Using the stability rules, rank the following isotopes in order of increasing nuclear stability.

Rank in order from 1 (least stable) to 3 (most stable).

Phosphorus-30 should be the least stable because it has odd numbers of both protons and neutrons. Calcium-39 has an even number of protons (20) and an odd number of neutrons. With an even number of protons and a "magic number" at that, it should be more stable than P-30.

The Belt of Stability. The principal factor for determining whether a nucleus is stable is the neutron to proton ratio. Figure 23.1 shows a plot of the number of neutrons versus the number of protons in various isotopes. Each dot represents a stable isotope. The stable nuclei are located in an area of the graph known as the belt of stability. In this figure we see that at low atomic numbers stable nuclei possess a neutron to proton ratio of about 1.0. Above Z = 20 the number of neutrons always exceeds the number of protons in stable isotopes. The n : p ratio increases to about 1.5 at the upper end of the belt of stability.

Figure 23.1 A graph showing all the stable isotopes when plotted as neutron number versus proton number. The shaded area represents the belt of stability. In order for isotopes with Z = 20 to be stable, the ratio of n : p must be about 1.1. For isotopes with Z = 40 to be stable, the ratio of n : p must be about 1.3 : 1.0. For isotopes with Z = 80 to be stable, the ratio of n : p must be about 1.5 : 1.0.

If you were given the symbol of a radioisotope, without any experience, it would be impossible to tell its mode of decay. But with knowledge of the belt of stability, you can make accurate predictions of the expected mode of decay.

Isotopes with too many neutrons lie above the belt of stability. The nuclei of these isotopes decay in such a way as to lower their n : p ratio. Thus one of its neutrons may decay into a proton and a beta particle.

The proton remains in the nucleus, and the beta particle is emitted from the atom. The loss of a neutron and the gain of a proton produces a new isotope with two important properties. It has a lower n : p ratio, and thus is more likely to be stable. Also, the daughter product has an atomic number that is one greater than the decaying isotope due to the additional proton. Consider the decay of carbon-14 for example. Carbon-14 is continually produced in the upper atmosphere by the interaction of cosmic rays with nitrogen. Carbon-14 has a higher n : p ratio than either of carbon's stable isotopes (C-12 and C-13), and decays by beta decay.

Note that the product isotope is one atomic number greater than carbon. Also, it is stable. Its n : p ratio is 1.0.

Isotopes with too many protons have a low n : p ratio and lie below the belt of stability. These isotopes tend to decay by positron emission because this process produces a new isotope with a higher n : p ratio. During positron emission a proton emits a positron, , and becomes a neutron.

The neutron remains in the nucleus, and the positron is ejected from the atom. Thus, a product nucleus will contain one less proton and one more neutron than the parent nucleus. The n : p ratio increases due to positron decay.

Electron capture accomplishes the same end, that is, a higher n : p ratio. Some nuclei decay by capturing an orbital electron of the atom.

Lanthanum-138, a naturally occurring isotope with an abundance of 0.089 percent, decays by electron capture.

Electron capture is accompanied by X-ray emission.

The only stable isotope of sodium is sodium-23. What type of radioactivity would you expect from sodium-25?

beta decay
positron emmission
electron capture

Sodium-23 has 11 protons and 12 neutrons and is in the belt of stability. Sodium-25 must have two more neutrons, and so has a higher n : p ratio than the stable isotope. Sodium-25 will decay by emission.

Nuclear Binding Energy. One of the important consequences of Einstein's theory of relativity was the discovery of the equivalence of mass and energy. The total energy content (E) of a system of mass m is given by Einstein's theory:

E = mc²

where c is the velocity of light (3.0 x 10⁸ m/s). Therefore, the mass of a nucleus is a direct measure of its energy content. It was discovered in the 1930s that the measured mass of a nucleus is always smaller than the sum of the separate masses of its constituent nucleons. This difference in mass is called the mass defect. When the mass defect is expressed as energy by applying Einstein's equation, it is called the binding energy of the nucleus. The binding energy is the energy required to break up a nucleus into its component protons and neutrons. The binding energy provides a quantitative measure of nuclear stability. The greater the binding energy the more stable the nucleus is toward decomposition.

In terms of the nucleus for instance, the binding energy (BE) is the energy required for the process

The mass defect m is equal to the total mass of the products minus the total mass of the reactants.

m = [8(proton mass) + 9(neutron mass)] – ( nuclear mass)

Here we can substitute atomic masses, which include the electrons, for the nuclear masses.

m = [8( atomic mass) + 9(neutron mass)] – ( atomic mass)

This works because the mass of 8 electrons in the 8 hydrogen atoms is canceled by the mass of the 8 electrons in the oxygen atom.

8( atomic mass) = 8(proton mass) + 8(electron mass)

( atom mass) = ( nuclear mass)+ 8(electron mass)]

Continuing by using atomic masses given in the text (Section 23.2) and in Table 23.3 below.

m = [8( atomic mass) + 9(neutron mass)] – ( atomic mass)

m = [8(1.007825 amu) + 9(1.008665 amu)] – (16.999131 amu) = 0.141454 amu

The eight protons and nine neutrons have more mass than the oxygen-17 nucleus. The binding energy is:

E = mc²

Calculate the nuclear binding energy of the light isotope of helium, helium-3. The atomic mass of is 3.01603 amu.
E = x 10^ J/nucleon

In comparing the stability of any two different nuclei, we must account for the different numbers of nucleons per nucleus. A satisfactory comparison of nuclear stabilities can be made by using the binding energy per nucleon, that is, the binding energy of each nucleus divided by the total number of nucleons in the nucleus. This is one of the most important properties of a nucleus. When the BE per nucleon is plotted as a function of the atomic mass we get the curve of binding energy as shown in Figure 23.2 of the text. Note that at first it rises rapidly with increasing atomic mass, and then it reaches a maximum at mass 56. Above mass 56, the binding energy drops slowly as atomic mass increases. Table 23.1 compares the total binding energy and the binding energy per nucleon for several isotopes.

Table 23.3 Binding Energies of Selected Isotopes

Complete the following questions to check your understanding of the material. Select the check button to see if you answered correctly.

1. For each pair of nuclei, predict which one is the more stable.
   1. 
      
   2. 
      
2. Fluorine has only one stable isotope, fluorine-19.
   1. The nucleus of fluorine-18 lies below the belt of stability. Write an equation for the decay of fluorine-18.
      
   2. The nucleus of fluorine-21 lies above the belt of stability. Write an equation for the decay of fluorine-21.
      
3. How many atomic mass units are in one kilogram?
   
4. Which has more mass?
   the nucleus of aluminum–27
   13 protons and 14 neutrons
5. What is the binding energy of ? Calculate its binding energy per nucleon. The atomic mass of aluminum-27 is 26.981541 amu.
   
6. How much energy is released when one Po-214 atom decays by alpha emission?
   
   Given the atomic masses: Pb-214 = 213.99519 amu, Po-210 = 209.98286 amu, and He-4 = 4.00260 amu.