The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of…
The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus
$$
=1.1 \text { fermi })\left[\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right]
$$
0.65 MeV
2.09 MeV
3.31 MeV
0.92 MeV
Solution
Potential energy due to two charges $=\frac{k q_1 q_2}{d}$.
For hydrogen atom, $q_1=q_2=1.6 \times 10^{-19} \mathrm{C}$
$
\begin{aligned}
U & =\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{2.2 \times 10^{-15}} \\
= & 10.47 \times 10^{-14} \mathrm{~J}=\frac{10.47 \times 10^{-4}}{1.6 \times 10^{-19}} \mathrm{eV} \\
& =6.5 \times 10^5 \mathrm{eV}=0.65 \mathrm{MeV}
\end{aligned}
$