A pure silicon crystal with $5 × 10^{28}$ atoms $m^{-3}$ has $n_i=1.5 × 10^{16} m^{-3}$. It is doped with a concentration of 1 in $10^5$ pentavalent atoms, the number density of holes (per $m^3$) in the doped semiconductor will be:

$4.5 × 10^8$

The correct answer is Option (2) → $4.5 × 10^8$

1 Atom of Si doped out of $10^6$ atoms

⇒ In $5×10^{28}$ atoms

Net Doned = $\frac{5×10^{28}}{10^6}=5×10^{22}$ atoms

The number of holes (p) is given by,

${n_i}^2=n×p$

$p=\frac{{n_i}^2}{n}=\frac{(1.5×10^{16})^2}{5×10^{22}}=4.5×10^8$