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A crystal of intrinsic silicon at room temperature has a carrier concentration of $1.6 \times 10^{16} / \mathrm{m}^3$. If the donor concentration level is $4.8 \times 10^{20} / \mathrm{m}^3$, then the concentration of holes in the semiconductor is
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Verified Answer
The correct answer is:
$5.3 \times 10^{11} / \mathrm{m}^3$
Given
$$
\begin{aligned}
& n_1=1.6 \times 10^{16} / \mathrm{m}^3 \\
& n_e=4.8 \times 10^{20} / \mathrm{m}^3 \\
& n_h=?
\end{aligned}
$$
The concentration of holes in the semiconductor
$$
\begin{aligned}
n_1^2 & =n_e \times n_h \\
\left(1.6 \times 10^{16}\right)^2 & =4.8 \times 10^{20} \times n_h \\
n_h & =\frac{2.56 \times 10^{32}}{4.8 \times 10^{20}}=5.3 \times 10^{11} / \mathrm{m}^3
\end{aligned}
$$
$$
\begin{aligned}
& n_1=1.6 \times 10^{16} / \mathrm{m}^3 \\
& n_e=4.8 \times 10^{20} / \mathrm{m}^3 \\
& n_h=?
\end{aligned}
$$
The concentration of holes in the semiconductor
$$
\begin{aligned}
n_1^2 & =n_e \times n_h \\
\left(1.6 \times 10^{16}\right)^2 & =4.8 \times 10^{20} \times n_h \\
n_h & =\frac{2.56 \times 10^{32}}{4.8 \times 10^{20}}=5.3 \times 10^{11} / \mathrm{m}^3
\end{aligned}
$$
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