Question

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

Answer 1

$F=-\frac{d U}{d r}$
$F=-\left[- \frac{10 \alpha }{r^{11}} + \frac{5 \beta }{r^{6}}\right]$
for equilibrium, $F=0$
$\frac{10 \alpha }{r^{11}}=\frac{5 \beta }{r^{6}}$
$\frac{2 \alpha }{\beta }=r^{5}$
$r=\left(\frac{2 \alpha }{\beta }\right)^{1 / 5}$
$a=1$