# If the potential energy between two molecules is given by

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:

1. $\left(\frac{B}{2 A}\right)^{\frac{1}{6}},-\frac{A^{2}}{2 B}$

2. $\left(\frac{B}{A}\right)^{\frac{1}{6}}, 0$

3. $\left(\frac{2 B}{A}\right)^{\frac{1}{6}},-\frac{A^{2}}{4 B}$

4. $\left(\frac{2 B}{A}\right)^{\frac{1}{6}},-\frac{A^{2}}{2 B}$

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,

$U\left(r=\left(\frac{2 B}{A}\right)^{1 / 6}\right)=-\frac{A}{2 B / A}+\frac{B}{4 B^{2} / A^{2}}$

$=\frac{-A^{2}}{2 B}+\frac{A^{2}}{4 B}=\frac{-A^{2}}{4 B}$