Question:
If the potential energy between two molecules is given by
$U=-\frac{A}{r^{6}}+\frac{B}{r^{12}}$, then at equilibrium, separation between
molecules, and the potential energy are:
Correct Option: , 3
Solution:
(3) Given : $U=\frac{-A}{r^{6}}+\frac{B}{r^{12}}$
For equilibrium,
$F=\frac{d U}{d r}=-\left(A\left(-6 r^{-7}\right)\right)+B\left(-12 r^{-13}\right)=0$
$\Rightarrow 0=\frac{6 A}{r^{7}}-\frac{12 B}{r^{13}} \Rightarrow \frac{6 A}{12 B}=\frac{1}{r^{6}}$
$\therefore$ Separation between molecules, $r=\left(\frac{2 B}{A}\right)^{1 / 6}$
Potential energy,
$U\left(r=\left(\frac{2 B}{A}\right)^{1 / 6}\right)=-\frac{A}{2 B / A}+\frac{B}{4 B^{2} / A^{2}}$
$=\frac{-A^{2}}{2 B}+\frac{A^{2}}{4 B}=\frac{-A^{2}}{4 B}$