A pure si crystal having 5×10^28 atoms m-3 is doped with 1ppm . . .

Question : A pure si crystal having 5×10²⁸ atoms m^-3 is doped with 1 ppm concentration of antimony. If the concentration of holes in the doped crystal is found to be 4.5×10⁹ m^-3, the concentration (in m^-3) of intrinsic charge carriers in Si crystal is about

(a) 1.2×10¹⁵ 

(b) 1.5×10¹⁶

(c) 3.0×10¹⁵

(d) 2.0×10¹⁶

[CBSE D 2024]

Doubt by Nevaeh

Solution : 

Number of silicon atoms = 5×10²⁸ m^-3

Doping concentration = 1 ppm (parts per million) of antimony

Hole concentration (n_h) =4.5×10⁹m⁻³

Antimony (Sb) is a pentavalent impurity, it will donate one free electron per atom.
so the electron concentration would be 
n_e=[5×10²⁸/10⁶] = 5×10²² m^-3

Now, using the mass action law
n_i²=n_e.n_h
n_i²=(5×10²²).(4.5×10⁹)
n_i²=22.5×10³¹

n_i²=2.25×10³²

n_i=1.5×10¹⁶

The concentration of intrinsic charge carrier is 1.5×10¹⁶ m^-3

Hence, (b) 1.5×10¹⁶ would be the correct option.