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 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?

Answer 1

(i) The proton separation energy is given by
`S_(pSn) =(M_(119.70) +M_(H) -M_(120.70))c^(2)`
`=(118.9058 + 1.0078252- 119.902199)c^(2)`
`=0.0114362c^(2)`
`"Similarly"" "S_(pSp) =(M_(120.70)+M_(H)-M_(121.70))c^(2)`
`=(119.902199+1.0078252-120.903822)c^(2)`
`=0.0059912c^(2)`
Since `S_(pSn) gtS_(pSb),` Sn nucleus is more stable than Sb nucleus
(ii) The existence of magic number indicates that the shell structure of nucleus similar to the shall structure of an atom. This also explains the peaks in binding energy `//`nucleon curve.