## Abhisar one year ago 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

1. Frostbite

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: |dw:1440933102863:dw| 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..

2. Frostbite

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.

3. Abhisar

Wow! Thanks c:

4. Frostbite

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

5. Abhisar

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