Question

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

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

Answer 1

Answer: (B)

$F=\frac{dU}{dr}=-\frac{d}{dr}\left[\frac{M}{r^3}-\frac{N}{R^{12}}\right]$
$=-\left[\frac{-6M}{r^2}+\frac{12N}{r^{13}}\right]$
In equilibrium position, $F=0$
$\therefore \frac{6M}{r^2}-\frac{12N}{r^{13}}=0$ or $,r^6=\frac{2N}{M}$
$\therefore$ Potential energy at equilibrium position
$U=\frac{M}{\left(2N/M\right)}=\frac{N}{{\left(2N/M\right)}^2}$
$=\frac{M^2}{2N}-\frac{M^2}{4N}=\frac{M^2}{4N}$