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(B) The potential energy of the gas molecule is given by $U = \frac{M}{r^6} - \frac{N}{r^{12}}$.At equilibrium,the force $F$ acting on the molecule is

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$M^2 / 4N$

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(B) The potential energy of the gas molecule is given by $U = \frac{M}{r^6} - \frac{N}{r^{12}}$.At equilibrium,the force $F$ acting on the molecule is zero.The force is related to potential energy by $F = -\frac{dU}{dr}$.Calculating the derivative: $F = -\frac{d}{dr} \left( \frac{M}{r^6} - \frac{N}{r^{12}} \right) = - \left( -\frac{6M}{r^7} + \frac{12N}{r^{13}} \right) = \frac{6M}{r^7} - \frac{12N}{r^{13}}$.Setting $F = 0$ for equilibrium: $\frac{6M}{r^7} = \frac{12N}{r^{13}}$.This simplifies to $r^6 = \frac{12N}{6M} = \frac{2N}{M}$.Now,substitute $r^6 = \frac{2N}{M}$ back into the potential energy expression:$U = \frac{M}{r^6} - \frac{N}{(r^6)^2} = \frac{M}{(2N/M)} - \frac{N}{(2N/M)^2}$.$U = \frac{M^2}{2N} - \frac{N}{4N^2/M^2} = \frac{M^2}{2N} - \frac{M^2}{4N}$.$U = \frac{2M^2 - M^2}{4N} = \frac{M^2}{4N}$.

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

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