Nuclear Stability and Neutron/Proton Ratio

It has been found that the stability of nucleus depends upon its neutron to proton (n/p) ratio. A plot of N (the neutron number) against Z (atomic number or number of protons) show that for stable nuclides upto Z = 20, N = 20 i.e., the relationship can be represented by a line with a slope of 45^o. Thus maximum stability is attained when N = Z.

fig 11.1 - A plot of number of neutrons (N) against the atomic number (Z) for a range of stable nuclei

From the plot it is evident that

(i) light nuclei (A < 20) have n/p ration close to unity.

(ii) for heavy nuclei (A > 20) the n/p ratio increases progressively due to dominance of number of neutrons. Thus stable nuclei (non-radioactive) have n/p ratio between 1 - 1.6. These lie in the shaded region of the plot which is also called stability belt or stability zone.

The elements whose nuclei do not fall within the stability zone are said to be unstable. The unstable nuclei, whose n/p ratio is either less than 1 or greater than 1.6, disintegrate giving out a, b, g rays in their attempt to attain stability. The process of disintegration continues till the n/p ratio falls within the stability limit.

To the right of the curve in the figure shown above, where the N/Z ratio is lower than that required for stability, a radioactive nuclide can decay by b+ emission or K - electron capture.

Example of positron emission

Example of K electron capture (this is not very common and occurs in case of nuclei with low n/p ratio and having insufficient energy for positron emission)

The daughter nucleus produced by b - emission or k - electron capture, has a ratio (N +1) / (Z - 1).

To the left of the curve, a radioactive nuclide would be neutron rich and would decay by b^- - emission to produce a daughter nucleus with a lower N/Z ratio of (N - 1) (Z + 1).

Example of b^- emission

When the value of Z becomes greater than 82, some nuclides attain greater stability (i.e., decay by a - emission) which reduces the initial N/Z value to (N - 2) / (Z - 2). Reduction of Z leads to the reduction of p - p repulsions.

Example of a - emission:

The stability linked to the n - p ratio is explained by strong n - p as well as p - p attractive forces operative at the level of nuclear distances. For heavier nuclides, p - p repulsions start to offset the attractive forces and an excess of neutrons over protons is required for stability.

Thus it appears that neutron - proton ratio plays a vital role in deciding the stability of nuclides as also the kinds of decay they undergo.