A Ge specimen is doped with Al. The concentra | vedclass

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(A) In a $p$-type semiconductor, the concentration of holes $(p_0)$ is approximately equal to the concentration of acceptor atoms $(N_A)$.Given, $N_A

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$10^{17}/m^3$

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(A) In a $p$-type semiconductor, the concentration of holes $(p_0)$ is approximately equal to the concentration of acceptor atoms $(N_A)$.Given, $N_A = p_0 = 10^{21} \, 	ext{atoms}/m^3$.The intrinsic carrier concentration is $n_i = 10^{19}/m^3$.According to the law of mass action for semiconductors, $n_i^2 = n_0 \cdot p_0$, where $n_0$ is the concentration of electrons.Substituting the values: $(10^{19})^2 = n_0 \cdot 10^{21}$.$10^{38} = n_0 \cdot 10^{21}$.$n_0 = \frac{10^{38}}{10^{21}} = 10^{17}/m^3$.

$A$ $Ge$ specimen is doped with $Al$. The concentration of acceptor atoms is $\sim 10^{21} \, \text{atoms}/m^3$. Given that the intrinsic concentration of electron-hole pairs is $\sim 10^{19}/m^3$, the concentration of electrons in the specimen is:

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