A system consists of two protons and one electron. They are separated by a distance of 1 Å from each other. If the potential energy at infinity is taken to be zero, the potential energy of the system is

-14.4 eV

The correct answer is Option (2) → -14.4 eV

Given:

Two protons and one electron, all separated by $r = 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m}$

Electrostatic potential energy between two charges $q_1$ and $q_2$ separated by distance $r$:

$U = \frac{k q_1 q_2}{r}$

Charges: proton: $+e$, electron: $-e$

Pairs in the system:

1. Proton-Proton: $U_{pp} = \frac{k e^2}{r}$

2. Proton-Electron (first proton): $U_{pe1} = \frac{-k e^2}{r}$

3. Proton-Electron (second proton): $U_{pe2} = \frac{-k e^2}{r}$

Total potential energy:

$U_{\text{total}} = U_{pp} + U_{pe1} + U_{pe2} = \frac{k e^2}{r} - \frac{k e^2}{r} - \frac{k e^2}{r} = - \frac{k e^2}{r}$

Substitute values: $k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2$, $e = 1.6 \times 10^{-19} \, \text{C}$, $r = 1 \times 10^{-10} \, \text{m}$

$U_{\text{total}} = - \frac{9 \times 10^9 \cdot (1.6 \times 10^{-19})^2}{1 \times 10^{-10}} = - 2.304 \times 10^{-18} \, \text{J}$

Answer: $U_{\text{total}} \approx -2.30 \times 10^{-18} \, \text{J}$