Ncert Exercises & Solutions

If one of the two electrons of an \(H_2\) molecule is removed, we get a hydrogen molecular ion \(H\) In the ground state of an \(H_2^+,\) the two protons are separated by roughly \(1.5~\mathring{\text{ A}},\) and the electron is roughly \(1~\mathring{\text{ A}}\) from each proton. Assuming the zero of potential energy is when all three particles (two protons and one electron) are infinitely far apart, determine the approximate potential energy of the \(H_2^+\) system in its ground state.
1. \(19.2~\text{eV}\)
2. \(-9.6~\text{eV}\)
3. \(-19.2~\text{eV}\)
4. \(9.6~\text{eV}\)

The system of two protons and one electron is represented in the given figure. - 1.6 x 10^-19C

Charge on proton 1, q₁ = 1.6 X 10^-19C

Charge on proton 2, q₂ = 1.6 X 10^-19C

Charge on electron, q₃ = -1.6 x 10^-10 m

Distance between protons 1 and 2, d₁ = 1.5 x 10^-10

Distance between proton 1 and electron, d₂ 1 x 10^-10 m Distance

between proton 2 and electron, d₃ = 1 x 10^-10 m

The potential energy at infinity is zero.

The potential energy of the system,

\begin{aligned} V = \frac{q_{1} q_{2}}{4 {πϵ}_{0} d_{1}} + \frac{q_{2} q_{3}}{4 {πϵ}_{0} d_{3}} + \frac{q_{3} q_{1}}{4 {πϵ}_{0} d_{2}} \\ Substituting \frac{1}{4 {πϵ}_{0}} = 9 \times 10^{\circ} {Nm}^{2} C^{- 2}, we obtain \end{aligned}

\begin{aligned} V & = \frac{9 \times 10^{9} \times 10^{- 19} \times 10^{- 19}}{10^{- 10}} [- (16)^{2} + \frac{(1 .6)^{2}}{1 .5} + - (1 .6)^{2}] \\ = - 30 .7 \times 10^{- 19} J \\ = - 19 .2 eV \end{aligned}

Therefore, the potential energy of the system is -19.2 ev.

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