Question

Potential energy function describing the interaction between two atoms of a diatomic molecule is $U\left(r\right)=\frac{a}{r^{12}}-\frac{b}{r^6}$. Force acting between them will be zero when the distance between them would be

• A. ${\left(\frac{2a}{b}\right)}^{\frac{1}{6}}$
• B. ${\left(\frac{b}{2a}\right)}^{\frac{1}{6}}$
• C. ${\left(\frac{a}{b}\right)}^{\frac{1}{6}}$
• D. ${\left(\frac{b}{a}\right)}^{\frac{1}{6}}$

Answer 1

The correct option is A ${\left(\frac{2a}{b}\right)}^{\frac{1}{6}}$

Given that,

Potential energy $U\left(r\right)=\frac{a}{r^{12}}-\frac{b}{r^6}$

Now, for equilibrium positions

$\frac{dU}{dr}=-F=0$

Now, put the value of U

$\frac{-12a}{r^{13}}-\frac{6b}{r^7}=0$

$\frac{b}{r^7}=\frac{2a}{r^{13}}$

$r^6=\frac{2a}{b}$

$r={\left(\frac{2a}{b}\right)}^{\frac{1}{6}}$

Hence, the distance is ${\left(\frac{2a}{b}\right)}^{\frac{1}{6}}$