# [PoM] Electrons Fermi level in a crystal

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1. Feb 3, 2017

### BRN

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. Feb 3, 2017

### TSny

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
3. Feb 5, 2017

### BRN

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. Feb 5, 2017

### TSny

Yes.

You will need to consider the allowed values of k for the 1D case. Then you can find g(k).

5. Feb 6, 2017

### BRN

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. Feb 6, 2017

### TSny

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.

7. Feb 6, 2017

### BRN

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. Feb 6, 2017

### TSny

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. Feb 7, 2017

### BRN

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