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← JEE Advanced 2016 · Paper 2
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JEE Advanced 2016 · Paper 2, Physics question 1

The electrostatic energy of ZZ protons uniformly distributed throughout a spherical nucleus of radius RR is given by
E=35 Z(Z−1)e24πε0RE=\frac35\,\frac{Z(Z-1)e^2}{4\pi\varepsilon_0R}
The measured masses of the neutron, 11H^1_1\mathrm H, 715N^{15}_{7}\mathrm N and 815O^{15}_{8}\mathrm O are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u, respectively. Given that the radii of both the 715N^{15}_{7}\mathrm N and 815O^{15}_{8}\mathrm O nuclei are same, 1 u=931.5 MeV/c21\ \mathrm u=931.5\ \mathrm{MeV}/c^2 (cc is the speed of light) and e2/(4πε0)=1.44e^2/(4\pi\varepsilon_0)=1.44 MeV fm. Assuming that the difference between the binding energies of 715N^{15}_{7}\mathrm N and 815O^{15}_{8}\mathrm O is purely due to the electrostatic energy, the radius of either of the nuclei is

(1 fm=10−151\ \mathrm{fm}=10^{-15} m)

Choose one option
Answer and worked solution

Official answer: C

Binding energies.

  • 15^{15}N: 7(1.007825)+8(1.008665)−15.000109=0.1239867(1.007825)+8(1.008665)-15.000109=0.123986 u.
  • 15^{15}O: 8(1.007825)+7(1.008665)−15.003065=0.1201908(1.007825)+7(1.008665)-15.003065=0.120190 u.

The difference is 0.003796×931.5=3.5360.003796\times931.5=3.536 MeV.

Coulomb energy difference. With Z(Z−1)=56Z(Z-1)=56 for O and 4242 for N:
ΔE=35⋅1.44R×(56−42)=12.096R MeV fm.\Delta E=\frac35\cdot\frac{1.44}{R}\times(56-42)=\frac{12.096}{R}\text{ MeV fm}.

Radius. R=12.0963.536=3.42R=\dfrac{12.096}{3.536}=3.42 fm.

(C)

Topics:Mass defect and binding energyPotential energy of charge systems
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