The potential energy of a diatomic molecule i | vedclass

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(B) At equilibrium,the potential energy $U$ is at a minimum,which implies that the force $F = -\frac{dU}{dr} = 0$.Given $U = Ar^{-12} - Br^{-6}$.Diffe

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$(\frac{2A}{B})^{1/6}$

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(B) At equilibrium,the potential energy $U$ is at a minimum,which implies that the force $F = -\frac{dU}{dr} = 0$.Given $U = Ar^{-12} - Br^{-6}$.Differentiating with respect to $r$:$\frac{dU}{dr} = -12Ar^{-13} - (-6)Br^{-7} = -\frac{12A}{r^{13}} + \frac{6B}{r^7}$.Setting $\frac{dU}{dr} = 0$ for equilibrium:$\frac{6B}{r^7} = \frac{12A}{r^{13}}$.Rearranging the terms:$r^{13-7} = \frac{12A}{6B}$.$r^6 = \frac{2A}{B}$.Therefore,the equilibrium distance is $r = (\frac{2A}{B})^{1/6}$.

The potential energy of a diatomic molecule is given by $U = \frac{A}{r^{12}} - \frac{B}{r^6}$,where $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is:

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