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Zero-Point Energy of a Debye Solid

  1. Apr 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Show that the zero point energy of a Debye solid is: (9/8) N k (Debye Temp)

    2. Relevant equations

    zero point energy = sum of (1/2 w hbar ) over all i

    3. The attempt at a solution

    I used the Debye spectrum g(w) = 9N w^2 / w(D)^3, for w < w(D)
    g(w) = 0 for w>w(D)

    Then I integrated g(w)dw from 0 to w(D) to get 3N modes of vibration for each N atoms

    Inputting 3N modes of vibration into the solution, I know that the mean zero-point energy for each of these 3N modes is (3/8) k (Debye Temp)

    Debye T = hbar w(D)/k --> so need to show that the zero-pt energy is (9/8) N Hbar w(D)

    So: .5 hbar integral w(i) from 0 to w(D) = (9/8) N hbar w(D)

    so: integral w(i) from 0 to w(D) = (.75 w(D)) (3N)

    I don't know where to go from here. Thank you!
    Last edited: Apr 7, 2008
  2. jcsd
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