Physics Group - II (3110018)

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:

                                                                                                 $\mathrm{f} \left(\mathrm{E}\right)=\frac{1}{1+{\mathrm{e}}^{\left({\displaystyle \frac{\mathrm{E}-{\mathrm{E}}_{\mathrm{F}}}{{\mathrm{K}}_{\mathrm{B}}\mathrm{T}}}\right)}}$

Where, f(E) is fermi distribution function, E_F is fermi energy, E is energy of level in which electron is present, T is absolute temperature and k_B 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 E_F 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 < E_F

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

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

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

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

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

b)   At T = 0 K, E > E_F 

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

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

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

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

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

Case - 2: At T > 0 K

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

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

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

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

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

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 K_BT and go to the higher vacant state. 

As a result, fermi function falls.

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