Question

On a plot of the number of neutrons versus the number of protons, a nucleus of aluminum-28 lies above the belt of stability. Write an equation for the radioactive decay of aluminum-28.

Answer 1

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.

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.

Now come to our problem. a nucleus of aluminum-28 lies above the belt of stability is given this means n/p ratio > 1, hence it loses a beta particle as follows