The potential energy of a diatomic molecule i | vedclass

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Question

At equilibrium $\frac{\mathrm{dU}}{\mathrm{dr}}=0$

$\frac{-12 A}{r^{13}}+\frac{6 B}{r^{7}}=0$

$r=\left(\frac{2 A}{B}\right)^{1 / 6}$

Vedclass · Accepted Answer

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

Vedclass · Answer

At equilibrium $\frac{\mathrm{dU}}{\mathrm{dr}}=0$

$\frac{-12 A}{r^{13}}+\frac{6 B}{r^{7}}=0$

$r=\left(\frac{2 A}{B}\right)^{1 / 6}$

The potential energy of a diatomic molecule is given by $U = \frac{A}{{{r^{12}}}} - \frac{B}{{{r^6}}}$ . $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is