If one of the two electrons of a H_{2} molecu | vedclass

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(A) The system consists of two protons $(q_1 = q_2 = +e)$ and one electron $(q_3 = -e)$.The potential energy $V$ of a system of point charges is given

Vedclass · Accepted Answer

$-19.2$

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(A) The system consists of two protons $(q_1 = q_2 = +e)$ and one electron $(q_3 = -e)$.The potential energy $V$ of a system of point charges is given by the sum of the potential energies of all pairs:$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1 q_2}{d_1} + \frac{q_1 q_3}{d_2} + \frac{q_2 q_3}{d_3} ight)$Given:$q_1 = q_2 = 1.6 	imes 10^{-19} \; C$$q_3 = -1.6 	imes 10^{-19} \; C$$d_1 = 1.5 	imes 10^{-10} \; m$$d_2 = d_3 = 1.0 	imes 10^{-10} \; m$Substituting the values:$V = (9 	imes 10^9) \left[ \frac{(1.6 	imes 10^{-19})^2}{1.5 	imes 10^{-10}} - \frac{(1.6 	imes 10^{-19})^2}{1.0 	imes 10^{-10}} - \frac{(1.6 	imes 10^{-19})^2}{1.0 	imes 10^{-10}} ight]$$V = (9 	imes 10^9) 	imes (1.6 	imes 10^{-19})^2 \left[ \frac{1}{1.5} - 1 - 1 ight] \; J$$V = (9 	imes 10^9) 	imes (2.56 	imes 10^{-38}) \left[ \frac{1 - 3}{1.5} ight] \; J$$V = 23.04 	imes 10^{-29} 	imes \left( -\frac{2}{1.5} ight) \; J$$V = -30.72 	imes 10^{-19} \; J$To convert to $eV$, divide by $1.6 	imes 10^{-19} \; J/eV$:$V = \frac{-30.72 	imes 10^{-19}}{1.6 	imes 10^{-19}} \; eV = -19.2 \; eV.$The zero of potential energy is chosen at infinity.

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