Question

The potential energy function for the force between two atoms in a diatomic molecule which is approximately given as,
$U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$. Where $a\&b$ are constants and $x$ is the distance between the atoms. The disassociation energy of the molecule, $D=\left[U\right(x=\infty )-U_{equilibrium}]$ is

• 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 $\frac{b^2}{4a}$

The potential energy of the diatomic molecule is given by,

$U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}....\left(1\right)$

$\Rightarrow Atx=\infty i.e{U}_{\infty }=0$

Using the relation between force and potential energy, we get,

$\Rightarrow F=-\frac{dU}{dx}=-\frac{d}{dx}[\frac{a}{x^{12}}-\frac{b}{x^6}]$

$=-[\frac{12a}{x^{13}}-\frac{6b}{x^7}]$

At equilibrium, i.e $F=0$

$\Rightarrow \frac{12a}{x^{13}}-\frac{6b}{x^7}=0$

$\Rightarrow \frac{12a}{x^{13}}=\frac{6b}{x^7}$

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

On putting this value in (1) we get,

$\therefore U_{equilibrium}=\frac{a}{{\left(\frac{2a}{b}\right)}^2}-\frac{b}{\left(\frac{2a}{b}\right)}$

$\therefore D=\left[U\right(x=\infty )-U_{equilibrium}]$

$\Rightarrow D=0-[\frac{a{b}^2}{4a^2}-\frac{b^2}{2a}]=\frac{b^2}{4a}$

Hence, $\left(C\right)$ is the correct answer.