Question

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

• A. Zero
• B. $N{M}^2/4$
• C. $M{N}^2/4$
• D. $M^2/4N$

Answer 1

The correct option is D $M^2/4N$

Given in question,

$U=\frac{M}{r^6}-\frac{N}{r^{12}}$

$\therefore F=\frac{-du}{dr}=\frac{-d}{dr}\left(\begin{array}{c}\frac{M}{r^6}-\frac{N}{r^{12}}\end{array}\right)$
$=-\left(\begin{array}{c}\frac{-6M}{r^7}+\frac{12N}{r^{13}}\end{array}\right)=\left(\begin{array}{c}\frac{6M}{r^7}-\frac{12N}{r^{13}}\end{array}\right)$
For equilibrium position, Force $F=0$
$\therefore \frac{6M}{r^7}=\frac{12N}{r^{13}}$ or $r^6=\frac{2N}{M}$
Hence, $U=\frac{M}{\left(2N/M\right)}-\frac{N}{(2N/M{)}^2}=\frac{M^2}{4N}$

So the potential energy at equilibrium must be $U=\frac{M^2}{4N}$