If the potential energy between two molecules | vedclass

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(C) The potential energy is given by $U(r) = -\frac{A}{r^6} + \frac{B}{r^{12}}$.At equilibrium,the force $F = -\frac{dU}{dr} = 0$.Calculating the deri

Vedclass · Accepted Answer

$\left(\frac{2B}{A}\right)^{1/6}, -\frac{A^2}{4B}$

Vedclass · Answer

(C) The potential energy is given by $U(r) = -\frac{A}{r^6} + \frac{B}{r^{12}}$.At equilibrium,the force $F = -\frac{dU}{dr} = 0$.Calculating the derivative: $\frac{dU}{dr} = -A(-6r^{-7}) + B(-12r^{-13}) = \frac{6A}{r^7} - \frac{12B}{r^{13}}$.Setting $\frac{dU}{dr} = 0$: $\frac{6A}{r^7} = \frac{12B}{r^{13}} \Rightarrow r^6 = \frac{12B}{6A} = \frac{2B}{A}$.Thus,the equilibrium separation is $r = \left(\frac{2B}{A}\right)^{1/6}$.Substituting this value into the potential energy expression:$U = -\frac{A}{(2B/A)} + \frac{B}{(2B/A)^2} = -\frac{A^2}{2B} + \frac{B \cdot A^2}{4B^2} = -\frac{A^2}{2B} + \frac{A^2}{4B} = -\frac{A^2}{4B}$.

If the potential energy between two molecules is given by $U = -\frac{A}{r^6} + \frac{B}{r^{12}}$,then at equilibrium,the separation between the molecules and the potential energy are:

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