Question

Write the Fermi Dirac distribution function and terms in it. What is the probability of an electron being thermally excited to the conduction band in Si at 30 ℃. The band gap energy is 1.12 eV.  

Answer 1

Each energy band in a crystal accommodates a large number of electron energy levels. According to Pauli’s exclusion principle any energy level can be occupied by two electrons only, one spin up and down . however, all the available energy states are not filled in an energy band. 
The separation between the consecutive energy level is very small around 10^-27eV due to which the 
energy states are not filled in an energy band.
FERMI DIRAC DISTRIBUTION FUNCTION.
The carrier occupancy of the energy states is represented by a continuous distribution function known as the Fermi-Dirac distribution function, given by

`f(E) = 1/(1+e((E-E_F)/(kT))`

This indicates the probability that a particular quantum state at the energy level E is occupied by an electron. Here k is Boltzmann’s constant and T is absolute temperature of the semiconductor.The energy E_F is called Fermi energy that corresponds to a reference level called Fermi level.

FERMI LEVEL
Fermi level is not an allowed energy level it is an imaginary reference level used to specify other energy levels. Fermi level is defined as the highest filled energy level in any solid at absolute zero temperature. 

Hence, at absolute zero temperature all energy levels below E_F  are empty for which the probability of occupancy can be written from Fermi-Dirac distribution function.

NUMERICAL:-
Given Data  :-  T = 30℃= 303 K , E_g = 1.12eV, 

   `K = 1.38 ×10^(-23) J/K =(1.38 ×10^(-23))/(1.6 ×10^(-19)
= 86.25 ×10^(-6)eV⁄K`

Formula :-` f(E_c ) =1/(1+e(E_C-E_F)/(kT))`

Calculation :- Si is an intrinsic semiconductor. Hence ,

`E_C - E_F= (E_g)/2 = 0.56 eV.`
`f(E_c ) =1/(+exp⁡[(0.56)/(86.25×10^(-6)×303)] ) = 4.9×10^(-10).`

Answer :- Probability = 4.9×10^-10.