Question

Hey, I'm working with the lennard Jones potential trying to convert it into a form which includes the point of equilibrium (r_0) from the form U(r)=A/r^12 - B/r^6 . Ive reached the point where I've expressed the binding energy in terms of constants A and B but not sure what to do next, any ideas?

Answer 1

The treatment that I've seen involved defining:
\[\sigma = \sqrt[6]{\frac{A}{B}}\]
\[\epsilon = \frac{B^2}{4A}\]
\[A = 4 \epsilon \sigma^{12}, B = 4 \epsilon \sigma^{6}\]

After that, define a "cut off" distance. We used 2.5 sigma:
\[V(2.5 \sigma) = 4 \epsilon ( (\frac{\sigma}{2.5 \sigma})^{12}-(\frac{\sigma}{2.5 \sigma})^6)\]

You then shift the potential by this value (in terms of epsilon), to make it zero at that point. The potential is then defined to be the AB form minus this amount for all r less than the cutoff, and 0 for r above this cutoff.

I think this is what you were asking about. If not, I can see what else I can come up with :)

Answer 2

Thanks man, I'm mot too sure if that's what they are on about though. The exact question is: given that binding energy= -epsilon (ie U(r_0)=-epsilon) show the lennard jones potential cam be written in the form \[U(r)=\epsilon((r_0/r)^{12}-2(r_0/r)^{6})\]