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Intrinsic semiconductor, carrier concentration

  1. Mar 12, 2017 #1
    1. The problem statement, all variables and given/known data
    An intrinsic semiconductor with a direct gap have valance band ##\epsilon_k = E_v-b|k|^2## and conduction band ##\epsilon_k = =E_c+a|k|^2##, with ##E_v=6.0##eV, ##E_c = 5.5eV##, ##a=5.0eV\cdot Å^2##, and ##b=3.0eV\cdot Å^2##.
    Calculate the chemical potential ##\mu## and the number of charge carriers ##n+p##.

    2. Relevant equations
    Effective mass:
    ##m^*_e = \frac{\hbar^2}{2a}##
    Intrinsic semiconductor ##n=p##:
    ##p=n = \sqrt{pn} = \frac{1}{\sqrt{2}}\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}(m_e^* m_n^*)^{3/4}e^{-E_g/(2k_BT)}##
    ##e^{(2\mu-E_g)/k_BT}= \left( \frac{m_h^*}{m_e^*}\right)^{3/2}##
    ##\mu = \frac{E_g}{2}+\frac{3}{4}k_BT\ln \frac{m_h^*}{m_e^*}##
    At ##T\approx 300##, ##k_BT \approx 25.7meV##.

    3. The attempt at a solution
    For the first part I get the correct answer just plugging in the numbers, with ##E_g = E_c-E_v = 0.5##eV and then shifting the energy by ##E_V## and noting that ##m_h/m_e=a/b##
    ##\mu = 5.5+0.25+25.7/1000 \cdot ln(5/3) \approx 5.76eV##.

    For the second part we can rewrite
    ##n=\frac{1}{\sqrt{2}}\left( \frac{k_BT}{2\pi \sqrt{ab}}\right)^{3/2}e^{-E_g/(2k_BT)}##
    plugging in the numbers here we get ##n \approx 1.45\cdot 10^{-9} Å^{-3}= 1.45\cdot 10^{21} m^{-3}##. However the answer claims I should get ##3.9\cdot 10^{23}m^{-3}##.

    Any obvious error I'm doing or is the answer incorrect here? The factor in front can change a bit from simply changing ##k_BT## and the answer may be by a factor 2 as well if they mean ##n+p## but the factor of 100 is troubling.

    Here is the numerical calculation in matlab
    Code (Text):

    eg=0.5;
    kbt = 25.7e-3;
    sqab=sqrt(15);
    c=1e30*(kbt/(2*pi*sqab))^(3/2)*exp(-eg/kbt/2)/sqrt(2)
     
     
  2. jcsd
  3. Mar 12, 2017 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Where did the "2 π" in the denominator come from? I got just π. This will change the value from 1.45 to about 4 but I agree with you that the power of 10 is +21 given your numbers.
     
  4. Mar 12, 2017 #3
    It's from the 2 in the effective masses ##\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}(m_e^*m_h^*)^{3/4} =\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}\left(\frac{\hbar^2 \hbar^2}{(2a)(2b)}\right)^{3/4}=\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}\left(\frac{\hbar^2}{2\sqrt{ab}}\right)^{3/2}##.
    Anyway I take your answer as confirmation that I'm using the correct method so despite the answer being different I think it can be considered solved! Thanks for looking it over!
     
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