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.
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.
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 $-19.2\; eV.$
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