Question

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 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 tightly bound proton from a nucleus of that nuclide. It is given by `S_P = (M_(z-1^' N) + M_H - M_(ZN))c^2`. 

Given ¹¹⁹In = 118.9058u, ¹²⁰Sn = 119.902199u, ¹²¹Sb = 120.903824u, ¹H = 1.0078252u.

(ii) What does the existance of magic number indicate?

Answer 1

(i) `S_(pSn) = (M_119.70 + M_H - M_120.70)  c^2`

= (118.9058 + 1.0078252 – 119.902199) c²

= 0.0114362 c²

`S_(pSb) = (M_120.70) + M_H - M_121.70)  c^2`

= (119.902199 + 1.0078252 – 120.903822) c²

= 0.0059912 c²

Since `S_(pSn) > S_(pSb)`, Sn nucleus is more stable than Sb nucleus.

(ii) It indicates the shell structure of nucleus similar to the shell structure of an atom. This also explains the peaks in BE/ nucleon curve.