# [PoM] Electrons Fermi level in a crystal

• BRN
In summary, for a hypothetical one-dimensional crystal of one-dimensional Cesium reticular with step a=300 pm (1 atom per cell), the Fermi level of the electrons is at -3.9935 [eV] and the work function is 3.9935 [eV]. The chemical potential remains constant, since the band dispersion is symmetric around the Fermi point. The allowed values of k for the 1D case are ##k=\pm n \frac{\pi}{L}## with ##n=\pm 1; \pm 2; \pm 3; ...## and the density of states is ##g(k)=\frac{2L}{\pi}##. The Fermi level
BRN

## 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?

BRN said:
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:
TSny said:
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##?

BRN said:
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).

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...

BRN said:
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.

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?

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##.

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.

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## 1. What is the Fermi level in a crystal?

The Fermi level in a crystal is the energy level at which electrons reside at 0 Kelvin (absolute zero). It represents the highest occupied energy level in a crystal at this temperature.

## 2. How does the Fermi level relate to the energy of electrons in a crystal?

The Fermi level is directly related to the energy of electrons in a crystal. It represents the energy at which there is a 50% probability of finding an electron at 0 Kelvin. At higher temperatures, the probability distribution of electrons changes and the Fermi level shifts accordingly.

## 3. What determines the position of the Fermi level in a crystal?

The position of the Fermi level in a crystal is determined by the number of electrons and the energy bands present in the crystal. It also depends on the temperature and the type of crystal (metallic, semiconducting, or insulating).

## 4. How does the Fermi level affect the electrical conductivity of a crystal?

The Fermi level plays a critical role in determining the electrical conductivity of a crystal. In metals, where the Fermi level is within the conduction band, there is a high number of delocalized electrons, making them good conductors of electricity. In semiconductors and insulators, the position of the Fermi level can be adjusted through doping or applying an external electric field, which affects their conductivity.

## 5. Can the Fermi level of a crystal be measured?

Yes, the Fermi level of a crystal can be measured using various techniques such as Hall effect measurements, photoemission spectroscopy, and capacitance-voltage measurements. These techniques provide information about the energy levels and electronic properties of a crystal, including the position of the Fermi level.

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