Question

Show that fermi energy level in intrinsic semiconductor is at the Centre of forbidden energy gap. 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

At any temperature T>0K in an intrinsic semiconductor a number of electrons are found in the conduction band and the rest of the valence electrons are left behind in the valence band.

N = n_c + n_V 

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

`f(E_v) =1 / ( 1+e -(E_C -E_F )/(kT))` ………………………(2)

`n_V = Nf(E_v) = N/( 1+e-(E_C-E_F)/(kT))`

`N = N /(+ 1+e-(E_C-E_F)/(kT)) +  N/(1+e (E_C -E_F )/(kT))`

`1= 1/( 1+e-(E_C-E_F)/(kT)) + 1/(1+e(E_C -E_F )/(kT)`

`1 =( (2+e(E_C -E_F )/(kT))+e( -(E_C -E_F ))/(kT))/([1+e(E_C - E_F )/(kT)][1+e-(E_C -E_F )/(kT))]` 

Solving above equation using cross multiplication method.
e^{(EC-2EF+EV)/kT} = 1 

`(E_C-2E_F+E_V )/(kT) `= 0

`E_F =(E_C+E_V_)/ 2`
Hence it is proved that fermi energy level in intrinsic semiconductor is at the Centre of forbidden energy gap.
NUMERICAL:-
Given Data :- T = 30℃ =  303K     E_g = 1.12_eV 

`K = 1.38^(-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)`

Calculations :- Si is an intrinsic semiconductor. Hence,

`E_C - E_F = (E_g)/2  = 0.56 eV `

`f(E_C ) = 1/( 1+exp ( (0.56)/( 86.25×10^(-6)×303)))= 4.9×10^(-10)`

probability = 4.9×10^-10.