The potential energy between two atoms in a m | vedclass

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Question

(D) For stable equilibrium,the net force on the system must be zero,which implies that the derivative of the potential energy with respect to distance

Vedclass · Accepted Answer

$x = \sqrt[6]{\frac{2a}{b}}$

Vedclass · Answer

(D) For stable equilibrium,the net force on the system must be zero,which implies that the derivative of the potential energy with respect to distance must be zero.$F = -\frac{dU}{dx} = 0$Given $U(x) = ax^{-12} - bx^{-6}$,we differentiate with respect to $x$:$\frac{dU}{dx} = -12ax^{-13} + 6bx^{-7} = 0$Rearranging the terms:$12ax^{-13} = 6bx^{-7}$$\frac{12a}{x^{13}} = \frac{6b}{x^7}$$\frac{2a}{b} = \frac{x^{13}}{x^7} = x^6$$x = \sqrt[6]{\frac{2a}{b}}$Thus,the atoms are in equilibrium at $x = \sqrt[6]{\frac{2a}{b}}$.

The potential energy between two atoms in a molecule is given by $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$,where $a$ and $b$ are positive constants and $x$ is the distance between the atoms. The atoms are in stable equilibrium when:

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