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. $NM^2/4$
• C. $MN^2/4$
• D. $M^2/4N$

Answer 1

Answer: (D)

Given, $U=\frac{M}{r^{6}}-\frac{N}{r^{12}}$
$\therefore F=\frac{-d u}{d r}=\frac{-d}{d r}\left(\frac{M}{r^{6}}-\frac{N}{r^{12}}\right)$
$=-\left(\frac{-6 M}{r^{7}}+\frac{12 N}{r^{13}}\right)$
$=\left(\frac{6 M}{r^{7}}-\frac{12 N}{r^{13}}\right)$
For equilibrium position, $F=0$
$\therefore \frac{6 M}{r^{7}}=\frac{12 N}{r^{13}}$
or $ r^{6}=\frac{2 N}{M}$
Hence, $ U=\frac{M}{(2 N / M)}-\frac{N}{(2 N / M)^{2}}=\frac{M^{2}}{4 N}$