Question

If the potential energy of two molecules is give by, $U=\frac{A}{r^{12}}-\frac{B}{r^6}$ then at equilibrium position, its potential energy is equal to?

• A. $\frac{A^2}{4B}$
• B. $-\frac{B^2}{4A}$
• C. $(\frac{2A}{B}{)}^{\frac{1}{6}}$
• D. $3A$

Answer 1

The correct option is C $(\frac{2A}{B}{)}^{\frac{1}{6}}$

for equilibrium positions $\frac{dU}{dR}=-F$ must be zero

$U=\frac{A}{r^{12}}-\frac{B}{r^6}$

$\frac{dU}{dR}=\frac{-12A}{r^{13}}+\frac{6B}{r^7}=0$

$\frac{B}{r^7}=\frac{2A}{r^{13}}$

$r^6=\frac{2A}{B}$

$r={\left(\frac{2A}{B}\right)}^{\frac{1}{6}}$