}_9^{19}\text{F}$ (atomic mass, 18.9984 amu)?

Answer:

7.810 MeV/nucleon