The number of silicon atoms per m³ is 5×1028. This is doped simultaneously with 5×1022 atoms per m³ of Arsenic . . .

Question : The number of silicon atoms per m³ is 5×10²⁸. This is doped simultaneously with 5×10²² atoms per m³ of Arsenic and 5×10²⁰ per m³ atoms of Indium. Calculate the number of electrons and holes. Given that n_i=1.5×10¹⁶ m^-3. Is the material n-type or p-type?

Doubt by Palak

Solution : 

Let us first recall some formulas
1) In an intrinsic semiconductor, n_e=n_h=n_i
where
n_e=number density of electron
n_h=number density of holes
n_i=intrinsic charge carrier concentration (No of thermally generated electrons)

2) At equilibrium in any semiconductor
n_en_h=n_i²

3) In an n-type semiconductor
N_D≃n_e>>n_h
where N_D=Number Density of Donor atoms

4) In p-type semiconductor
N_A≃n_h>>n_e
where N_A=Number Density of Acceptor atoms

Here

Number of Atoms of Arsenic Per m³ (N_D) =5×10²² m^-3
Number of Atoms of Indium Per m³ (N_A) = 5×10²⁰ m^-3
n_i=1.5×10¹⁶ m^-3

Number Density of electrons / Number of electron Per m³ (n_e)
=N_D-n_i
=5×10²² - 1.5×10¹⁶
=5×10²²-0.0000015×10²²
= 4.9999985×10²²
n_e = 4.99 × 10²² m^-3

Also, by using the formula

n_en_h= n_i²
n_h= n_i²/n_e
n_h = (1.5×10¹⁶)²/4.99 × 10²²
n_h = 0.4509×10¹⁰
n_h = 4.51×10⁹ m^-3

Clearly,
n_e>n_h

∴ The material is a n-type semiconductor.