Solution :
The distance between electron-proton of a hydrogen atom, $d= 0.53\;A$
Charge on an electron, $q_1 = −1.6 ×10^{−19}\; C$
Charge on a proton,$ q_2 = +1.6 \times 10^{−19}\; C $
Potential at infinity is zero
Potential energy of the system, p-e = Potential energy at infinity − Potential energy at distance, d
$\qquad= 9- \large\frac{q_1q_2}{4 \pi \in_0 d}$
Where,
$\in_0$ is the permittivity of free space
$\large\frac{1}{4\pi \in_0}$$=9 \times 10^{9} Nm^{2} C^{-2}$
Potential energy $=0 - \large\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{0.53 \times 10^{19}}$$=43 \times 10^{-19}\;J$
Since $1.6 \times 10^{-19}\;J= 1\;eV$
Potential energy $=-43.7 \times 10^{-19}=\frac{-43.7 \times 10^{19}}{1.6 \times 10^{-19}}= -27.2\;eV$