InterviewSolution
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InterviewSolution
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A `Ge` specimen is dopped with `Al`. The concentration of acceptor atoms is `~10^(21) at oms//m^(3)`. Given that the intrinsic concentration of electron hole pairs is `~10^(19)//m^(3)`, the concentration of electron in the speciman isA. `10^(15)//m^(3)`B. `10^(17)//m^(3)`C. `10^(4)//m^(3)`D. `10^(2)//m^(3)` |
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Answer» Correct Answer - B From law of mass-action `n_(i)^(2)=n_(e)xxn_(h)` Where `n_(i)` is concentration of electron-hole pair and `n_(h)` is concentration of acceptor or holes. Given, `n_(i)=10^(19) per m^(3),n_(h)=10^(21) per m^(3)` `(10^(19))^(2)=n_(e)xx10^(21)` `implies n_(e)=(10^(38))/(10^(21))=10^(17) per m^(3)` |
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