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 $1.5\mathring{A}$ , and the electron is roughly $1\mathring{A}$ from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Answer 1

The system of two protons and one electron is represented in the given figure.

Charge on proton 1, $q_1=1.6\times {10}^{-19}C$

Charge on proton 2, $q_2=1.6\times {10}^{-19}C$

Charge on electron, $q_3=-1.6\times {10}^{-19}C$

Distance between protons 1 and 2, $d_1=1.5\times {10}^{-10}m$

Distance between proton 1 and electron, $d_2=1\times {10}^{-10}m$

Distance between proton 2 and electron, $d_3=1\times {10}^{-10}m$

The potential energy at infinity is zero.

Potential energy of the system,

$V=\frac{q_1{q}_2}{4\pi {\in }_0{d}_1}+\frac{q_2{q}_3}{4\pi {\in }_0{d}_3}+\frac{q_3{q}_1}{4\pi {\in }_0{d}_2}$

Substituting $\frac{1}{4\pi {\in }_0}=9\times {10}^{9}N{m}^2{C}^{-2}$, we obtain

$V=\frac{9\times {10}^9\times {10}^{-19}\times {10}^{-19}}{{10}^{-19}}[-(16{)}^2+\frac{(1.6{)}^2}{1.5}+-\left(1.6{)}^2\right]$

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

$=-19.2eV$

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