Question

The mutual electrostatic potential energy between two protons which are at a distance of $ 9\times 10^{-15}\,m, $ in $ _{92}U^{235} $ nucleus is

• A. $ 1.56\times 10^{-14}J $
• B. $ 5.5\times 10^{-14}J $
• C. $ 2.56\times 10^{-14}J $
• D. $ 4.56\times 10^{-14}J $

Answer 1

Answer: (C)

Electrostatic potential energy
$U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}$
Here, $q_{1}=q_{2}=1.6 \times 10^{-19} C$
and $r=9 \times 10^{-15} m$
$\therefore U = \frac{9\times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{9\times 10^{-15}}$
$ = 2.56 \times 10^{-14} J$