Question

The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of radius $R$ is given by
$E=\frac{3}{5}\frac{Z(Z-1)e^2}{4\pi {\epsilon }_{0}R}$
The measured masses of the neutron, ${}_1^{1}H,{}_7^{15}N$ and ${}_8^{15}O$ are $1.008665u,1.007825u,15.000109u$ and $15.003065u$ respectively. Given that the radii of both the ${}_7^{15}N$ and ${}_8^{15}O$ nuclei are same, $1u=931.5MeV/c^2$ ($c$ is the speed of light) and $e^2/\left(4\pi {\epsilon }_0\right)=1.44MeVfm$. 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 $(1fm={10}^{-15}m)$

• A. $2.85fm$
• B. $3.03fm$
• C. $3.42fm$
• D. $3.80fm$

Answer 1

The correct option is C $3.42fm$

For ${}_7^{15}N$ : $Z_1=7$ $N_1=8$
For ${}_8^{15}O$ : $Z_2=8$ $N_2=7$

Difference in the electrostatic energy, $\Delta E=E_2-E_1$
where, $E=\frac{3}{5}\frac{Z(Z-1)e^2}{4\pi {\epsilon }_{0}R}$
so, $\Delta E=\frac{3}{5}(\frac{8(8-1)}{R}-\frac{7(7-1)}{R})\times 1.44=\frac{12.096}{R}$ $MeV$ $fm$

Mass defect, $\Delta M=Z{M}_H+N{M}_n-M_{atom}$
$\therefore$ For ${}_7^{15}N$ $\Delta M_1=7\left(1.007825\right)+8\left(1.008665\right)-15.000109=0.12399$ u

For ${}_8^{15}O$, $\Delta M_2=8\left(1.007825\right)+7\left(1.008665\right)-15.003065=0.12019$ u

Difference in binding energy, $\Delta B.E=[\Delta M_1-\Delta M_2]\times 931.5$ $MeV$
$⟹$ $\Delta B.E=[0.12399-0.12019]\times 931.5=3.54$ $MeV$

According to the question, $\frac{12.096}{R}=3.54$ $⟹R=3.42$ $fm$