# Physics Group - II (3110018)

BE | Semester-1   Winter-2019 | 02-01-2020

## Q2) (c)

#### Explain the dependence of Fermi level on temperature.

The highest energy level which an electron can occupy at absolute zero temperature is called fermi level.

The energy possessed by electron in the fermi level at absolute zero temperature is called fermi energy.

Here, the fermi function gives the probability of occupancy of energy levels by fermions in terms of fermi energy, temperature and Boltzmann constant.

Consider a system of electrons in thermal equilibrium at temperature (T).The probability function f(E) of an electron occupying any particular energy level is given by:

Where, f(E) is fermi distribution function, EF is fermi energy, E is energy of level in which electron is present, T is absolute temperature and kB is Boltzmann constant.

For a filed energy level f(E) = 1 and for an unfilled level, f(E) = 0.

The dependence of fermi function under different conditions of temperature and EF is as follows:

Case - 1: At T = 0 K

At absolute zero electrons occupy the lowest energy level first, followed by next higher levels as per Pauli’s principle.

a)   At T = 0 K, E < EF

For energy levels E, lying below EF, E - EF takes a negative value

$\mathrm{f}\left(\mathrm{E}\right)=\frac{1}{1+{\mathrm{e}}^{-\infty }}$

$⇒\mathrm{f}\left(\mathrm{E}\right)=\frac{1}{1+0}$

$⇒\mathrm{f}\left(\mathrm{E}\right)=1$

This implies that all energy levels below EF are completely filled.

b)   At T = 0 K, E > EF

For energy levels E, lying below EF, E - EF takes a positive value

$\mathrm{f}\left(\mathrm{E}\right)=\frac{1}{1+{\mathrm{e}}^{\infty }}$

$⇒\mathrm{f}\left(\mathrm{E}\right)=\frac{1}{1+\infty }$

$⇒\mathrm{f}\left(\mathrm{E}\right)=0$

This implies that all energy levels below EF are completely vacant.

Case - 2: At T > 0 K

At room temperature, the probability starts reducing from 1 for values of E close to EF, but larger than EF

At T > 0 K, for E = EF , the argument in the exponential function becomes zero.

$\mathrm{f}\left(\mathrm{E}\right)=\frac{1}{1+{\mathrm{e}}^{0}}$

$⇒\mathrm{f}\left(\mathrm{E}\right)=\frac{1}{1+1}$

$⇒\mathrm{f}\left(\mathrm{E}\right)=\frac{1}{2}=0.5$

This implies that probability of occupancy of an electron is 50 % at any temperature above 0 K.

Case - 3: At T = Very high temperature

At very high temperatures, it can be seen from the figure 1 that the transition between completely filled states and completely empty states is rather gradual than abrupt. i.e., f (E) changes from 1 to 0 more gradually.

As temperature T increases, electron may get an energy of an order of KBT and go to the higher vacant state.

As a result, fermi function falls.

For the energies below EF, such that (E - EF) >> KBT, the value of f(E) will be unity.