Question

The interaction between atoms of a diatomic molecule is described by the relation $P.E\left(x\right)=\frac{\alpha }{x^{12}}-\frac{\beta }{x^6}$, where P.E. is the potential energy. If the two atoms enjoy stable equilibrium, the distance between them would be :

• A. ${\left(\frac{\alpha }{\beta }\right)}^{1/6}$
• B. ${\left(\frac{2\alpha }{\beta }\right)}^{1/6}$
• C. ${\left(\frac{2\beta }{\alpha }\right)}^{1/6}$
• D. ${\left(\frac{\beta }{\alpha }\right)}^{1/6}$

Answer 1

Answer: (B)

${\left(\frac{2\alpha }{\beta }\right)}^{1/6}$

At stable equilibrium PE is minimum
$\Rightarrow \frac{d}{dx}[P.E.(x\left)\right]=0$ and at equilibrium point $\frac{d^2\left(PE\right)}{d{x}^2}>0$
i.e.$\frac{d}{dx}[\frac{\alpha }{x^{12}}-\frac{\beta }{x^6}]=0$
i.e. $\frac{-12\alpha }{x^{13}}+\frac{6\beta }{x^7}=0$
i.e. $[-2\alpha +\beta x^6]\frac{6}{x^{13}}=0$
i.e. $(-2\alpha +\beta x^6)=0$
i.e. $x={\left(\frac{2\alpha }{\beta }\right)}^{1/6}$