Given potential energy

$U(r)=-\frac{A}{r^6}+\frac{B}{r^{12}}$

At equilibrium separation, the force must be zero.

Since $F=-\frac{dU}{dr},$ equilibrium requires $\frac{dU}{dr}=0.$

Differentiate $U(r)$ with respect to $r$

$U(r)=-Ar^{-6}+Br^{-12}$

Therefore, $\frac{dU}{dr}=(-A)(-6)r^{-7}+B(-12)r^{-13}$ $\frac{dU}{dr}=\frac{6A}{r^7}-\frac{12B}{r^{13}}$

Set this equal to zero: $\frac{6A}{r^7}-\frac{12B}{r^{13}}=0$

Solve for equilibrium separation

Multiply by $r^{13}$: $6Ar^6-12B=0$ $6Ar^6=12B$ $Ar^6=2B$ $r^6=\frac{2B}{A}$

Hence, $r=\left(\frac{2B}{A}\right)^{1/6}$

Find potential energy at equilibrium

Use $U=-\frac{A}{r^6}+\frac{B}{r^{12}}$

Since $r^6=\frac{2B}{A},$ we get $\frac{1}{r^6}=\frac{A}{2B}$ and $\frac{1}{r^{12}}=\left(\frac{A}{2B}\right)^2=\frac{A^2}{4B^2}$

Substitute: $U=-A\left(\frac{A}{2B}\right)+B\left(\frac{A^2}{4B^2}\right)$ $U=-\frac{A^2}{2B}+\frac{A^2}{4B}$ $U=-\frac{A^2}{4B}$

Match with options

Equilibrium separation: $\left(\frac{2B}{A}\right)^{1/6}$

Potential energy at equilibrium: $-\frac{A^2}{4B}$

This matches Option A.

Comparison with stored answer

Stored correct answer: A

Derived answer: A

So they agree.