The band gap energy is 1.12 eV.

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.

$e^((E_C-2E_F+E_V)/kT)=1$

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

$E_F= (E_C+E_V)/2$

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 $E_g=1.12eV$

$K = 1.38 ×10^{(-23)} J/K = (1.38 ×10^{(-23)})/(1.6 ×10^{(-19)} ) = 86.25×10^{(-6)} eV/K4$

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 = $4.9×10^{(-10)}.$