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The number of silicon atoms per $\mathrm{m}^3$ is $5 \times 10^{28}$. This is doped simultaneously with $5 \times 10^{22}$ atoms per $\mathrm{m}^3$ of Arsenic and $5 \times 10^{20}$ per $\mathrm{m}^3$ atoms of Indium. Calculate the number of electrons and holes. Given that $n_i=1.5 \times 10^{16} \mathrm{~m}^{-3}$. Is the material n-type or p-type?
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Here, $\mathrm{n}_{\mathrm{e}}=5 \times 10^{22}-5 \times 10^{20}=(5-0.05) \times 10^{22}$ $=4.95 \times 10^{22} / \mathrm{m}^3$;
$$
\mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{e}}}=\frac{\left(1.5 \times 10^{16}\right)^2}{4.95 \times 10^{22}}=4.54 \times 10^9 / \mathrm{m}^3 \text {. }
$$
As $\mathrm{n}_{\mathrm{e}}>\mathrm{n}_{\mathrm{h}} \quad \therefore \mathrm{n}$-type semiconductor.
$$
\mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{e}}}=\frac{\left(1.5 \times 10^{16}\right)^2}{4.95 \times 10^{22}}=4.54 \times 10^9 / \mathrm{m}^3 \text {. }
$$
As $\mathrm{n}_{\mathrm{e}}>\mathrm{n}_{\mathrm{h}} \quad \therefore \mathrm{n}$-type semiconductor.
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