MadMatSci
- 4
- 0
Homework Statement
The total energy for a Van der Waals bonded solid is:
[itex]E=-\frac{A}{r^6}+B*e^{-\frac{r}{\rho}}[/itex]
where the 1st and 2nd terms are for attraction and repulsion respectively, A and B are constants, r is the inter-atomic distance, and [itex]\rho[/itex] is the range of interaction (i.e. characteristic length)
Given [itex]E=.9*E_{attr.}[/itex] and Eqm Spacing, [itex]r_o[/itex]= 1.5 angstroms, find [itex]\rho[/itex] in angstroms
Homework Equations
The Attempt at a Solution
The first piece of information, [itex]E=.9*E_{attr.}[/itex], gives me:
[itex]E=-\frac{A}{r^6}+B*e^{-\frac{r}{\rho}}=.9*E_{attr.}=.9*(-\frac{A}{r^6})[/itex]
Rearranging yields:
[itex].1*\frac{A}{r^6}=B*e^{\frac{-r}{\rho}}[/itex]
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:
[itex]6\frac{A}{1.5^7}=\frac{B}{\rho}e^{\frac{-1.5}{\rho}}[/itex]
From here I am confused. I first tried to solve [itex].1*\frac{A}{r^6}=B*e^{\frac{-r}{\rho}}[/itex] for A and then substitute that into [itex]6\frac{A}{1.5^7}=\frac{B}{\rho}e^{\frac{-1.5}{\rho}}[/itex].
Seeing no way out, I set the r's in the equation for A to 1.5. This led to:
[itex]\frac{60}{1.5}B*e^{\frac{-1.5}{\rho}}=\frac{B}{\rho}*e^{\frac{-1.5}{\rho}}[/itex] which leads to:
[itex]\rho=\frac{1.5}{60}=.025[/itex]
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 [itex]\rho[/itex] 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?
Last edited: