91Ó°ÊÓ

The equation of potential energy of diatomic molecule is,

$U=\frac{A}{r^{12}}-\frac{B}{r^6}$

Use the equation of force relating to the potential energy to determine${\mathit{r}}_{\mathbf{e}\mathbf{q}}$. Use the condition for minima to check whether the potential energy at${\mathit{r}}_{\mathbf{eq}}$is minimum or maximum. Then, state the nature oftheforce directly from the property of minima.

Formula is as follow:

$F=-\frac{dU}{dr}$

The equation for the force relating with potential energy function is,

$F=-\frac{dU}{dr}$

So, applying this equation for$r=r_{eq}$

$$\begin{gathered} F=-\frac{12A}{r^{13}}+\frac{6B}{r^7} \\ =0 \end{gathered}$$

Solving this equation,

$r_{eq}=1.12{\left(\frac{A}{B}\right)}^{\frac{1}{6}}$

Hence,the distance of equilibrium separation between the atoms is, $r_{eq}=1.12{\left(\frac{A}{B}\right)}^{\frac{1}{6}}$.

Now, check whether the value of$r_{eq}$has minimum or maximum potential energy.

So, to check it, take the double derivative ofthepotential energy function,

${\left[\frac{d^{2}U}{d{r}^2}\right]}_{r=r_{eq}}=\frac{dF}{dr}=\frac{12×13A}{r^{14}}-\frac{6×7B}{r^8}>0$

This proves that, the equilibrium separation has minimum potential energy.

So, for the$r<r_{eq}$ the force will be positive or repulsive.

As calculated in part b, as $r=r_{eq}$is minimum, the force for $r>r_{eq}$will be negative or attractive.

Therefore, the equation for force relating potential energy and the condition of minima can be used to solve this problem.