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

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$10^{17} / m^{3}$

$10^{15} / m^{3}$

$10^{4} / m^{3}$

$10^{2} / m^{3}$

answer is A.

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Detailed Solution

$n_{i}^{2} = n_{h} n_{e} \Rightarrow {(10^{19})}^{2} = 10^{21} \times n_{e} \Rightarrow n_{e} = 10^{17} / m^{3} .$

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