The potential energy function for the force b | vedclass

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(D) The potential energy function is $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$.At equilibrium,the force $F = -\frac{dU}{dx} = 0$.$F = -(\frac{-12a}{x^

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The potential energy function is $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$.At equilibrium,the force $F = -\frac{dU}{dx} = 0$.$F = -(\frac{-12a}{x^{13}} + \frac{6b}{x^7}) = \frac{12a}{x^{13}} - \frac{6b}{x^7} = 0$.$\frac{12a}{x^{13}} = \frac{6b}{x^7} \Rightarrow x^6 = \frac{2a}{b} \Rightarrow x = (\frac{2a}{b})^{1/6}$.At $x = \infty$,$U(\infty) = 0$.At equilibrium,$U_{	ext{equilibrium}} = \frac{a}{(2a/b)^2} - \frac{b}{(2a/b)} = \frac{ab^2}{4a^2} - \frac{b^2}{2a} = \frac{b^2}{4a} - \frac{2b^2}{4a} = -\frac{b^2}{4a}$.The dissociation energy $D = U(\infty) - U_{	ext{equilibrium}} = 0 - (-\frac{b^2}{4a}) = \frac{b^2}{4a}$.

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