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Homework Statement
The total energy for a Van der Waals bonded solid is:
E=-\frac{A}{r^6}+B*e^{-\frac{r}{\rho}}
where the 1st and 2nd terms are for attraction and repulsion respectively, A and B are constants, r is the inter-atomic distance, and \rho is the range of interaction (i.e. characteristic length)
Given E=.9*E_{attr.} and Eqm Spacing, r_o= 1.5 angstroms, find \rho in angstroms
Homework Equations
The Attempt at a Solution
The first piece of information, E=.9*E_{attr.}, gives me:
E=-\frac{A}{r^6}+B*e^{-\frac{r}{\rho}}=.9*E_{attr.}=.9*(-\frac{A}{r^6})
Rearranging yields:
.1*\frac{A}{r^6}=B*e^{\frac{-r}{\rho}}
The second piece of information leads me to assume that at r=1.5 the derivative of the expression for total energy=0. This gives me the following equation:
6\frac{A}{1.5^7}=\frac{B}{\rho}e^{\frac{-1.5}{\rho}}
From here I am confused. I first tried to solve .1*\frac{A}{r^6}=B*e^{\frac{-r}{\rho}} for A and then substitute that into 6\frac{A}{1.5^7}=\frac{B}{\rho}e^{\frac{-1.5}{\rho}}.
Seeing no way out, I set the r's in the equation for A to 1.5. This led to:
\frac{60}{1.5}B*e^{\frac{-1.5}{\rho}}=\frac{B}{\rho}*e^{\frac{-1.5}{\rho}} which leads to:
\rho=\frac{1.5}{60}=.025
It does not seem valid to just sub in r=1.5. This was a shot in the dark as I am out of ideas.
I spoke to my instructor about this and how I did not see a way to solve for \rho because it looks as though there are 2 equations and 3 unknowns. His response was that this is a simple problem and that I do not need to know the constants A and B. I must be over thinking this. Does anyone see what I am missing?
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