Section 10.2 gives the energy and approximate proton separation of the ${\mathbf{\text{H}}}_{\mathbf{2}}^{\mathbf{+}}$ molecule. What is the energy of the electron alone?

The distance between two protons is $0.11\text{nm}$.

The potential energy between the two protons is given by a relation, $\mathbf{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{a}}$.

Here, ${\mathbf{\epsilon }}_{\mathbf{0}}$ is permittivity of the free space,$\mathbf{e}$ is charge on each proton, and role="math" localid="1658384048724" $\mathit{a}$is the distance between the two protons.

Substitute$8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^{\text{2}}{\text{/Nm}}^{\text{2}}$ for${\epsilon }_0,1.6\times {10}^{-19}\text{\hspace{0.17em}C}$for$e,0.11\text{\hspace{0.17em}nm}$for$a$ in the equation$\mathrm{U}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{e}}^2}{\mathrm{a}}$and solve for$U$.

$\begin{array}{c}\mathrm{U}=\frac{{(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}^2}{\left(4\mathrm{\pi }\right)(8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^{\text{2}}{\text{/Nm}}^{\text{2}})(0.11\text{\hspace{0.17em}nm})}\\ \mathrm{U}=\frac{{(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}^2}{\left(4\mathrm{\pi }\right)(8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^{\text{2}}{\text{/Nm}}^2)(0.11\times {10}^{-9}\text{\hspace{0.17em}m})}\\ \mathrm{U}=2.09\times {10}^{-18}\text{J}\end{array}$

Convert joules to electron volte.

$\begin{array}{c}\mathrm{U}=2.09\times {10}^{-18}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ \mathrm{U}=13.1\text{eV}\end{array}$

The energy of the electron is given by the difference of the total energy and the proton's energy, ${\mathrm{E}}_{\mathrm{e}}={\mathrm{E}}_{\mathrm{T}}-\mathrm{U}$.

Here, ${\mathrm{E}}_{\mathrm{e}}$ is the energy of the electron, ${\mathrm{E}}_{\mathrm{T}}$ is the total energy, and $\mathrm{U}$ is the energy of the proton.

Substitute$-16.3\text{eV}$for${\mathrm{E}}_{\mathrm{T}}$and$13.1\text{eV}$for$\mathrm{U}$in the equation${\mathrm{E}}_{\mathrm{e}}={\mathrm{E}}_{\mathrm{T}}-\mathrm{U}$and solve for${\mathrm{E}}_{\mathrm{e}}$.

$\begin{array}{c}{\mathrm{E}}_{\mathrm{e}}=-16.3\text{eV}-13.1\text{eV}\\ {\mathrm{E}}_{\mathrm{e}}=-29.4\text{eV}\end{array}$

Therefore, the energy of the electron in ${\text{H}}_2^{+}$ is role="math" $-29.4\text{eV}$.