[PoM] Electrons Fermi level in a crystal

  • #1
BRN
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Homework Statement


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?

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?
 

Answers and Replies

  • #2
TSny
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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##
In this exercise, you are dealing with a hypothetical one-dimensional crystal. So, what is the dimension of the corresponding k-space?
 
Last edited:
  • #3
BRN
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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?
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##?
 
  • #4
TSny
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k-space is one-dimensional too, I think...
Yes.

But, at this point, I can not understand how to proceed...
In one-demensional case how to calculate ##k_F##?
You will need to consider the allowed values of k for the 1D case. Then you can find g(k).
 
  • #5
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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...
 
  • #6
TSny
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In 1-D case are allowed values of ##k=\pm n \frac{\pi}{a}##
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.

but, if I derive it with respect to k, I lose the dependence on k...
For the 1D case, g(k) is independent of k.
 
  • #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?
 
  • #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##.
 
  • #9
BRN
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6
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:

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