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Pure Si at 300 K has equal electron (n_(e)) and hole (n_(h)) concentration of 1.5xx10^(16) m^(-3) . Doping by indium increases n_(h) to 4.5xx10^(22)m^(-3) . Calculate n_(e) in the doped silicon. |
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Answer» Solution :Here `n_(i)=1.5xx10^(16)m^(-3),n_(H)=4.5xx10^(22)m^(-3)` But `n_(E)n_(h)=n_(1)^(2)` `therefore n_(e)=(n_(1)^(2))/(n_(h))=((1.5xx10^(16))^(2))/(4.5xx10^(22))=5xx10^(9)m^(-3)` |
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