If the potential energy of a gas molecule is& | vedclass

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Question

$\mathrm{F}=\frac{\mathrm{d} \mathrm{U}}{\mathrm{dr}}=-\frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{\mathrm{M}}{\mathrm{r}^{6}}-\frac{\mathrm{N}}{\mathrm

Vedclass · Accepted Answer

$\frac {M^2}{4N}$

Vedclass · Answer

$\mathrm{F}=\frac{\mathrm{d} \mathrm{U}}{\mathrm{dr}}=-\frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{\mathrm{M}}{\mathrm{r}^{6}}-\frac{\mathrm{N}}{\mathrm{r}^{12}}ight]$

$=-\left[-\frac{6 \mathrm{M}}{\mathrm{r}^{7}}+\frac{12 \mathrm{N}}{\mathrm{r}^{13}}ight]$

In equilibrium position, $F=0$

$	herefore \quad \frac{6 \mathrm{M}}{\mathrm{r}^{7}}-\frac{12 \mathrm{N}}{\mathrm{r}^{13}}=0$

Or         $r^{6}=\frac{2 N}{M}$

Potential energy at equilibrium position,

$\mathrm{U}=\frac{\mathrm{M}}{(2 \mathrm{N} / \mathrm{M})}-\frac{\mathrm{N}}{(2 \mathrm{N} / \mathrm{M})^{2}}$

$=\frac{M^{2}}{2 N}-\frac{M^{2}}{4 N}=\frac{M^{2}}{4 N}$

If the potential energy of a gas molecule is $U = \frac{M}{{{r^6}}} - \frac{N}{{{r^{12}}}},M$ and $N$ being positive constants, then the potential energy at equilibrium must be

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