# What is the equilibrium bond length for a Lennard-Jones potential?

If I have $$\epsilon$$ and $$\sigma$$ can I calculate equilibrium distance $$r_e$$ in one run? What I have tried is to put $$V = \epsilon$$ and bring out the $$r$$ from the formula, but it seems not solvable. So I am clueless. I need to calculate the equilibrium distance for some project of mine (game-like) with atom interactions.

• This question showed up in the review queue as having been flagged (by someone other than me) for being "low quality" either because of its length or its content. Could you please add what research you've tried to do in order to figure this out yourself, or why you are stumped by this question? Aug 23 at 16:51
• I edited the question. Aug 23 at 16:59
• I noticed about 11 seconds before you wrote the comment, thanks! Aug 23 at 17:00

"The potential minimum is at $$r=r_m = 2^{1/6}\sigma$$."
You can obtain this by calculating the derivative of the Lennard-Jones potential with respect to $$r$$ and finding which $$r$$ value makes that derivative equal to 0.