If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion H₂⁺. In the ground state of a H₂⁺, the two protons are separated by roughly 1.5Å, 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

Let us consider the distance between two protons p₁and p₂as r₁₂and,
Distance of electron(e^-) from each proton p₁ and p₂be r₁₃and r23.

Given,

r_{12} = 1.5 Å = 1.5 \times 10^{- 10} m

r_{13} = r_{23} = 1 Å = 10^{- 10} m C harge on proton, q_{1} = q_{2} = 1.6 \times 10^{- 19} C C harge on electron, q_{3} = - 1.6 \times 10^{- 19} C

Now, the potential energy of the system

P . E = \frac{1}{4 {πε}_{0}} (\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{1} q_{3}}{r_{13}} + \frac{q_{2} q_{3}}{r_{23}})

= 9 \times 10^{9} [\frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{1.5 \times 10^{- 10}} + \frac{1.6 \times 10^{- 19} \times (- 1.6 \times 10^{- 19})}{10^{- 10}} + \frac{1.6 \times 10^{- 19} \times (- 1.6 \times 10^{- 19})}{10^{- 10}}]

= \frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19} \times 9 \times 10^{9}}{10^{- 10}} [\frac{1}{1.5} - \frac{1}{1} - \frac{1}{1}] = 1.6 \times 1.6 \times 9 \times 10^{- 19 - 19 + 9 + 10} [\frac{2}{3} - 2] = - 2.56 \times 9 \times 10^{- 19} \times \frac{4}{3} = - 2.56 \times 12 \times 10^{- 19} J = \frac{- 2.56 \times 12 \times 10^{- 19}}{1.6 \times 10^{- 19}} eV = - 1.6 \times 12 eV P . E = - 19.2 e V

The zero Potential Energy is taken to be at infinity.

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