# 2.19  If one of the two electrons of a H2 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 $\dot{A}$, and the electron is roughly 1 $\dot{A}$ from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Given,

Distance between proton 1 and 2

$d_{p_1-p_2}=1.5*10^{-10}m$

Distance between proton 1 and electron

$d_{p_1-e}=1*10^{-10}m$

Distance between proton 2 and electron

$d_{p_2-e}=1*10^{-10}m$

Now,

The potential energy of the system :

$V=\frac{kp_1e}{d_{p_1-e}}+\frac{kp_2e}{d_{p_2-e}}+\frac{kp_1p_2}{d_{p_1-p_2}}$

Substituting the values, we get

$V=\frac{9*10^{9}*10^{-19}*10^{-19}}{10^{-10}}\left [ -(16)^2+\frac{(1.6)^2}{1.5} -(1.6)^2\right ]=-19.2eV$

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