Potential energy as a function of $$r$$ is given by $$U = \frac{A}{r^{10}} - \frac{B}{r^5}$$, where $$r$$ is the interatomic distance, $$A$$ and $$B$$ are positive constants. The equilibrium distance between the two atoms will be :

Given the potential energy function $$U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$$ where $$A$$ and $$B$$ are positive constants, the force is the negative derivative of potential energy with respect to $$r$$, so $$F = -\frac{dU}{dr}$$. Computing the derivative gives $$\frac{dU}{dr} = \frac{d}{dr}\left(\frac{A}{r^{10}}\right) - \frac{d}{dr}\left(\frac{B}{r^{5}}\right) = -\frac{10A}{r^{11}} + \frac{5B}{r^{6}}$$. Therefore, $$F = -\frac{dU}{dr} = \frac{10A}{r^{11}} - \frac{5B}{r^{6}}$$.

At equilibrium, the net force is zero ($$F = 0$$), leading to $$\frac{10A}{r^{11}} = \frac{5B}{r^{6}}$$. Multiplying both sides by $$r^{11}$$ yields $$10A = 5B \cdot r^{5}$$, so $$r^{5} = \frac{10A}{5B} = \frac{2A}{B}$$ and hence $$r = \left(\frac{2A}{B}\right)^{\frac{1}{5}}$$. The equilibrium distance between the two atoms is $$\left(\frac{2A}{B}\right)^{1/5}$$. The correct answer is Option C.