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# Fermi Energy

Top
 Sub Topics The Fermi energy of a substance is a result of the Pauli exclusion principle when the temperature of a material is lowered to absolute zero. By this principle, only one electron can inhabit a given energy state at a given time. At the temperature of absolute zero, all electrons in the solid attempt to get into the lowest available energy level. As a result, they form a sea of energy states known as the Fermi sea. The high energy level of this sea is called the Fermi energy or Fermi level.

## Definition

At absolute no electrons have enough energy to occupy any energy levels above the Fermi level. In metals the Fermi level sits between the valence and conduction bands. The size of the band gap between the Fermi level and the conduction band determines if the metal is a conductor, insulator or semiconductor. If the temperature of a given material is increased beyond the absolute zero, this Fermi energy can use to find out the probability of electrons having a specified energy level.

## Formula

The probability that an electron is in a state with energy E is given by the distribution function. This equation is called the Fermi-Dirac distribution function. It provides a physical meaning of Fermi energy:an electron has 50% probability of having the energy E = Ef at any temperature. The mathematical representation of Fermi energy is given by,

$E_{f}$ = $\frac{h^{2}}{2m_{e}}(\frac{3}{8\pi }n_{e})^{\frac{2}{3}}$

where $E_{f}$ is the Fermi energy, h is the Planck's constant, $m_{e}$ is the mass of the electron, $n_{e}$ is the electron density.

## Semiconductor

Semiconductors have the same type of band structure as an insulator, but the energy gap is much smaller. on the order of 1eV. The Fermi level is located near the middle of the gap for a semiconductor and Eg is small, appreciable numbers of electrons are thermally excited from the valence band to the conduction band. Because of the many empty levels above the thermally filled levels in the conduction band, a small applied potential difference can easily raise the energy of the electrons in the conduction band, resulting in a moderate current.

## Table

Fermi energy and Fermi temperature of some of the elements are listed below:

 Elements Fermi Energy (Ef)in eV Fermi Temperature (Tf)$\times 10^{4}$ K Li 4.767 5.527 Na 3.249 3.767 K 2.123 2.463 Cs 1.593 1.847 Cu 7.060 8.186 Ag 5.508 6.386 Au 5.533 6.415 Al 11.681 13.543

## What is Fermi Temperature?

The Fermi energy can be used to define the Fermi temperature.The ratio of the Fermi level of an assembly of fermions to the Boltzmann constant is known as the Fermi temperature. It is represented by the letter $T_{f}$ and is given by

$T_{f}$ = $\frac{E_{f}}{k_{B}}$
Fermi energy plays an important role in determining the energy an electron needs to begin moving.

## Fermi Function

The distribution of the energies of a large number of particles and its change with temperature can be calculated by means of statistical considerations. The kinetic energy of an electron gas is governed by Fermi-Dirac statistics.The probability function is given by,

F(E) = $\frac{1}{e^{\frac{(E-E_{f})}{k_{B}T}}+1}$

The function F(E) is called the Fermi distribution function. The name is apt, for F(E) describes the way that fermions are distributed over the various single-particle states. A shorter name for F(E) is simply the Fermi function. Although the notation F(E) displays only the dependence on energy E, the Fermi function depends also on the temperature T and the chemical potential $\mu$, which itself is a function of temperature.

## Fermi Velocity

The Fermi velocity is defined as 'at absolute zero the average velocity of an electron'. This average velocity corresponds to the average energy ($\frac{3}{5}$$E_{f}). A typical value for the Fermi velocity at 0K is -10^{6} m/s or 1000 km/s. In other words, Fermi velocity is the velocity of the electrons at the Fermi energy. ## Calculation Back to Top Some of the solved problems related to the Fermi energy is given below: ### Solved Examples Question 1: Calculate the Fermi energy of the conduction electrons in Ga at room temperature (300K) in the units of eV? Assuming that 3 conduction electrons per atom at 0K. The density of Ga atom is 10^{28} atoms/m3. Solution: The given parameters are, nGa = 10^{28} atoms/m3 ne = 3\timesnGa = 3\times10^{28} atoms/m3 The formula for Fermi energy is, E_{f} = \frac{h^{2}}{2m_{e}}$$($$\frac{3}{8\pi }$$n_{e})^{\frac{2}{3}}$

$E_{f}$ = $\frac{(6.626\times10^{-34})^2}{2\times9.1\times10^{-31}}$$($$\frac{3}{8\pi }$$3\times10^{28})^{\frac{2}{3}} E_{f} = 2.4123\times10^{-68}\times10^{31} \times$$($$\frac{3}{8\pi }$$30\times10^{27})^{\frac{2}{3}}$

$E_{f}$ = 2.4123$\times10^{-37} \times 2.341\times10^{18}$

$E_{f}$ = 5.6482$\times10^{-19}$ J

In the units of ev:

$E_{f}$ = $\frac{5.6482\times10^{-19}}{1.6\times10^{-19}}$ = 3.530 eV

Question 2: Calculate the Fermi energy of the conduction electrons in Na at room temperature (300K) in the units of eV? The electron density of Na atom is 2.53$\times10^{28}$ atoms/m3.

Solution:
The given parameters are,
ne = 2.53$\times10^{28}$ atoms/m3
The formula for Fermi energy is,

$E_{f}$ = $\frac{h^{2}}{2m_{e}}$$($$\frac{3}{8\pi }$$n_{e})^{\frac{2}{3}} E_{f} = \frac{(6.626\times10^{-34})^2}{2\times9.1\times10^{-31}}$$($$\frac{3}{8\pi } 2.53\times10^{28})^{\frac{2}{3}} E_{f} = 2.4123\times10^{-68}\times10^{31}\times$$($$\frac{3}{8\pi }$$25.3\times10^{27})^{\frac{2}{3}}$

$E_{f}$ = 2.4123$\times10^{-37} \times 2.090\times10^{18}$
$E_{f}$ = 5.0417$\times10^{-19}$ J

In the units of eV:

$E_{f}$ = $\frac{5.0417\times10^{-19}}{1.6\times10^{-19}}$ = 3.151eV