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 are :

• A. ${\left(\frac{B}{A}\right)}^{1/6},0$
• B. ${\left(\frac{B}{2A}\right)}^{1/6},-\frac{A^2}{2B}$
• C. ${\left(\frac{2B}{A}\right)}^{1/6},-\frac{A^2}{4B}$
• D. ${\left(\frac{2B}{A}\right)}^{1/6},-\frac{A^2}{2B}$

Answer 1

Answer: (C)

$U=\frac{-A}{r^6}+\frac{B}{r^{12}}$
$F=-\frac{dU}{dr}=-\left(A\left(-6r^{-7}\right)\right)+B\left(-12r^{-13}\right)$
$0=\frac{6A}{r^7}-\frac{12B}{r^{13}}$
$\frac{6A}{12B}=\frac{1}{r^6}\Rightarrow r={\left(\frac{2B}{A}\right)}^{1/6}$
$U\left(r={\left(\frac{2B}{A}\right)}^{1/6}\right)=-\frac{A}{2B/A}+\frac{B}{4B^2/A^2}$
$=\frac{-A^2}{2B}+\frac{A^2}{4B}=\frac{-A^2}{4B}$