Question

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$, where $a$ and $b$ are constants and $x$ is the distance between the atoms. lf the dissociation energy of the molecule is $D=\left[U\right(x=\infty )-U_{atequilibrium}],$ then $D$ is equal to

• A. $\frac{b^2}{2a}$
• B. $\frac{b^2}{12a}$
• C. $\frac{b^2}{4a}$
• D. $\frac{b^2}{6a}$

Answer 1

The correct option is C $\frac{b^2}{4a}$

$U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$

$U(x=\infty )=0$

As, $F=-\frac{dU}{dx}=-[\frac{12a}{x^{13}}+\frac{6b}{x^7}]$ at equilibrium, $F=0$

$\therefore x^6=\frac{2a}{b}$

$\therefore U_{atequilibrium}=\frac{a}{(\frac{2a}{b}{)}^2}-\frac{b}{\left(\frac{2a}{b}\right)}=\frac{-b^2}{4a}$
$.\cdot .D=\left[U\right(x=\infty )-U_{atequilibrium}]=\frac{b^2}{4a}$