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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.
$-19.2\; eV.$
$-13.6\; eV.$
$-9.8\; eV.$
$-8.6\; eV.$
Solution
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,
$V=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} d_{1}}+\frac{q_{2} q_{3}}{4 \pi \epsilon_{0} d_{3}}+\frac{q_{1} q_{1}}{4 \pi \epsilon_{0} d_{2}}$
Substituting $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \,N\,m ^{2} \,C ^{-2},$ we obtain
$V =\frac{9 \times 10^{9} \times 10^{-19} \times 10^{-19}}{10^{-10}}\left[-(16)^{2}+\frac{(1.6)^{2}}{1.5}+-(1.6)^{2}\right]$
$=-30.7 \times 10^{-19} \,J$
$=-19.2\, eV$
Therefore, the potential energy of the system is $-19.2\; eV.$
