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(B) The force $F$ is given by the negative gradient of potential energy: $F = -\frac{dU}{dr}$.$F = -\frac{d}{dr} \left[ \frac{M}{r^6} - \frac{N}{r^{12

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

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(B) The force $F$ is given by the negative gradient of potential energy: $F = -\frac{dU}{dr}$.$F = -\frac{d}{dr} \left[ \frac{M}{r^6} - \frac{N}{r^{12}} \right] = -\left[ -\frac{6M}{r^7} + \frac{12N}{r^{13}} \right] = \frac{6M}{r^7} - \frac{12N}{r^{13}}$.At equilibrium,the net force is zero: $F = 0$.$\frac{6M}{r^7} = \frac{12N}{r^{13}} \implies r^6 = \frac{12N}{6M} = \frac{2N}{M}$.Substitute $r^6 = \frac{2N}{M}$ into the expression for potential energy $U$:$U = \frac{M}{(r^6)} - \frac{N}{(r^6)^2} = \frac{M}{(2N/M)} - \frac{N}{(2N/M)^2}$.$U = \frac{M^2}{2N} - \frac{N \cdot M^2}{4N^2} = \frac{M^2}{2N} - \frac{M^2}{4N} = \frac{M^2}{4N}$.

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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