The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by E=35ZZ−1e24πε0R The measured masses of the neutron, 11H, 715N and 815O are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u, respectively. Given that the radii of both the 715N and 815O nuclei are same, 1u=931.5 MeV/c2 (c is the speed of light) and e2/4πε0=1.44 MeV fm. Assuming that the difference between the binding energies of 715N and 815O is purely due to the electrostatic energy, the radius of either of the nuclei is 1fm=10−15
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a
2.85 fm
b
3.03 fm
c
3.42 fm
d
3.80 fm
answer is C.
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Detailed Solution
Binding energy of nitogen atom=8×1.008665+7×1.007825−15.000109931.5Binding energy of oxygen atom=7×1.008665+8×1.007825−15.003065931.5∴Difference in BE=0.0037960×931.5=3.54 MeVNow, Difference in electrostatic energy=EO−EN=358×7R−7×6R1.44MeV fm=12.096RMeV fmBoth differences are equal, 12.096RMeV fm=3.54 MeV⇒R=3.42fm