Question

If the potential energy of two molecules is given by \(\sf U=\frac{A}{r^{12}} - \frac{B}{r^6}\)then at equilibrium position, its potential energy is equal to

Answer 1

We define the equilibrium position as the position in which the attractive force and the repulsion force is equal to 0. that is:
\[\Large F_{total}=F_{repulsion}+F_{attractive}=0\]
The expression you want to evaluate is therefore:
\[\large F=\frac{ dV }{ dr }=0\]
I have just noted \(V\) as the potential energy.
For your function we then get:
\[\large \frac{ d }{ dr }\left( \frac{ A }{ r^n }-\frac{ B }{ r^m } \right)=Bmr ^{-m-1}-Anr ^{-n-1}=0\]
Solve for the the distance \(r\):
\[\large r=\exp \left( \frac{ -\log(A)+\log(B)+\log(m)-\log(n) }{ m-n } \right)\]
Insert the values for \(n\) and \(m\) we can evaluate the distance to:
\[\Large r=\left( \frac{ 2a }{ b } \right)^{\frac{ 1 }{ 6 }}\]
I have discarded the negative solution as it does not make since to talk about negative distances. Or it kinda does if you think about it like this:


Insert the result into your function and evaluate the potential energy:
\[\Large U=\frac{ A }{ \left( \sqrt[6]{\frac{ 2A }{ B }} \right)^{12} }-\frac{ B }{ \left( \sqrt[6]{\frac{ 2A }{ B }} \right)^{6} }\]
I have tried to evaluate the expression to get:
\[\Large U=\frac{ -B^2 }{ 4A }\]

Unsure if I've got the last evaluation right, but I am kinda confident.
Hope this is what you were looking for..

Answer 2

When I wrote
\[\Large r=\left( \frac{ 2a }{ b } \right)^{\frac{ 1 }{ 6 }}\]
I meant:
\[\Large r=\left( \frac{ 2A }{ B } \right)^{\frac{ 1 }{ 6 }}\]
Sorry.

Answer 3

Wow!
Thanks c:

Answer 4

No problem. the values of A and B I can obviously not help with. :)

Answer 5

No, actually I wanted in terms of A and B only...