Nuclei with magic no. of proton Z = 2, 8, 20, 28, 50, 52 and magic no. of neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found to be very stable
(i) Verify this by calculating the proton separation energy Sp for Sn120 (Z = 50) and Sb121 = (Z = 51). The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by Sp = (MZ–1, N + MH – MZ,N) c2. Given In119 = 118.9058u, Sn120 = 119.902199u, Sb121 = 120.903824u, H1 = 1.0078252u
(ii) What does the existence of magic number indicate?
(i) The proton separation energy is SpSn = (M119.70 + Mh – M120.70)c2 = 0.0114362 c2
Similarly SpSp = (M120.70 + Mh – M121.70)c2 = 0.0059912 c2
Since SpSn > SpSp, Sn nucleus is more stable than Sb nucleus.
(ii) The magic numbers indicate that the shell structure of the nucleus is similar to the shell structure of the atom. This explains the peaks in the binding energy.