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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 19 of 37
Marks: +1, -0
If one of the two electrons of a H2\mathrm{H}_2 molecule is removed, we get a hydrogen molecular ion H2+\mathrm{H}_2^{+} . In the ground state of an H2+\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:  
It is a system of three point charges and the potential energy stored is this system of charges is
U' = 14πε0[q1q2r12+q1q3r13+q2q3r23]\frac{1}{4\pi\varepsilon_0} \left[ \frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}} \right]
U = 9 × 10910^9 [e×−e1+e×e1.5+e×−e1]\left[ \frac{e \times -e}{1}+\frac{e \times e}{1.5} + \frac{e \times -e}{1} \right] × 110−10 m\frac{1}{10^{-10}\,\text{m}}
or U = 9 × 109910^{99} × (1.6×10−19)2\left(1.6 \times 10^{-19}\right)^2 × 1010[−1+10.5−1]10^{10}\left[-1+\frac{1}{0.5}-1\right] J
or U = 9×1019×(1.6×10−19)21.6×10−19\frac{9 \times 10^{19} \times (1.6 \times 10^{-19})^2}{1.6 \times 10^{-19}} × (−21.5)\left(-\frac{2}{1.5}\right) eV or U = - 19.2 eV
with zero potential energy at infinity
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