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2 equations, 3 unknowns?

  1. Sep 3, 2015 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations


    3. 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: Sep 3, 2015
  2. jcsd
  3. Sep 3, 2015 #2

    mfb

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    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.
     
  4. Sep 3, 2015 #3
    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!!
     
  5. Sep 3, 2015 #4

    haruspex

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    Why do you say that answer is clearly not correct? The method is sound, and I can see no numerical errors.
     
  6. Sep 3, 2015 #5
    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.
     
  7. Sep 3, 2015 #6

    haruspex

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    The equation in question is valid for all r. In particular, it is valid at r=1.5 A.
     
  8. Sep 3, 2015 #7
    Thank you very much for your help!
     
  9. Sep 3, 2015 #8

    mfb

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    The equation that says the total energy is 90% of the attractive potential energy? No, that is not correct everywhere.
     
  10. Sep 3, 2015 #9

    haruspex

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    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.
     
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