If the potential energy between two molecules is given by U-Ar6-Br12- then at equilibrium- separation between molecules and the potential energy respectively are-

Given : U= Ar^6+Br^12 For equilibrium, F=dUdr=0 ⇒ (A( 6r^ 7))+B( 12r^ 13)=0 ⇒ 0=6Ar^7 12Br^13⇒6A12B=1r^6 ∴ Separation between molecules, r=(2BA)^1/6 Also, Po ...

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Standard XII

Physics

Gravitational Potential Energy of a Two Mass System

If the potent...

Question

If the potential energy between two molecules is given by $U = - \frac{A}{r^{6}} + \frac{B}{r^{12}},$ then at equilibrium, separation between molecules and the potential energy respectively are:

{(\frac{B}{2 A})}^{\frac{1}{6}}, - (\frac{A^{2}}{2 B})

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{(\frac{B}{A})}^{\frac{1}{6}}, 0

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{(\frac{2 B}{A})}^{\frac{1}{6}}, - (\frac{A^{2}}{4 B})

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{(\frac{2 B}{A})}^{\frac{1}{6}}, (\frac{A^{2}}{2 B})

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Solution

The correct option is C ${(\frac{2 B}{A})}^{\frac{1}{6}}, - (\frac{A^{2}}{4 B})$
Given : $U = - \frac{A}{r^{6}} + \frac{B}{r^{12}}$
For equilibrium,
$F = \frac{d U}{d r} = 0$
$\Rightarrow - (A (- 6 r^{- 7})) + B (- 12 r^{- 13}) = 0$
$\Rightarrow 0 = \frac{6 A}{r^{7}} - \frac{12 B}{r^{13}} \Rightarrow \frac{6 A}{12 B} = \frac{1}{r^{6}}$
$∴$ Separation between molecules, $r = {(\frac{2 B}{A})}^{1 / 6}$

Also, Potential energy,
$U at (r = {(\frac{2 B}{A})}^{1 / 6}) = - \frac{A}{2 B / A} + \frac{B}{4 B^{2} / A^{2}}$
$= \frac{- A^{2}}{2 B} + \frac{A^{2}}{4 B} = - (\frac{A^{2}}{4 B})$

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If the potential energy between two molecules is given by U=−Ar6+Br12, then at equilibrium, separation between molecules and the potential energy respectively are:

The correct option is C (2BA)16,−(A24B)Given : U=−Ar6+Br12 For equilibrium, F=dUdr=0 ⇒ −(A(−6r−7))+B(−12r−13)=0 ⇒0=6Ar7−12Br13⇒6A12B=1r6 ∴ Separation between molecules, r=(2BA)1/6 Also, Potential energy, U at (r=(2BA)1/6)=−A2B/A+B4B2/A2 =−A22B+A24B=−(A24B)

If the potential energy between two molecules is given by $U = - \frac{A}{r^{6}} + \frac{B}{r^{12}},$ then at equilibrium, separation between molecules and the potential energy respectively are: