Question

If the potential energy of two molecules is given by, U=A/r^12–B/r^6 then at equilibrium position, its potential energy is equal to :

Answer 1

Answer:

-B^2/4A

Explanation:

Refer to the material.

Answer 2

Answer:

At equilibrium position, potential energy of two molecules is equal to $U=-\dfrac{B^2}{4A}$ .

Explanation:

Given :

$U=\dfrac{A}{r^{12}}-\dfrac{B}{r^6}$           ...... ( 1 )

We know , force is given by :

$F=-\dfrac{dU}{dr}\\\\F=-\dfrac{d(\dfrac{A}{r^{12}}-\dfrac{B}{r^6})}{dr}\\\\F=-(-12\dfrac{A}{r^{13}}+6\dfrac{B}{r^7})\\\\F=12\dfrac{A}{r^{13}}-6\dfrac{B}{r^7}$

Now , for equilibrium , F = 0

So , $12\dfrac{A}{r^{13}}-6\dfrac{B}{r^7}=0$

Therefore , $r^6=\dfrac{2A}{B}$ and $r^{12}=\dfrac{4A^2}{B^2}$ .

Putting value of $r^6$ and $r^{12}$ in equation 1 .

We get :

$U=-\dfrac{B^2}{4A}$

Hence , this is the required solution .