Question

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

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

Answer 1

The correct option is B $\frac{M^2}{4N}$

$F$ =$\frac{dU}{dr}$

$=-\frac{d}{dr}\left[\frac{M}{r^6}-\frac{N}{R^{12}}\right]$

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

In equilibrium position F=0

$\therefore$ $\frac{6M}{r^7}$ = $\frac{12N}{r^{13}}$ = 0

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

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

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

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