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Pure Si at $500 \mathrm{~K}$ has equal number of electron $\left(n_e\right)$ and hole $\left(n_h\right)$ concentrations of $1.5 \times 10^{16} \mathrm{~m}^{-3}$. Doping by indium increases $n_h$ to $4.5 \times 10^{22} \mathrm{~m}^{-3}$. The doped semiconductor is of
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The correct answer is:
p-type having electron concentration $n_e=5 \times 10^9 \mathrm{~m}^{-3}$
$$
\begin{aligned}
& n_i^2=n_e n_h \\
& n_e=\frac{\left(n_i\right)^2}{n_h} \\
& n_e=\frac{\left(1.5 \times 10^{16}\right)^2}{\left(4.5 \times 10^{22}\right)} \\
& n_e=5 \times 10^9 \mathrm{~m}^{-3}
\end{aligned}
$$
So, $n_{f_1} \gg n_e$ semiconductor is $p$-type.
\begin{aligned}
& n_i^2=n_e n_h \\
& n_e=\frac{\left(n_i\right)^2}{n_h} \\
& n_e=\frac{\left(1.5 \times 10^{16}\right)^2}{\left(4.5 \times 10^{22}\right)} \\
& n_e=5 \times 10^9 \mathrm{~m}^{-3}
\end{aligned}
$$
So, $n_{f_1} \gg n_e$ semiconductor is $p$-type.
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