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Photoelectric absorption in semiconductors

  1. Feb 3, 2012 #1
    Hi everybody. I'm new here and, first of all, sorry for my bad english :-D

    I'm studying photoelectric absorption in semiconductors.
    The book (and professor too) says that, in the conservation law:

    ki + kph = kf

    (where ki and kf are wave vectors of initial and final electron state, and kph is the wave vector of incoming photon) we can neglect kph because it is ≈ 2π/λ, whereas ki and kf are ≈ 2π/a, and λ>>a. (a is the length of unitary cell in real space).
    But I think that this assumption is good only if electron is at the edge of Brillouin Zone; if the initial and final electrons are near [itex]\Gamma[/itex]-point, they should have a very little wave vector, comparable with kph, making the approximation not valid.

    what is wrong in my words?
  2. jcsd
  3. Feb 3, 2012 #2


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    Just do the math once and sketch the photon dispersion (energy versus k) and the electron dispersion (crystal electron or for simplicity even a free one) into the same graph with the correct dimensions. This is pretty instructive and will give you a good argument for why the approximation your professor gave you is a very good one pretty much everywhere.
  4. Feb 3, 2012 #3
    So, I think having understood from your words, the key is that the conservation of momentum has to be combined with the conservation of energy?
  5. Feb 3, 2012 #4


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    Actually, that conservation law is valid only for the in-plane momentum, i.e. parallel to the surface of the material. The out-of-plane momentum is way more complicated than that.

  6. Feb 3, 2012 #5
    I tried to make the following math.
    from energy conservation: Eph≈Egap
    (assuming that initial and final electrons are in proximity of, respectively, the maximum of VB and minimum of CB)

    We know that kph = ω/c = Eph/([itex]\hbar[/itex]c)

    So, from momentum conservation law:

    |kf - ki | = |kph|≈Egap/([itex]\hbar[/itex]c)

    Doing the calc (i assumed 1 eV for Egap):
    |kf - ki | ≈ 10-4 angstrom-1, which is about 1 part of thousand of tipical size of Brillouin zone.

    Are my reasoning correct to justify the assumpion kph=0 ?
    Last edited: Feb 3, 2012
  7. Feb 3, 2012 #6
    However, I think I made the math more complex than necessary:

    ki - kf = kph = [itex]\frac{2π}{λ}\widehat{k}_{ph}[/itex]

    (ki - kf) / (size of Brillouin zone) = [itex]\frac{2π}{λ}\frac{a}{2π}\widehat{k}_{ph}[/itex] << 1 [itex]\cdot[/itex] [itex]\widehat{k}_{ph}[/itex]

    Thanks to all for the reply!
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