Q 11-05-160JEE MainJEE Main 2020 (6 Sep, Shift 1)Medium
If the potential energy between two molecules is given by $U = -\dfrac{A}{r^6} + \dfrac{B}{r^{12}}$, then at equilibrium, the separation between the molecules and the potential energy are:
Answer: (C) $\left(\dfrac{2B}{A}\right)^{1/6},\ -\dfrac{A^2}{4B}$
At equilibrium the force $-\dfrac{dU}{dr}$ is zero:
$$\frac{dU}{dr} = \frac{6A}{r^7} - \frac{12B}{r^{13}} = 0 \Rightarrow r^6 = \frac{2B}{A} \Rightarrow r = \left(\frac{2B}{A}\right)^{1/6}$$
Substituting $r^6 = \dfrac{2B}{A}$:
$$U = -\frac{A\cdot A}{2B} + \frac{B\cdot A^2}{4B^2} = -\frac{A^2}{2B} + \frac{A^2}{4B} = -\frac{A^2}{4B}$$
Solution by Sreeraj P, M.Sc Physics