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

Option: 1

\mathrm{10^{17}/m^{3}}


Option: 2

\mathrm{10^{15}/m^{3}}


Option: 3

\mathrm{10^{4}/m^{3}}


Option: 4

\mathrm{10^{2}/m^{3}}


Answers (1)

best_answer

\mathrm{n_i^2=n_h n_e \Rightarrow\left(10^{19}\right)^2=10^{21} \times n_e \Rightarrow n_e=10^{17} / \mathrm{m}^3}

Posted by

Gaurav

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