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Fermi-Dirac Statistics

  1. Oct 21, 2004 #1
    Hey kids,

    The question I'm having trouble with (this time) is as follows:

    Show that the Fermi-Dirac distribution function,

    [tex] f_{FD}(E)=\frac{1}{e^{(\frac{E-E_f}{kT})}+1} [/tex]

    Has the following functional form at T= 0K
    (see attachment)


    Now, the first thing that screamed at me was the division by T in the exponential bit. If T=0, what is going on!?

    The obvious things are:

    E>Ef then f(E) = 0

    and

    E<Ef then f(E) = 1.

    I'm just really confused at how I can show that the function has that form at T=0K

    Any ideas?

    Cheers
     

    Attached Files:

  2. jcsd
  3. Oct 21, 2004 #2

    Galileo

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    I don't understand your question. Didn't you just show the function has that form?
    [tex]\lim_{T \rightarrow 0}f_{FD}(E)=\left\{ \begin{array}{ll}1 & \mbox{if} E<E_f\\ \frac{1}{2} & \mbox{if} E=E_f \\0 & \mbox{if} E>E_f[/tex]
     
  4. Oct 22, 2004 #3
    I'm glad somebody else doesn't understand the question either.

    They want me to 'show' that the distribution has the (attached pic) form at T=0.

    The real problem I'm having is how do I "show" that it has that form? Via two lines of maths? That's it?
     
  5. Oct 22, 2004 #4

    Galileo

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    You HAVE just shown it. By taking the limits.
    So yeah, that's it. :)
     
  6. Oct 22, 2004 #5
    Well, just use the definition of the Fermi-level... It is the the maximum energy-level at T = 0 K. Just fill up all the available energy-levels with all available electrons. The last electron is placed at the highest energylevel which is called the Fermi-level. Ofcourse all this is done at zero Kelvin. That is why the distribution function has the drawn form. All levels are filled up (probability one) till the Fermi-level. Above this level there are no filled levels since the Fermi-level is the highest. It is just by QM-definition of the Fermi-level


    regards
    marlon
     
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