The potential energy function for the force between two atoms in a diatomic molecule is approximately given by

$U=\frac{a}{x^{12}}-\frac{b}{x^6}$
If Force is denoted by $F$, then

$F=-\frac{dU}{dx}=-\frac{d}{dx}\left(\frac{a}{x^{12}}-\frac{b}{x^6}\right)$$=-\left[\frac{-12a}{x^{13}}+\frac{6b}{x^7}\right]=\left[\frac{12a}{x^{13}}-\frac{6b}{x^7}\right]$

At equilibrium, $F=0.$
$\therefore \frac{12a}{x^{13}}-\frac{6b}{x^7}=0 \mathrm{or} x^6=\frac{2a}{b}$

By using the above equation we can write$U_{\mathrm{at} \mathrm{equilibrium} }=\frac{a}{{\left(\frac{2a}{b}\right)}^2}-\frac{b}{\left(\frac{2a}{b}\right)}$
$=\frac{a{b}^2}{4a^2}-\frac{b^2}{2a}=\frac{b^2}{4a}-\frac{b^2}{2a}=-\frac{b^2}{4a}$

Since, $U_{(x=\infty )}=0$

Therefore, th dissociation energy is given by
$D=$[$U_{(x=\infty )}-U_{\mathrm{at} \mathrm{equilibrium}}$
$=\left[0-\left(-\frac{b^2}{4a}\right)\right]=\frac{b^2}{4a}$