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Probability that energy level is occupied

  1. Jun 4, 2009 #1
    The probability that an electron is in an excited state is given:

    P = e^-(energy difference)/kt

    and the probability that an energy level is occupied in semi conductors is:

    p = 1/ (e^(E-Ef)kt + 1)

    I know both are related to fermi-dirac statistics, which one is the appropraite one to use?
  2. jcsd
  3. Jun 4, 2009 #2
    Both distributions,

    FD: p = 1 / (exp[(e - u)/kT] + 1)


    BE: p = 1 / (exp[(e - u)/kT] - 1)


    MB: p = exp[(u - e)/kT]

    at high temperatures (i.e. in classical limit).
    So it depends on the temperature.
    Does it answer your question? :)

    Last edited by a moderator: Aug 6, 2009
  4. Jun 4, 2009 #3
    uhh, im afraid i dont get you =x
  5. Jun 4, 2009 #4
    FD and BE differ by (+/-)1 factor.
    This can be neglected if

    exp(e - u) >> 1

    which happens in classical regime,
    that is at high temperatures
    or at low concentrations.

    You can try some rough values for semiconductors,
    Setting zero at the top of valence band,

    Eg ~ 1eV (energy gap)
    Ef ~ 0.5eV (midway between val. and cond. band)

    so E in cond. band is about 1eV

    also kT ~ 0.025eV

    so one gets

    1/[exp(20) + 1] ~ 1/exp(20) = exp(-20)

    since exp(20) is huge.

    Last edited by a moderator: Aug 6, 2009
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