Potential energy as a function of r is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}},$ where $r$ is the interatomic distance, $A a n d B$ are positive constants. The equilibrium distance between the two atoms will be :

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${(\frac{2 A}{B})}^{\frac{1}{5}}$

${(\frac{B}{A})}^{\frac{1}{5}}$

${(\frac{A}{B})}^{\frac{1}{5}}$

${(\frac{B}{2 A})}^{\frac{1}{5}}$

answer is C.

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Detailed Solution

At equilibrium,

$\begin{array}{l} \vec{F} = 0 \\ \Rightarrow \frac{- d u}{d r} = 0 \Rightarrow \frac{- d}{d r} (\frac{A}{r^{10}} - \frac{B}{r^{5}}) = 0 \\ \Rightarrow + \frac{A}{r^{11}} \times 10 - \frac{5 B}{r^{6}} = 0 \Rightarrow \frac{1}{r^{6}} [\frac{10 A}{r^{5}} - \frac{5 B}{1}] = 0 \\ _{\Rightarrow \frac{10 A}{r^{5}} = 5 B \Rightarrow r^{5} = \frac{10 A}{5 B} \Rightarrow r = {(\frac{2 A}{B})}^{\frac{1}{5}}} \end{array}$