Question

The potential energy$\left(\mathrm{U}\right)$ of a diatomic molecule is a function dependent on r (interatomic distance) as $\mathrm{U}=\frac{\mathrm{\alpha }}{{\mathrm{r}}^{10}}-\frac{\mathrm{\beta }}{{\mathrm{r}}^5}-3$. Where, $\mathrm{a}$ and $\mathrm{b}$ are positive constants. The equilibrium distance between two atoms will ${\left(\frac{2\mathrm{\alpha }}{\mathrm{\beta }}\right)}^{\raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{b}$}\right.}$. Where $\mathrm{a}$ =_______

Answer 1

Step 1: Given data

• Potential energy $\mathrm{U}=\frac{\mathrm{\alpha }}{{\mathrm{r}}^{10}}-\frac{\mathrm{\beta }}{{\mathrm{r}}^5}-3$
• Equilibrium distance=${\left(\frac{2\mathrm{\alpha }}{\mathrm{\beta }}\right)}^{\raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{b}$}\right.}$

Step 2: Computing the force

We know that the change in potential energy with respect to distance applies a force on the object. So,

$\begin{array}{rcl}\mathrm{F}& =& -\frac{\mathrm{dU}}{\mathrm{dr}}\\ & =& -\left[-10{\mathrm{\alpha r}}^{-11}+5{\mathrm{\beta r}}^{-6}\right]\end{array}$

For the equilibrium $\mathrm{F}$ must be zero. So,

$\begin{array}{rcl}-\left[-10{\mathrm{\alpha r}}^{-11}+5{\mathrm{\beta r}}^{-6}\right]& =& 0\\ 10{\mathrm{\alpha r}}^{-11}& =& 5{\mathrm{\beta r}}^{-6}\\ \frac{2\mathrm{\alpha }}{\mathrm{\beta }}& =& {\mathrm{r}}^5\\ \mathrm{r}& =& {\left(\frac{2\mathrm{\alpha }}{\mathrm{\beta }}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\\ & & \end{array}$

Hence, the value of $\mathrm{a}$ will be $1$.