The potential energy funtions for the force between two along in a distance molecule is approximately given by
`U(x) = (a)/(x^(12)) - (b)/(x^(6)) ` where `a` and `b` are constant and `x` is the distance between the aloms , if the discision energy of the molecale is `D = [U(x = oo) - U` atequlibrium ] , D is

Options

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

Answer

B

Explanation

`U_((x))-(a)/(x^(12))-(b)/(x^(6))`
`F=-(dU(x))/(dx)=-(12a)/(x^(13))+(6b)/(x^(7))`
At equilibrium distance between the molecules,
`F=0`
`:. -(12a)/(x^(13))+(6b)/(x^(7))=0` or `a=((2a)/(b))^(1//6)`
From `(i)` `U_((oo))=(a)/(oo^(12))-(b)/(oo^(6))=0`
Dissociation energy of molecule,
`D=U_((x=oo))-U_("at equilibrium")`
`=0-((a)/(x^(12))-(b)/(x^(6)))` where`x=((2a)/(b))^(1//6)`
`D=(-a)/(x^(12))+(b)/(x^(6))=-(a)/((2a//b)^(12//6))+(b)/((2a//b)^(6//6))`
`=-(b^(2))/(4a)+(b^(2))/(2a)=(b^(2))/(4a)`