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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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The correct answer is:
$M^{2} / 4 N$
Given, $U=\frac{M}{r^{6}}-\frac{N}{r^{12}}$
$\begin{array}{l}
\therefore \mathrm{F}=\frac{-\mathrm{du}}{\mathrm{dr}}=\frac{-\mathrm{d}}{\mathrm{dr}}\left(\frac{\mathrm{M}}{\mathrm{r}^{6}}-\frac{\mathrm{N}}{\mathrm{r}^{12}}\right) \\
=-\left(\frac{-6 \mathrm{M}}{\mathrm{r}^{7}}+\frac{12 \mathrm{~N}}{\mathrm{r}^{13}}\right)=\left(\frac{6 \mathrm{M}}{\mathrm{r}^{7}}-\frac{12 \mathrm{~N}}{\mathrm{r}^{13}}\right)
\end{array}$
For equilibrium position, $\mathrm{F}=0$
$\therefore \frac{6 \mathrm{M}}{\mathrm{r}^{7}}=\frac{12 \mathrm{~N}}{\mathrm{r}^{13}} \text { or } \mathrm{r}^{6}=\frac{2 \mathrm{~N}}{\mathrm{M}}$
Hence, $U=\frac{M}{(2 N / M)}-\frac{N}{(2 N / M)^{2}}=\frac{M^{2}}{4 N}$
$\begin{array}{l}
\therefore \mathrm{F}=\frac{-\mathrm{du}}{\mathrm{dr}}=\frac{-\mathrm{d}}{\mathrm{dr}}\left(\frac{\mathrm{M}}{\mathrm{r}^{6}}-\frac{\mathrm{N}}{\mathrm{r}^{12}}\right) \\
=-\left(\frac{-6 \mathrm{M}}{\mathrm{r}^{7}}+\frac{12 \mathrm{~N}}{\mathrm{r}^{13}}\right)=\left(\frac{6 \mathrm{M}}{\mathrm{r}^{7}}-\frac{12 \mathrm{~N}}{\mathrm{r}^{13}}\right)
\end{array}$
For equilibrium position, $\mathrm{F}=0$
$\therefore \frac{6 \mathrm{M}}{\mathrm{r}^{7}}=\frac{12 \mathrm{~N}}{\mathrm{r}^{13}} \text { or } \mathrm{r}^{6}=\frac{2 \mathrm{~N}}{\mathrm{M}}$
Hence, $U=\frac{M}{(2 N / M)}-\frac{N}{(2 N / M)^{2}}=\frac{M^{2}}{4 N}$
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