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 stable. (i) Verify this by calculating the proton separation energy `S_(p)` for `.^(120)Sn (Z=50)` and `.^(121)Sb =(Z=51)`.
The proton separation energy for a nuclide is the minimum energy required to separated the least tightly bound proton form a nucleus of that nuclide. It is given by
`S_(p)=(M_(z-1,N)+M_(H)-M_(Z,N))c^(2)`.
given `.^(119)Sn_49 =118.9058u, .^(120) Sn_50 =119.902199u, .^(121)Sb_51=120.903824u, .^(1)H=1.0078252u`
(ii) what does the existence of magic number indicate?

Explanation

To solve the problem, we will calculate the proton separation energy \( S_p \) for the isotopes \( ^{120}Sn \) (Z=50) and \( ^{121}Sb \) (Z=51) using the given formula:

\[ S_p = (M_{Z-1,N} + M_H - M_{Z,N})c^2 \]

### Step 1: Identify the values needed for the calculation

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