Question

The figure below is the plot of potential energy versus internuclear distance (𝑑) of $H_2$ molecule in the electronic ground state. What is the value (only magnitude) of the net potential energy $E_0$ (as indicated in the figure) in $kJmo{l}^{-1}$, for $d=d_0$ at which the electron-electron repulsion and the nucleus-nucleus repulsion energies are absent? As reference, the potential energy of $H$ atom is taken as zero when its electron and the nucleus are infinitely far apart. Use Avogadro constant as $6.023\times {10}^{23}\text{mol}{}^{-1}$

Answer 1

Potential Energy = $2\times \text{Total Energy}$

Total energy, $E=-13.6\frac{Z^2}{n^2}$ ev/atom

So, potential Energy = $2\times -13.6\frac{Z^2}{n^2}$ ev/atom
As one $H_2$ contains $2H$ atoms, energy, $E=–2\times 13.6\frac{Z^2}{n^2}$ ev/atom $\times 2\text{atom}$

$=4\times 13.6\times 1.6\times {10}^{-19}\times 6.023\times {10}^{23}J$
$=–5242.42\text{kJ/mol}$
Hence, the magnitude of the net potential energy $E_0$ is $5242.42$