The potential energy (U) of a diatomic molecu | vedclass

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(B) For a system to be in equilibrium,the net force must be zero,which implies that the derivative of potential energy with respect to distance must b

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$1$

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(B) For a system to be in equilibrium,the net force must be zero,which implies that the derivative of potential energy with respect to distance must be zero: $\frac{dU}{dr} = 0$.Given $U = \alpha r^{-10} - \beta r^{-5} - 3$.Differentiating with respect to $r$:$\frac{dU}{dr} = -10\alpha r^{-11} + 5\beta r^{-6} = 0$.Rearranging the terms:$5\beta r^{-6} = 10\alpha r^{-11}$.Dividing both sides by $5\beta r^{-11}$:$r^5 = \frac{10\alpha}{5\beta} = \frac{2\alpha}{\beta}$.$r = \left(\frac{2\alpha}{\beta}\right)^{\frac{1}{5}}$.Comparing this with the given form $\left(\frac{2\alpha}{\beta}\right)^{\frac{a}{b}}$,we find $\frac{a}{b} = \frac{1}{5}$.Thus,$a = 1$.

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