The electrostatic energy of \(Z\) protons uniformly distributed throughout a spherical nucleus of radius \(R\) is given by \[ E=\frac{3}{5} \frac{Z(Z-1) e^{2}}{4 \pi \varepsilon_{0} R} \] The measured masses of the neutron, \({ }_{1}^{1} \mathrm{H},{ }_{7}^{15} \mathrm{~N}\) and \({ }_{8}^{15} \mathrm{O}\) are \(1.008665 \mathrm{u}, 1.007825 \mathrm{u}\), 15.000109 u and 15.003065 u , respectively. Given that the radii of both the \({ }_{7}^{15} \mathrm{~N}\) and \({ }_{8}^{15} \mathrm{O}\) nuclei are same, \(1 \mathrm{u}=931.5 \mathrm{MeV} / c^{2}\) ( \(c\) is the speed of light) and \(e^{2} /\left(4 \pi \varepsilon_{0}\right)=1.44 \mathrm{MeV} \mathrm{fm}\). Assuming that the difference between the binding energies of \({ }_{7}^{15} \mathrm{~N}\) and \({ }_{8}^{15} \mathrm{O}\) is purely due to the electrostatic energy, the radius of either of the nuclei is ( \(1 \mathrm{fm}=10^{-15} \mathrm{~m}\) )

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Modern Physics

The electrostatic energy of

Z

protons uniformly distributed throughout a spherical nucleus of radius

R

is given by

E = \frac{3}{5} \frac{Z ( Z - 1 ) e ^{2}}{4 π ε _{0} R}

The measured masses of the neutron,

_{1}^{1} H,_{7}^{15} N

and

_{8}^{15} O

are

1.008665 u, 1.007825 u

, 15.000109 u and 15.003065 u , respectively. Given that the radii of both the

_{7}^{15} N

and

_{8}^{15} O

nuclei are same,

1 u = 931.5 MeV / c^{2}

(

c

is the speed of light) and

e^{2} / (4 π ε_{0}) = 1.44 MeV fm

. Assuming that the difference between the binding energies of

_{7}^{15} N

and

_{8}^{15} O

is purely due to the electrostatic energy, the radius of either of the nuclei is
(

1 fm = 1 0^{- 15} m

)

Answer: (C)

The electrostatic energy difference between

_{8}^{15} O

and

_{7}^{15} N

is:

Δ E = \frac{3}{5} \times \frac{1.44}{R} \times 7 \times 2 MeV

For

_{7}^{15} N

: mass defect =

8 \times 1.008665 + 7 \times 1.007825 - 15.000109 = 0.1239864

u

Binding energy =

0.1239864 \times 931.5

MeV

For

_{8}^{15} O

: mass defect =

7 \times 1.008665 + 8 \times 1.007825 - 15.003065 = 0.12019044

u

Binding energy =

0.12019044 \times 931.5

MeV

The difference in binding energies =

0.0037960 \times 931.5

MeV

Equating this to the electrostatic energy difference:

\frac{3}{5} \times \frac{1.44 \times 14}{R} = 0.0037960 \times 931.5

Solving:

R = 3.42

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option