Nuclear Instability

Instability for a nucleus arises when there is a modification for it that has a higher binding energy. When there are alternate modifications the mode of change is the one which leads to the higher binding energy.

Binding Energy

There are two components of the binding energy for a nucleus. One is the loss in potential energy when its component protons and neutrons are brought together. The other is the energy equivalent of its mass deficit. Its mass deficit is the difference between the sum of the masses of its components and its measured mass. This later component is available for 2931 nuclides. The total binding energy is available for only one nuclide, the deuteron. There is a a good chance that there is a strong enough correlation between total binding energy and mass deficit binding energy that the latter will serve as a suitable proxy for the former. In the rest of this study binding energy will stand for mass deficit binding energy used as a proxy for total binding energy.

Modes of Nuclear Decay

The following beautiful display from Wikipedia shows the nature of their instabilities.

As can be seen from the display, up to proton number 82 overwhelming the mode of decay is either the ejection of an electron or a positron. The conversion of a neutron into a proton and an electron releases energy. The positron ejection accompanies the conversion of a proton into a neutron, but this conversion requires an input of energy.

Above proton number 82 the dominant, but not exclusive, mode of decay is the ejection of an alpha particle.

In the following, the proton number and the neutron number will be denoted by p and n, respectively.

Examples

What is plotted in the following graph are the binding energies for a complete sequence of decay products as a function of the number of neutrons in the decay product.

The binding energy reaches a maximum at p=56 and n=81. This is Ba₁₃₇ which is a stable isotope of Barium.

Consider the isotope of Neodymium which is an element of the above beta decay sequence. Its numbers are p=60 and n=77. Its decay mode is the ejection of a positron. Its binding energy is 1138.34 million electron volts (MeV). Its decay product with p=61 and n=78 has a binding energy of 1142.82 MeV. On the other hand if it were to decay by the ejection of an alpha particle the binding energies of it decay products, which are the nuclide with p=58 and n=75 and an alpha particle, are 1110.5 MeV and 28.295674 MeV, respectively, for a total of 1138.795674 MeV. This make its decay by alpha emission inferior energywise to beta decay.

Now consider the isotope of Uranium (p=92) with n=150. It has a binding energy of 1822.67 MeV. If it were to decay by the ejection of an electron the binding energy of its decay product would be 1823.09 MeV. For alpha decay its decay products would have binding energies of 1798.2 and 28.295674 MeV for total of 1826.495674 MeV. Thus its preferred mode of decay is the ejection of an alpha particle.

Conclusion

From the limited sample it appears that the mode of decay of unstable nuclides can be explained by comparing the binding energies of the alternate decay products.