Question

The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of radius $R$ is given by
$E=\frac{3}{5} \frac{Z(Z-1) e^{2}}{4 \pi \varepsilon_{0} R}$
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 $e ^{2} /\left(4 \pi \varepsilon_{0}\right)=1.44\, MeV fm$. 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 $\left(1 \,fm =10^{-15} m \right)$

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

Answer 1

Answer: (C)

$E =\frac{3}{5} \frac{ Z ( Z -1) e ^{2}}{4 \pi \varepsilon_{0} R }$
$B.E. \left({ }_{7}^{15} N \right)=\left[7 m _{ p }+8 m _{ n }- M _{ N }\right] c ^{2}$
$B.E. \left({ }_{8}^{15} O \right)=\left(8 m _{ p }+7 m _{ n }- M _{ O }\right) c ^{2}$
$B.E. \left({ }_{7}^{15} N \right)-$ B.E. $\left({ }_{8}^{15} O \right)=\left[\left( m _{ n }- m _{ p }\right)+ M _{0}- M _{ N }\right] c ^{2}$
$=(1.008665-1.007825+15.003065-15.000109) \times 931.5\, MeV$
$\Delta B.E. \Rightarrow 0.003796 \times 931.5\, MeV$
$\Delta B.E. =3.535974\, MeV$
$E _{1}- E _{2}=\Delta B \cdot E$
$\frac{3}{5} \frac{Z_{1}\left(Z_{1}-1\right) e ^{2}}{4 \pi \varepsilon_{0} R }-\frac{3}{5} \frac{ Z _{2}\left( Z _{2}-1\right) e ^{2}}{4 \pi \varepsilon_{0} R }$
$=3.535974$
$\frac{3}{5}[8(7)-7(6)] \frac{ e ^{2}}{4 \pi \varepsilon_{0} R }$
$=3.535974$
$R =\frac{3}{5} \times \frac{14 \times 1.44}{3.535974}$
$=3.42\, fm$