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[PoM] Electrons Fermi level in a crystal

  1. Feb 3, 2017 #1

    BRN

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    1. The problem statement, all variables and given/known data
    The conduction band of a hypothetical crystal of one-dimensional Cesium reticular with step a=300 pm (1 atom per cell) is characterized by the ε dispersion law
    ##\epsilon (k) = V_0 + \frac{\hbar^2}{m_e}(\frac{1}{2}k^2 - \frac{a}{3\pi}|k|^3##
    where ##V_0 = -4 eV##, is set so that the energy of a stationary electron in a vacuum on the outside of the crystal appears ε = 0. At zero temperature, determine the position of the Fermi level of the electrons of this crystal (one electron per atom in the conduction band) and the value of the work function. A finite temperature (but small compared to the Fermi temperature) the chemical potential is greater or less than the Fermi level?

    3. The attempt at a solution
    The Fermi energy level is the highest occupied level in an fermions system at T=0. The dispersion law describes the shape of the conduction band and the Fermi level corresponds to ##\epsilon (k_F)##.
    Then, I determine density of states ##g(k)## differentiating with respect k the number of states with wave vector ≥ k (which corresponds to the volume of a sphere of radius
    ##|\vec{n}|=\frac{V^{1/3}}{2\pi}|\vec{k}|##

    ##g(k)=g_s\frac{d}{dk}(\frac{4}{3}\pi \frac{V}{8\pi^3}k^3=g_s\frac{V}{2\pi}k^2##

    Total number of electrons is:

    ##N=\int_{0}^{k_F}g(k) dk=\int_{0}^{k_F}g_s\frac{V}{2\pi}k^2 dk=g_s\frac{V}{6\pi^2}k_F^3##

    with ##g_s=2##, I have:

    ##k_F=(3\pi^2\frac{N}{V})^{1/3}##

    Crystaline structure is an simple cubic cell, then

    ##\frac{N}{V}=\frac{1/8*8}{a^3}=\frac{1}{a^3}##

    so, ##k_F=1.0312*10^{10}[m^{-1}]##

    and ##\epsilon (k_F)=-4.1704*10^{-19}[J]=-2.6023[eV]##

    Work function is the minimum energy for extract an electron from metal and is the opposite of Fermi energy:

    ##W=-\epsilon (k_F)=2.6023[eV]##

    Solutions: ##\epsilon_F = -3.304 eV##; W = 3.304 eV; μ remains constant (neither increases nor decreases), since the band dispersion happens to be symmetric around the Fermi point.

    What is wrong?
     
  2. jcsd
  3. Feb 3, 2017 #2

    TSny

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    In this exercise, you are dealing with a hypothetical one-dimensional crystal. So, what is the dimension of the corresponding k-space?
     
    Last edited: Feb 3, 2017
  4. Feb 5, 2017 #3

    BRN

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    k-space is one-dimensional too, I think...

    But, at this point, I can not understand how to proceed...
    In one-demensional case how to calculate ##k_F##?
     
  5. Feb 5, 2017 #4

    TSny

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    Yes.

    You will need to consider the allowed values of k for the 1D case. Then you can find g(k).
     
  6. Feb 6, 2017 #5

    BRN

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    In 1-D case are allowed values of ##k=\pm n \frac{\pi}{a}##
    Then:

    ##|n|=\frac{a}{\pi}k##

    but, if I derive it with respect to k, I lose the dependence on k...
     
  7. Feb 6, 2017 #6

    TSny

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    This is not quite correct. Instead of ##a## , shouldn't the relevant distance be the length ##L## of the crystal?

    Also, it is important to specify the type of boundary conditions that you are using. Since you are allowing ##k## to be both positive and negative, it appears that you are using periodic boundary conditions. You have to be very careful not to be off by a factor of 2 somewhere.

    For the 1D case, g(k) is independent of k.
     
  8. Feb 6, 2017 #7

    BRN

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    Sorry but, I'm lost...

    I have a dispersion law depends on k and Fermi level is ##\epsilon_F(k=k_F)##.
    If in 1-D case the density of states is independent of k, then g(k) what would be the use?
     
  9. Feb 6, 2017 #8

    TSny

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    Even though g(k) is independent of k, you can still formally use g(k) to find kF just as you would for 3D (see your integral for N in your first post ). However, in the 1D case, it is easy to find ##k_F## without introducing g(k) once you have the correct allowed values for ##k## in terms of the integer ##n## and the length ##L## of the crystal. Note that ##L = Na##, which is helpful if you want to express ##k_F## in terms of ##a##.
     
  10. Feb 7, 2017 #9

    BRN

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    Ok, maybe I understood.

    In 1-D case are allowed value of ##k=n\frac{\pi}{L}## with ##n=\pm 1; \pm 2; \pm 3; ...##
    I determine the density of states g(k) differentiating with respect k the number of states with wave vector ≥ k, that corresponds to portion of the length

    ##n=\frac{L}{\pi}k##

    ##g(k)=g_s\frac{d}{dk}(\frac{L}{\pi}k)=g_s\frac{2L}{\pi}##

    with ##g_s=2 \rightarrow## 2 values of spin allower for each k.

    Now, number of states within in Fermi region is:

    ##N=\int_{0}^{k_F}g(k) dk=\int_{0}^{k_F}\frac{2L}{\pi} dk=\frac{2L}{\pi}k_F##

    ##\Rightarrow k_F=\frac{N}{L}\frac{\pi}{2}##

    There is one atom for cell, so (##L=Na##):

    ##\frac{N}{L}=\frac{N}{Na}=\frac{1}{a}##

    then,

    ##k_F=\frac{\pi}{2a}##

    From dispersion law, Fermi level is:

    ##\epsilon(k=k_F)=V_0+ \frac{\hbar^2}{m_e}(\frac{1}{2}k_F^2 - \frac{a}{3\pi}|k_F|^3)=-6.3999*10^{-19}[J]=-3.9935 [eV]##

    and the work function is:

    ##W=-\epsilon(k_F)=3.9935 [eV]##

    I'm closer to the solution.
     
    Last edited: Feb 7, 2017
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