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Q.
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
Kinetic Theory
Solution:
$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} $