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Nuclei with magic no. of proton Z=2, 8,2...

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.
Verify this by the calculating the proton separation energy `S_p` for `"^120Sn(Z=50)` and `"^121Sb=(Z=51)`. The proton separation energy for a nuclide is the minimum energy required to separate the least rightly bound proton from a nucleus of that nuclide. It is given by
`S_p=(M_(z-1,N)+M_H-M_(Z,N))c^2`
Given `"^119ln=118.9058u` , `"^120Sn=119.902199u`, `"^121Sb=120.903824u`, `"^1H=1.0078252u`.

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