# Homework Help: 2 equations, 3 unknowns?

1. Sep 3, 2015

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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?

Last edited: Sep 3, 2015
2. Sep 3, 2015

### Staff: Mentor

I didn't check the numbers, but the approach is right.
The condition E=0.9 EAttr is satisfied at r0 only.

You have three unknowns for two equations, but it is neither possible nor necessary to find A and B separately. You can find the ratio A/B, as both equations depend on this and ρ only.

3. Sep 3, 2015

Wow. Thank you. I did not realize that E=.9 EAttr is valid at r0 only. I guess it does make sense though since at equilibrium it is possible to determine what fraction of the total energy is attractive and repulsive. Thanks again!!

4. Sep 3, 2015

### haruspex

Why do you say that answer is clearly not correct? The method is sound, and I can see no numerical errors.

5. Sep 3, 2015

I did not think that I could just assume r=1.5. I thought it had to be valid for all r until mfb indicated otherwise.

6. Sep 3, 2015

### haruspex

The equation in question is valid for all r. In particular, it is valid at r=1.5 A.

7. Sep 3, 2015

Thank you very much for your help!

8. Sep 3, 2015

### Staff: Mentor

The equation that says the total energy is 90% of the attractive potential energy? No, that is not correct everywhere.

9. Sep 3, 2015

### haruspex

I'm fairly sure MadMatSci was referring to the general equation quoted first off: $E=−\frac A{r^6}+B∗e^{−rρ}$. The value of 1.5 for r had already been substituted in the 90% equation.