Ncert Exercises & Solutions

Nuclei with magic number of proton Z = 2, 8, 20, 28, 50, 52 and magic number 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:

$S_{p} = (M_{z - 1},_{N} + M_{H} - M_{Z, N}) c^{2} .$

Given,

$^{119} \ln = 118.9058 u,^{120} Sn = 199.902199 u,$
$121 Sb = 120.903824 u,^{1} H = 1.0078252 u .$

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

Hint: The stability of a nucleus depends on the separation energy of the proton.

Step 1: Find the separation energy of Sn and Sb.

(i) The proton separation energy is given by:

S_{pSn} = (M_{119.70} + M_{H} - M_{121.70}) c^{2}

= (118.9058 + 1.0078252 - 119.902199) c^{2}

= 0.0114362 c^{2}

Similarly,

S_{pSb} = (M_{120.70} + M_{H} - M_{121.70}) c^{2}

= (119.902199 + 1.0078252 - 120.903822) c^{2}

= 0.0059912 c^{2}

Since,

S_{pSn} > S_{pSb}

, Sn nucleus is more stable than Sb nucleus.

Step 2: Find the significance of the existence of magic numbers.

(ii) The existence of magic numbers indicates that the shell structure of the nucleus similar to the shell structure of an atom. This also explains the peaks in the binding energy/nucleon curve.

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