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Atom distance and repulsive energy

  1. Jun 4, 2013 #1
    I'm had this class yesterday that I did'nt understand anything, I have exam on two days and I started to study by myself about potential energy so, I found this equation:

    Ue(r) = q1 q2/4πε0r + repulsive energy (same as van der Waals)

    where q1 = -q2 and each q = + or - 1.6E-19 C
    and ε0 = permittivity of free space = 8.854E-12 C^2/J m. (C = coulomb)

    I did the following calculation for measure the distance of separation at the minimun energy
    Ue(r)=(1.6E-19 C)(-1.6E-19 C)/4*3.1416* 8.854E-12 C^2/J m
    =1.95738 Jm

    So, I am wondering what would be the repulsive energy that the book mention and how could I find the distance of equilibrium?

    Thanks
     
  2. jcsd
  3. Jun 4, 2013 #2
    From your post, it is not clear exactly what problem you are trying to solve, but in general:

    (a) You need an explicit expression for the potential energy in terms of the internuclear distances or lattice constant in your problem;
    (b) You then find the configuration of minimum energy (corresponding to dU/dr = 0 for a diatomic, or dU/da = 0 for a cubic crystal with lattice constant a, etc.).
     
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