Question

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$,where a and b are constant and x is the distance between the atoms,if the disassociation energy of the molecule is $D=\left[U\right(x=\infty )-U_{atequilibrium}]$,D is

• A. $\frac{b^2}{6a}$
• B. $\frac{b^2}{2a}$
• C. $\frac{b^2}{12a}$
• D. $\frac{b^2}{4a}$

Answer 1

The correct option is D $\frac{b^2}{4a}$

$U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$

At $x=\infty ;U\left(\infty \right)=0$

At equilibrium :$\frac{du}{dx}=0$

$-12a{x}^{-13}+6b{x}^{-7}=0$

$\frac{-12a}{x^{13}}=\frac{-6b}{x^7}$

$x=\sqrt{\frac{2a}{b}}$

$U$ at equilibrium ;$U=\frac{a}{x^6.x^6}-\frac{b}{x^6}$

$=\frac{a-b\left(\frac{2a}{b}\right)}{(\frac{2a}{b}{)}^2}$

$=\frac{-b^2}{4a}$

$D=[0+\frac{b^2}{4a}]=\frac{b^2}{4a}$