JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Work, Power & Energy

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 and B are positive constants. The equilibrium distance between the two atoms will be :

(A) $(\frac{A}{B})^{\frac{1}{5}}$
(B) $(\frac{B}{A})^{\frac{1}{5}}$
(C) $(\frac{2 A}{B})^{\frac{1}{5}}$ ✓ Correct
(D) $(\frac{B}{2 A})^{\frac{1}{5}}$

Answer: (C)

Solution

At equilibrium, force

F = - \frac{d U}{d r} = 0

. Taking the derivative:

\frac{d U}{d r} = - \frac{10 A}{r ^{11}} + \frac{5 B}{r ^{6}} = 0

. Solving:

\frac{10 A}{r ^{11}} = \frac{5 B}{r ^{6}}

, which gives

r^{5} = \frac{10 A}{5 B} = \frac{2 A}{B}

. Therefore,

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

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option

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