# Fermi Radius for 2D Rectangular Lattice?

• ThereIam
In summary, the conversation is about finding the free electron Fermi radius for a 2D metal with a rectangular primitive cell. The equations for calculating the Fermi energy and the number of electrons in the system are discussed, as well as the volume element and periodicity of the wavefunctions. It is clarified that for a 2D system, the volume element should only include the two periodic dimensions and the total number of electrons should be equal to the number of valence electrons. The expression for the Fermi energy for a rectangular lattice is also mentioned.
ThereIam

## Homework Statement

Hi all,

This is Kittel 9.2. This problem has been asked about before, but the people asking found solutions.

I'm trying to find the free electron Fermi radius for a 2D metal with a rectangular primitive cell (a = 2 Ang, b = 4 Ang).

## Homework Equations

Please forgive my lack of latex skills:

E_f = hbar^2/(2m) * k_f^2
N = 2 * "volume" of Fermi sphere / volume element size.

For 3D this is N = V / (3 pi^2) * k_f^3 = 4 pi k_f^3 / 3 *(L /2 pi) ^3

## The Attempt at a Solution

Since this is a 2D problem I take the volume of the Fermi sphere to be pi*k_f^2 (the area of a Fermi circle.

I am confused about the volume element. My lattice has different periodicities in each direction. I would assume based on the derivation of the volume element for phonons that this means my volume element should reflect this. I am tempted to take it to be (4 pi^2)/(a*b) = (2 pi^2)/(a^2). This is just the area of the first Brillouin zone.

I am not sure if this makes sense. I'm requiring that the wavefunctions be periodic like traveling plane waves in the respective directions k_x, k_y, as on page 137. Do I need to modify my expression for the Fermi energy to reflect my rectangular lattice in any way?

Anyway, I then get:
N = V * 1/(volume element) = a^2*k_f^2/pi

Then I assume N = 1, for valency 1? Though I just calculated the area of the occupied states...

Hello!

Based on your equations, it looks like you are on the right track. However, there are a few things that need to be clarified.

Firstly, for a 2D system, the Fermi energy should be calculated using the area of the Fermi circle, as you have correctly done. However, the volume element should be the area of the first Brillouin zone, which in this case would be (2pi/a)*(2pi/b) = 4pi^2/(ab). This is because in a 2D system, the wavefunctions are not periodic in all three dimensions, so the volume element should only include the two dimensions that are periodic.

Secondly, the N value that you are calculating is the total number of electrons in the system. This value should be equal to the number of valence electrons in the system, not necessarily 1. So if the system has multiple valence electrons, you would need to modify your equation accordingly.

Finally, for a rectangular lattice, the expression for the Fermi energy would be slightly different than for a square lattice. It would be E_f = hbar^2/(2m)*(k_x^2 + k_y^2), where k_x and k_y are the components of the wavevector in the x and y directions, respectively.

I hope this helps clarify things for you. Let me know if you have any further questions. Good luck with your calculations!

## 1. What is the Fermi radius and why is it important in scientific research?

The Fermi radius, also known as the Fermi energy, is a measure of the maximum energy that an electron in a system can have. It is crucial in research as it helps scientists understand the behavior of electrons in different materials and systems, and it can also be used to calculate other important quantities such as the Fermi velocity and Fermi temperature.

## 2. How is the Fermi radius calculated?

The Fermi radius is calculated using the Fermi-Dirac distribution function, which describes the probability of finding an electron with a given energy in a system at thermal equilibrium. The formula for the Fermi radius is given by rF = ħ/(2πm*EF), where ħ is the reduced Planck's constant, m is the electron mass, and EF is the Fermi energy.

## 3. Can the Fermi radius be observed or measured directly?

No, the Fermi radius cannot be directly observed or measured. It is a theoretical concept used to describe the behavior of electrons in a system. However, it can be indirectly inferred from other experimental measurements such as the electronic density of states or the specific heat capacity of a material.

## 4. How does the Fermi radius play a role in the study of semiconductors?

In semiconductors, the Fermi radius determines the energy gap between the valence and conduction bands. This energy gap is crucial in understanding the electrical and optical properties of semiconductors and is directly related to the Fermi energy. Additionally, the Fermi radius can help researchers predict the behavior of charge carriers in semiconductors and design new electronic devices.

## 5. Are there any practical applications of the Fermi radius?

Yes, the Fermi radius has several practical applications in various scientific fields. In addition to its role in understanding the properties of semiconductors, it is also used in the study of metals, superconductors, and plasmas. It is also a crucial concept in quantum mechanics and is used in the development of new theories and models in physics.

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