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प्रश्न
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 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?
उत्तर
(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.
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