# If the potential energy between two molecules

Question:

If the potential energy between two molecules

is given by $\mathrm{U}=\frac{\mathrm{A}}{\mathrm{r}^{6}}+\frac{\mathrm{B}}{\mathrm{r}^{12}}$, then at equilibrium,

separation between molecules, and the potential energy are :

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

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

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

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

Correct Option: , 3

Solution:

$U=\frac{-A}{r^{6}}+\frac{B}{r^{12}}$

$\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dr}}=-\left(\mathrm{A}\left(-6 \mathrm{r}^{-7}\right)\right)+\mathrm{B}\left(-12 \mathrm{r}^{-13}\right)$

$0=\frac{6 \mathrm{~A}}{\mathrm{r}^{7}}-\frac{12 \mathrm{~B}}{\mathrm{r}^{13}}$

$\frac{6 \mathrm{~A}}{12 \mathrm{~B}}=\frac{1}{\mathrm{r}^{6}} \Rightarrow \mathrm{r}=\left(\frac{2 \mathrm{~B}}{\mathrm{~A}}\right)^{1 / 6}$

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

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