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. If the dissociation energy of the molecule is $D=[U_{x=\infty }-U_{\text{at equilibrium}}]$, $D$ 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 $\frac{b^2}{4a}$

$U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$
At $x\to \infty ;U\left(\infty \right)=0$

At equilibrium: $\frac{dU}{dx}=0$
$-12a{x}^{-13}+6b{x}^{-7}=0$
$\frac{-12a}{x^{13}}=\frac{-6b}{x^7}$
$x^6=\frac{2a}{b}$

At equilibrium, $U=\frac{a}{x^6.x^6}-\frac{b}{x^6}$
$=\frac{a-b\left(\frac{2a}{b}\right)}{{\left(\frac{2a}{b}\right)}^2}$
$=\frac{-b^2}{4a}$

$D=[U_{\infty }-U_{\text{at equilibrium}}]$
$=[0+\frac{b^2}{4a}]=\frac{b^2}{4a}$