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Question

The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by
E=35Z(Z1)e24πε0R
The measured masses of the neutron, 11H,157N and 158O are 1.008665 u,1.007825 u,15.000109 u and 15.003065 u respectively. Given that the radii of both the 157N and 158O nuclei are same, 1u=931.5MeV/c2 (c is the speed of light) and e2/(4πε0)=1.44MeVfm. Assuming that the difference between the binding energies of 157N and 158O is purely due to the electrostatic energy, the radius of either of the nuclei is (1fm=1015m)

A
2.85 fm
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B
3.03 fm
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C
3.42 fm
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D
3.80 fm
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Solution

The correct option is C 3.42 fm
For 157N : Z1=7 N1=8
For 158O : Z2=8 N2=7

Difference in the electrostatic energy, ΔE=E2E1
where, E=35Z(Z1)e24πε0R
so, ΔE=35(8(81)R7(71)R)×1.44=12.096R MeV fm

Mass defect, ΔM=ZMH+NMnMatom
For 157N ΔM1=7(1.007825)+8(1.008665)15.000109=0.12399 u

For 158O, ΔM2=8(1.007825)+7(1.008665)15.003065=0.12019 u

Difference in binding energy, ΔB.E=[ΔM1ΔM2]×931.5 MeV
ΔB.E=[0.123990.12019]×931.5=3.54 MeV

According to the question, 12.096R=3.54 R=3.42 fm


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