For stable equilibrium, the net force acting on the particles must be zero, and the potential energy must be at a minimum. The force $F$ is related to the potential energy $U(x)$ by the equation: $F = -\frac{dU}{dx}$ Given the potential energy function $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6} = ax^{-12} - bx^{-6}$, we differentiate it with respect to $x$: $\frac{dU}{dx} = -12ax^{-13} - (-6)bx^{-7} = -\frac{12a}{x^{13}} + \frac{6b}{x^7}$ For equilibrium, the force $F = 0$, which means $\frac{dU}{dx} = 0$: $-\frac{12a}{x^{13}} + \frac{6b}{x^7} = 0$ $\frac{6b}{x^7} = \frac{12a}{x^{13}}$ $x^6 = \frac{12a}{6b} = \frac{2a}{b}$ $x = \sqrt{\frac{2a}{b}}$ To confirm it is a point of stable equilibrium, the second derivative $\frac{d^2U}{dx^2}$ must be positive (indicating a potential energy minimum). $\frac{d^2U}{dx^2} = 156ax^{-14} - 42bx^{-8}$ Substituting $x^6 = \frac{2a}{b}$ into this derivative yields a positive value, confirming stable equilibrium. Thus, the atoms are in stable equilibrium when $x = \sqrt{\frac{2a}{b}}$.