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Fermi Dirac probabilty/density confusion

  1. Oct 6, 2012 #1
    -Admin, If possible, please delete the similar looking question posted in the introductory physics category


    I'm having trouble understanding the concept of the Fermi Dirac probability. What I understand so far, its based upon two variables, cubic length and energy level. In addition, I understand that

    [itex]fF(E)=\frac{4\pi*(2*m*n)(3/2)}{h3}\sqrt{E}[/itex]



    [itex]g(E)=\frac{1}{1+e\frac{E-EF}{kT}}[/itex]

    and that

    [itex]N(E)=\intfF(E)*g(E)dE[/itex]

    I end up having trouble when trying to integrate this. Using a textbook example of intrinsic silicon at T=300 Kelvin, (solving for the density of carriers in the conduction band) it mentions the answer as [itex]N(E)=1.102*1010 cm-3 [/itex].

    If I recall correctly:
    [itex]\int\sqrt{x}=\frac{2}{3}*(x)(3/2)[/itex] Since the Energy variable is being integrated, I approach this problem in this fashion

    [itex]N(E)=\frac{1}{1+e\frac{E-EF}{kT}*(\frac{2}{3}*(ΔE)(3/2))[/itex] (within the textbook the limits are Ec and Ec+1ev)

    Since I already know the Boltzmann constant, the temperature, the Energy gap (1.12eV) and the Fermi Energy (given as Energy/2), I'm able to plug in most of the values. What is troubling me is when I preform the math, I end up with a value of 2.073*1012cm-3 and not 1.102*1010
    cm-3

    the values I'm using for e is (Ec)-(Ec-Eg/2), and (300k)*(Boltzmann constant) resulting in
    [itex]e\frac{0.56eV}{0.0259eV}[/itex]

    I do convert the delta E into Joules




    what I don't understand is, why am getting the wrong answer? The textbook doesn't fully explain nor is the professor approachable
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 7, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    For some reason I'm not getting your itex decoded.
     
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