The number of silicon atoms per $m^{3}$ is $5 \times 10^{28}$. This is doped simultaneously with $5 \times 10^{22}$ atoms per $m^{3}$ of Arsenic and $5 \times 10^{20}$ atoms per $m^{3}$ of Indium. Calculate the number of electrons and holes. Given that $n_{i} = 1.5 \times 10^{16} \; m^{-3}$. Is the material $n$-type or $p$-type?

(A) Number of silicon atoms,$N = 5 \times 10^{28} \; \text{atoms}/m^{3}$.
Number of arsenic atoms (donor),$n_{D} = 5 \times 10^{22} \; \text{atoms}/m^{3}$.
Number of indium atoms (acceptor),$n_{A} = 5 \times 10^{20} \; \text{atoms}/m^{3}$.
Intrinsic carrier concentration,$n_{i} = 1.5 \times 10^{16} \; m^{-3}$.
Since $n_{D} > n_{A}$,the net donor concentration is $n_{e} \approx n_{D} - n_{A} = 5 \times 10^{22} - 0.05 \times 10^{22} = 4.95 \times 10^{22} \; \text{electrons}/m^{3}$.
Using the mass action law,$n_{e} n_{h} = n_{i}^{2}$.
$n_{h} = \frac{n_{i}^{2}}{n_{e}} = \frac{(1.5 \times 10^{16})^{2}}{4.95 \times 10^{22}} = \frac{2.25 \times 10^{32}}{4.95 \times 10^{22}} \approx 4.55 \times 10^{9} \; \text{holes}/m^{3}$.
Since $n_{e} > n_{h}$,the material is an $n$-type semiconductor.