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If one of the two electrons of a $\mathrm{H}_2$ molecule is removed, we get a hydrogen molecular ion $\mathrm{H}_2^{+}$. In the ground state of an $\mathrm{H}_2^{+}$, the two protons are separated by roughly $1.5 Å$, and the electron is roughly $1 Å$ from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Solution:
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Verified Answer
By formula, potential energy of the system $U=$ P.E. of first proton and electron system + P.E. of second proton and electron system + P.E. of proton and proton system
Applying values, we get
$$
\begin{aligned}
U=9 \times 10^9 \times \frac{\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{1 \times 10^{-10} \times 1.6 \times 10^{-19}} \\
&+9 \times 10^9 \times \frac{\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{1 \times 10^{-10} \times 1.6 \times 10^{-19}} \\
&+9 \times 10^9 \times \frac{\left(1.6 \times 10^{-19}\right)^2}{1.5 \times 10^{-10} \times 1.6 \times 10^{-19}} \\
=& 9 \times 10^9 \times \frac{1.6 \times 10^{-19}}{10^{-10}}\left(-\frac{1}{1}-\frac{1}{1}+\frac{1}{1.5}\right) \\
=&-19.2 \mathrm{eV} .
\end{aligned}
$$
(zero of potential energy is taken to be at infinite)
Applying values, we get
$$
\begin{aligned}
U=9 \times 10^9 \times \frac{\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{1 \times 10^{-10} \times 1.6 \times 10^{-19}} \\
&+9 \times 10^9 \times \frac{\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{1 \times 10^{-10} \times 1.6 \times 10^{-19}} \\
&+9 \times 10^9 \times \frac{\left(1.6 \times 10^{-19}\right)^2}{1.5 \times 10^{-10} \times 1.6 \times 10^{-19}} \\
=& 9 \times 10^9 \times \frac{1.6 \times 10^{-19}}{10^{-10}}\left(-\frac{1}{1}-\frac{1}{1}+\frac{1}{1.5}\right) \\
=&-19.2 \mathrm{eV} .
\end{aligned}
$$
(zero of potential energy is taken to be at infinite)
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