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(D) At $d = d_0$, the electron-electron repulsion and nucleus-nucleus repulsion are absent. The potential energy is primarily due to the attraction be

Vedclass · Accepted Answer

$2623.249$

Vedclass · Answer

(D) At $d = d_0$, the electron-electron repulsion and nucleus-nucleus repulsion are absent. The potential energy is primarily due to the attraction between the proton and electron in each $H$ atom.The potential energy $(P.E.)$ for one $H$ atom is given by:$P.E. = \frac{-K q_1 q_2}{r} = \frac{-(9 	imes 10^9) 	imes (1.6 	imes 10^{-19})^2}{0.529 	imes 10^{-10}} \ J$$P.E. = -4.355 	imes 10^{-18} \ J$To convert this to $kJ \ mol^{-1}$ for one mole of $H$ atoms:$E_0 = (-4.355 	imes 10^{-18} \ J) 	imes (6.023 	imes 10^{23} \ mol^{-1}) 	imes 10^{-3} \ kJ/J$$E_0 = -2623.249 \ kJ \ mol^{-1}$Since the question asks for the magnitude of the potential energy $E_0$ as indicated in the figure, the value is $2623.249 \ kJ \ mol^{-1}$.

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