Work Energy Power

JEE Main level, single correct, medium

133 of 183

Question 133 of 183

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by U(x)=ax12−bx6U(x)=\frac{a}{x^{12}}-\frac{b}{x^6}, where a and bb are constants and xx is the distance between the atoms. If the dissociation energy of the molecule is D=[U(x=∞)−Uat equilbrium ],DD=\left[U(x=\infty)-U_{\text {at equilbrium }}\right], D is

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Answer: (C) b24a\frac{b^2}{4 a}

U(x)=ax12−bx6U(x=∞)=0\begin{aligned} & U(x)=\frac{a}{x^{12}}-\frac{b}{x^6} \\ & U(x=\infty)=0 \end{aligned}
as, F=−dUdx=−[12ax13+6 bx7]\quad \mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}=-\left[\frac{12 \mathrm{a}}{\mathrm{x}^{13}}+\frac{6 \mathrm{~b}}{\mathrm{x}^7}\right] at equilibrium, F=0\quad F=0
∴x6=2ab∴Uat equiltrium =a(2ab)2−b(2ab)=−b24a∴D=[U(x=∞)−Uat equilibium ]=b24a\begin{array}{ll}\therefore & x^6=\frac{2 a}{b} \\ \therefore & U_{\text {at equiltrium }}=\frac{a}{\left(\frac{2 a}{b}\right)^2}-\frac{b}{\left(\frac{2 a}{b}\right)}=\frac{-b^2}{4 a} \\ \therefore & D=\left[U(x=\infty)-U_{\text {at equilibium }}\right]=\frac{b^2}{4 a}\end{array}

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