Nuclei with magic number of proton $Z=2,8,20,28,50,52$ and magic number 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 $S_p$ for ${ }^{120} \operatorname{Sn}(Z=50)$ and ${ }^{121} \mathrm{Sb}(Z=51)$.
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
$$\begin{aligned} S_p & =\left(M_{Z-1, N}+M_{\mathrm{H}}-M_{Z, N}\right) c^2 . \\ \text{Given,}\quad { }^{119} \mathrm{In} & =118.9058 \mathrm{u},{ }^{120} \mathrm{Sn}=199.902199 \mathrm{u}, \\ { }^{121} \mathrm{Sb} & =120.903824 \mathrm{u},{ }^1 \mathrm{H}=1.0078252 \mathrm{u} . \end{aligned}$$
(ii) What does the existence of magic number indicate?
(i) The proton separation energy is given by
$$\begin{aligned} S_{p S n} & =\left(M_{119.70}+M_H-M_{120,70}\right) c^2 \\ & =(118.9058+1.0078252-119.902199) c^2 \\ & =0.0114362 c^2 \\ \text{Similarly}\quad S_{p S p} & =\left(M_{120,70}+M_H-M_{121,70}\right) c^2 \\ & =(119.902199+1.0078252-120.903822) c^2 \\ & =0.0059912 c^2 \end{aligned}$$
Since, $S_{p S_n}>S_{p S b}$, Sn nucleus is more stable than Sb nucleus.
(ii) The existence of magic numbers indicates that the shell structure of nucleus similar to the shell structure of an atom. This also explains the peaks in binding energy/nucleon curve.