1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

  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.).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook