Question:
The potential energy (U) of a diatomic molecule is a function dependent on $r$ (interatomic distance)
as $\mathbf{U}=\frac{\alpha}{r^{10}}-\frac{\beta}{r^{5}}-3$ Where, $\mathbf{a}$ and $\mathbf{b}$ are positive constants. The equilibrium distance between two
atoms will $\left(\frac{2 \alpha}{\beta}\right)^{\frac{a}{b}} \cdot$ Where $\mathbf{a}=$
Solution:
(1)
$F=-\frac{d U}{d r}$
$F=-\left[-\frac{10 \alpha}{r^{11}}+\frac{5 \beta}{r^{6}}\right]$
for equilibrium, $\mathrm{F}=0$
$\frac{10 \alpha}{r^{11}}=\frac{5 \beta}{r^{8}}$
$\frac{2 \alpha}{\beta}=r^{5}$
$r=\left(\frac{2 \alpha}{\beta}\right)^{1 / 5}$
$a=1$
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