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A Ge specimen is doped with $\mathrm{Al}$. The concentration of acceptor atoms is $\sim 10^{21}$ atoms $/ \mathrm{m}^3$. Given that the intrinsic concentration of electron-hole pairs is $\sim 10^{19} / \mathrm{m}^3$, the concentration of electrons in the specimen is
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Verified Answer
The correct answer is:
$10^{17} / \mathrm{m}^3$
When Ge specimen is doped with $\mathrm{Al}$, then concentration of acceptor atoms is also called concentration of holes.
Using formula, $n_i^2=n_0 p_0$, where
$$
\begin{aligned}
n_i= & \text { concentration of electron-hole pair }=10^{19} / \mathrm{m}^3 \\
& n_0=\text { concentration of electron } \\
& p_0=\text { concentration of holes }=10^{21} \text { atom } / \mathrm{m}^3 \\
\Rightarrow \quad & \left(10^{19}\right)^2=10^{21} \times n_0 \Rightarrow n_0=10^{17} / \mathrm{m}^3 .
\end{aligned}
$$
Using formula, $n_i^2=n_0 p_0$, where
$$
\begin{aligned}
n_i= & \text { concentration of electron-hole pair }=10^{19} / \mathrm{m}^3 \\
& n_0=\text { concentration of electron } \\
& p_0=\text { concentration of holes }=10^{21} \text { atom } / \mathrm{m}^3 \\
\Rightarrow \quad & \left(10^{19}\right)^2=10^{21} \times n_0 \Rightarrow n_0=10^{17} / \mathrm{m}^3 .
\end{aligned}
$$
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