What is the Missing Piece in Solving for ρ in this Simple Problem?

[MadMatSci]

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?

 

[mfb]

I realize that is not correct.

I didn't check the numbers, but the approach is right.
The condition E=0.9 E_Attr 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.

 

[MadMatSci]

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.

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!

 

[haruspex]

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

 

[MadMatSci]

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

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.

 

[haruspex]

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.

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

 

[MadMatSci]

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

Thank you very much for your help!

 

[mfb]

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

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

 

[haruspex]

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

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.