Q80 / 90 · JEE Main 2010 · Physics
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U(x)=\frac{a}{x^{12}}-\frac{b}{x^6}$, where a and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $D=\left[U(x=\infty)-U_{\text {at equilbrium }}\right], D$ is
- A$\frac{b^2}{2 a}$
- B$\frac{b^2}{12 a}$
- Correct answerC$\frac{b^2}{4 a}$
- D$\frac{b^2}{6 a}$
Solution
$$
\begin{aligned}
& U(x)=\frac{a}{x^{12}}-\frac{b}{x^6} \\
& U(x=\infty)=0
\end{aligned}
$$
as, $\quad \mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}=-\left[\frac{12 \mathrm{a}}{\mathrm{x}^{13}}+\frac{6 \mathrm{~b}}{\mathrm{x}^7}\right]$ at equilibrium, $\quad F=0$
$\begin{array}{ll}\therefore & x^6=\frac{2 a}{b} \\ \therefore & U_{\text {at equiltrium }}=\frac{a}{\left(\frac{2 a}{b}\right)^2}-\frac{b}{\left(\frac{2 a}{b}\right)}=\frac{-b^2}{4 a} \\ \therefore & D=\left[U(x=\infty)-U_{\text {at equilibium }}\right]=\frac{b^2}{4 a}\end{array}$
Hence, the correct answer is option: 'C'.