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

Properties of Matter PYQ — JEE Main 2022 · 24 Jun · Shift 2 · Q55 with Solution

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Correct answer: C

For equilibrium separation, the potential energy must be minimum, so

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

Given,

$U=\frac{A}{r^{10}}-\frac{B}{r^5}$

$\frac{dU}{dr}=A\frac{d}{dr}(r^{-10})-B\frac{d}{dr}(r^{-5})$

$\frac{dU}{dr}=-10Ar^{-11}+5Br^{-6}$

Set this equal to zero:

$-10Ar^{-11}+5Br^{-6}=0$

Multiply by $r^{11}$ :

$-10A+5Br^5=0$

$5Br^5=10A$

$Br^5=2A$

$r^5=\frac{2A}{B}$

Hence,

$r=\left(\frac{2A}{B}\right)^{1/5}$

Second derivative:

$\frac{d^2U}{dr^2}=110Ar^{-12}-30Br^{-7}$

At equilibrium, using $Br^5=2A$ :

$\frac{d^2U}{dr^2}=\frac{1}{r^{12}}\left(110A-30Br^5\right)$

$=\frac{1}{r^{12}}\left(110A-30(2A)\right)=\frac{50A}{r^{12}}>0$

So the point is indeed a stable equilibrium.

$r=\left(\frac{2A}{B}\right)^{1/5}$

This corresponds to Option C.