Q. 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 $D = \left\lbrack U(x = \infty) - U_{atequilibrium} \right\rbrack,D$ is
A) $\frac{b^{2}}{2a}$
B) $\frac{b^{2}}{12a}$
C) $\frac{b^{2}}{4a}$
D) $\frac{b^{2}}{6a}$
Explanation
(c) U$(x) = \frac{a}{x^{12}} - \frac{b}{x^{6}}$
$U(x = \infty) = 0$
As $F = - \frac{dU}{dx} = - \left\lbrack \frac{12a}{x^{13}} + \frac{6b}{x^{7}} \right\rbrack$
At equilibrium, $F = 0$
$X^{6} = \frac{2a}{b}$
$\therefore U_{at\ equilibrium}\ \ = \frac{a}{\left( \frac{2a}{b} \right)^{2}}\ –\frac{b}{\left( \frac{2a}{b}\ \right)} = - \frac{b^{2}}{4a}$
$\therefore D = \left\lbrack U(x - \infty) - U_{at\ equilibrium}\ \right\rbrack = \frac{b^{2}}{4a}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.