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Stat mech qual question

  1. Jan 9, 2007 #1
    Hey folks,

    studying for the qualifying exam and i wanted to run this by the forum. A model system has a density of states
    g(E) = X for 0 < E < A
    g(E) = 0 for A <= E <= B
    g(E) = Y for B < E

    the total number of particles in the system is given by N = X A.
    Now this looks like a very simple model of a band gap. The first part of the problem asks us to consider particles that obey fermi-dirac statistics.
    The first question asks for expressions for the total number of particles and the total energy. so we have

    [tex]
    N = \int_0^{\infty} g(E) f(E) dE
    [/tex]

    [tex]
    N = \int_0^{A} \frac{X dE}{e^{(E - \mu)/kT} + 1} +
    \int_B^{\infty} \frac{Y dE}{e^{(E - \mu)/kT} + 1}
    [/tex]

    [tex]
    U = \int_0^{\infty} E g(E) f(E) dE
    [/tex]

    [tex]
    U = \int_0^{A} \frac{X E dE}{e^{(E - \mu)/kT} + 1} +
    \int_B^{\infty} \frac{Y E dE}{e^{(E - \mu)/kT} + 1}
    [/tex]

    Im fine with this much. Next part asks for the fermi energy as the temperature goes to 0. I know that the chemical potential at T=0 is equal to the fermi energy. and in this case it seems obvious that the fermi energy is A given that N = X A.

    The third part is what im having a bit of trouble with. We are asked to calculate the chemical potential at low but non-zero temperature. Now remember these problems are meant to be done without the use of calculators or integral tables. I see that I need to evaluate the integral for N and solve for the chemical potential. I can see a way to do it if i assume that T is low that

    [tex]
    e^{(E - \mu) / kT} >> 1
    [/tex]

    then the 1 in the denominator can be neglected but that is like switching to a classical distribution which doesnt make sense in a low temperature limit. Any help would be appreciated.
     
    Last edited: Jan 9, 2007
  2. jcsd
  3. Jan 10, 2007 #2
    hey mod

    can you move this thread to the problem help section? It might get some more attention there.
    thanks
     
    Last edited: Jan 10, 2007
  4. Jan 16, 2007 #3
    Found the answer and figured Id post it here in case anyone was interested. It is indeed not correct disregard the 1 in the denominator and use the classical distribution function. Instead one makes a power series expansion in the variable epsilon - mu. The is called the Sommerfeld Expansion if anyone wants to look at references.
     
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