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 $\left(1fm=10^{-15}m\right)$

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

Answer 1

Answer: (C)

$E=\frac{3}{5}\frac{Z(Z-1)e^2}{4\pi {\epsilon }_{0}R}$
$B.E.\left({}_7^{15}N\right)=\left[7m_p+8m_n-M_N\right]c^2$
$B.E.\left({}_8^{15}O\right)=\left(8m_p+7m_n-M_O\right)c^2$
$B.E.\left({}_7^{15}N\right)-$ B.E. $\left({}_8^{15}O\right)=\left[\left(m_n-m_p\right)+M_0-M_N\right]c^2$
$=(1.008665-1.007825+15.003065-15.000109)\times 931.5MeV$
$\Delta B.E.\Rightarrow 0.003796\times 931.5MeV$
$\Delta B.E.=3.535974MeV$
$E_1-E_2=\Delta B\cdot E$
$\frac{3}{5}\frac{Z_1\left(Z_1-1\right)e^2}{4\pi {\epsilon }_{0}R}-\frac{3}{5}\frac{Z_2\left(Z_2-1\right)e^2}{4\pi {\epsilon }_{0}R}$
$=3.535974$
$\frac{3}{5}[8(7)-7(6)]\frac{e^2}{4\pi {\epsilon }_{0}R}$
$=3.535974$
$R=\frac{3}{5}\times \frac{14\times 1.44}{3.535974}$
$=3.42fm$