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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 Sn120(Z = 50) and Sb121 = (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 Sp = (MZ-1, N + MH – MZ,N)c2. Given In ln119 = 118.9058u, Sn120= 119.902199u, Sb121 = 120.903824u, H1 = 1.0078252u

(ii) What does the existence of a magic number indicate?

Answers (1)

i) Here, from the given data, the energy required for the separation of proton is:

Sp.Sn = (M119.70 + Mh – M120.70)c2 = 0.0114362 c2

Similarly, Sp.Sp  = (M120.70 + Mh – M121.70)c2 = 0.0059912 c2

Here, we can clearly observe, Sp.Sn > Sp.Sp

Therefore, the Sn nucleus will be more stable than Sb nucleus.

ii) From the magic numbers, we can understand that the shell structure of the nucleus and the shell structure of the atom are similar. Peaks in the Binding energy can also be observed from it.

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