Question

If one of the two electrons of a $H_2$ Molecule is removed, We get a hydrogen molecular ion $H_2^{+}$. In the ground state of an $H_2^{+}$, the two protons are separated by roughly $1.5A^{\circ }$ and the electron is roughly $1A^{\circ }$ from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Answer 1

Given, Charge of the $1st$ proton,
$q_{p1}=1.6\times {10}^{-19}C$
Charge of the $2nd\text{proton}$,
$q_{p2}=1.6\times {10}^{-19}C$
Charge of the electron,
$q_e=-1.6\times {10}^{-19}C$

Distance between the $1st\text{and}2nd\text{proton}$,
$d_1=1.5A^{\circ }=1.5\times {10}^{-10}m$
Distance between the $1st\text{proton}$ and the electron,
$d_2=1A^{\circ }=1\times {10}^{-10}m$
Distance between the $2nd\text{proton}$ and the electron,
$d_3=1A^{\circ }=1\times {10}^{-10}m$
The potential energy at infinty is zero.
Therefore, the potential energy of the system is

$V=\frac{1}{4\pi {\epsilon }_0}(\frac{q_{p1}{q}_{p2}}{d_1}+\frac{q_{p1}{q}_e}{d_2}+\frac{q_{p2}{q}_e}{d_3})$

Where,
$\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^2$

Then,
$V=9\times {10}^9\times (1.6\times {10}^{-19}{)}^2\times (\frac{1}{1.5}-1-1)\times {10}^{10}$

$V=30.7\times {10}^{-19}J$

$V=30.7\times {10}^{-19}\times \frac{1}{1.6\times {10}^{-19}}eV$

$V=-19.2eV$

Therefore, the potential energy of the system is $-19.2eV$.

Final Answer: $-19.2eV$, The potential energy will be zero at infinity.