With increase in the atomic number, the proton-proton electrostatic repulsion begins to overcome attractive forces involving proton and neutrons in heavier nuclides. The stability relationship can be represented by a line with a slope of 45$$^\circ$$, i.e., the maximum stability is attained when N = Z. Right of the curve a radioactive nuclide would be neutron rich and would decay by $$\beta$$-emission to produce a daughter nucleus with a lower n/p ratio. For heavier nuclides, p-p repulsions start to offset the attractive forces and an excess of neutrons over protons, is required for stability.