The potential energy function for the force b | vedclass

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Question

(C) At equilibrium,the force is zero,which means the derivative of the potential energy with respect to distance is zero: $\frac{dU(x)}{dx} = 0$.Given

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$\frac{b^2}{4a}$

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(C) At equilibrium,the force is zero,which means the derivative of the potential energy with respect to distance is zero: $\frac{dU(x)}{dx} = 0$.Given $U(x) = ax^{-12} - bx^{-6}$,we differentiate:$\frac{dU}{dx} = -12ax^{-13} + 6bx^{-7} = 0$.$12ax^{-13} = 6bx^{-7} \Rightarrow \frac{2a}{x^6} = b \Rightarrow x^6 = \frac{2a}{b}$.Now,substitute $x^6 = \frac{2a}{b}$ into the potential energy function to find $U_{	ext{at equilibrium}}$:$U_{	ext{at equilibrium}} = \frac{a}{(x^6)^2} - \frac{b}{x^6} = \frac{a}{(2a/b)^2} - \frac{b}{(2a/b)} = \frac{a}{4a^2/b^2} - \frac{b^2}{2a} = \frac{b^2}{4a} - \frac{2b^2}{4a} = -\frac{b^2}{4a}$.Since $U(x = \infty) = 0$,the dissociation energy $D$ is:$D = U(\infty) - U_{	ext{at equilibrium}} = 0 - (-\frac{b^2}{4a}) = \frac{b^2}{4a}$.

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