The potential energy function for the force between two atoms in a diatomic molecule is approximate given by `U(r) = (a)/(r^(12)) - (b)/(r^(6))`, where `a` and `b` are constants and `r` is the distance between the atoms. If the dissociation energy of the molecule is `D = [U (r = oo)- U_("at equilibrium")],D` is

Options

A. `(b^(2))/(2a)`
B. `(b^(2))/(12 a)`
C. `(b^(2))/(4a)`
D. `(b^(2))/(6a)`

Answer

C

Explanation

`U = (a)/(x^(12)) -(b)/(x^(6))`
Force, `F = -(dU)/(dx) = -(d)/(dx) ((a)/(x^(12)) -(b)/(x^(6)))`
=`-[(-12a)/(x^(13)) + (6b)/(x^(7))]=[(12a)/(x^(13)) -(6b)/(x^(7))]`
At equilibrium, `F = 0`
`:. (12 a)/(x^(13)) -(6b)/(x^(7)) = 0` or `x^(6) = (2a)/(b)`
`U_("at equilibrium") = (a)/(((2a)/(b))^(2)) -(b)/(((2a)/(b)))`
=`(ab^(2))/(4a^(2)) -(b^(2))/(2a) =(b^(2))/(4a) -(b^(2))/(2a) =-(b^(2))/(4a)`
`U_((x==infty)) =0`
`D =U_((x=infty)) - U_("at equilibrium")`
=`[0-(-(b^(2))/(4a))] = (b^(2))/(4a)`.