A Ge specimen is doped with $\mathrm{Al}$. The concentration of acceptor atoms is $\sim 10^{21}$ atoms $/…
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
$10^{17} / \mathrm{m}^3$
$10^{15} / \mathrm{m}^3$
$10^4 / \mathrm{m}^3$
$10^2 / \mathrm{m}^3$.
Solution
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
$\begin{aligned}
n_i= & \text { concentration of electron-hole pair }=10^{19} / \mathrm{m}^3 \\
& n_0=\text { concentration of electron } \\
& p_0=\text { concentration of holes }=10^{21} \text { atom } / \mathrm{m}^3 \\
\Rightarrow \quad & \left(10^{19}\right)^2=10^{21} \times n_0 \Rightarrow n_0=10^{17} / \mathrm{m}^3 .
\end{aligned}$