L-J Potential-Distance of equilibrium

[Pahoo]

LJ-Potential

Homework Statement

Find distance of equilibrium for:
1.
Uvdw^(r)=-A/r⁶+B/r¹²
A=10^-77 Jm⁶
B=10^-134 Jm¹²

Homework Equations

Uvdw^(r)=-A/r⁶+B/r¹²

The Attempt at a Solution

It is suppose that r is the distance, but I have not any idea how to calculate this.

appreciate your help

 

[berkeman]

Homework Statement

Find distance of equilibrium for:
1.
Uvdw^(r)=-A/r⁶+B/r¹²
A=10^-77 Jm⁶
B=10^-134 Jm¹²

Homework Equations

Uvdw^(r)=-A/r⁶+B/r¹²

The Attempt at a Solution

It is suppose that r is the distance, but I have not any idea how to calculate this.

appreciate your help

This article may help: http://en.wikipedia.org/wiki/Lennard-Jones_potential

This isn't really an engineering question, though. Would you like me to move this to the HH/Advanced Physics forum?

 

[Pahoo]

Homework Statement

Find distance of equilibrium for:
1.
Uvdw(r)=-A/r⁶+B/r¹²
A=10^-77 Jm⁶
B=10^-134 Jm¹²

Homework Equations

Uvdw(r)=-A/r⁶+B/r¹²

The Attempt at a Solution

It is suppose that r is the distance, I suppose that r=(2A/B)^(1/6)
and 0=2A-Br⁶.
However I`m totally lost

appreciate your help

 

[Pahoo]

Thanks

 

[mfb]

The equilibrium distance is the minimum of the potential. How do you find a minimum?

 

[Pahoo]

ok, finally I figure out how to calculate the minimun which is r = (2^(1/6) B^(1/6))/A^(1/6). Am I right?

 

[mfb]

Looks good. I would express it as $\displaystyle \left(\frac{2B}{A}\right)^{\frac{1}{6}}$