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

u=−Arb+Br12F=−dudrF=−6Ar7+12Br13At equilibrium F=06r7(−A+2Br6)=0R=(2BA)1/6Substituting in given equationU=−A24B

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

u=−Arb+Br12F=−dudrF=−6Ar7+12Br13At equilibrium F=06r7(−A+2Br6)=0R=(2BA)1/6Substituting in given equationU=−A24B

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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 are

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

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

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

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

answer is C.

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Detailed Solution

$u = - \frac{A}{r^{b}} + \frac{B}{r^{12}}$

$F = - \frac{d u}{d r}$

$F = - \frac{6 A}{r^{7}} + \frac{12 B}{r^{13}}$

At equilibrium $F = 0$

$\frac{6}{r^{7}} (- A + \frac{2 B}{r^{6}}) = 0$

$R = {(\frac{2 B}{A})}^{1 / 6}$

Substituting in given equation

$U = \frac{- A^{2}}{4 B}$

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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 are

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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 are