Pure Si at 300 K has equal electron (n_(e)) and hole (n_(h)) concentration of 1.5xx10^(16)m6(-3). Doping by indium increases into 4.5xx10^(22)m^(-3). Calculate n_(e) in the doped silicon .

Pure Si at 300 K has equal electron (n_(e)) and hole (n_(h)) concentra

1.	Pure Si at 300 K has equal electron (n_(e)) and hole (n_(h)) concentration of 1.5xx10^(16)m6(-3). Doping by indium increases into 4.5xx10^(22)m^(-3). Calculate n_(e) in the doped silicon .
Answer» Solution :Here `n_(i)=1.5xx10^(16)m^(-3),n_(H)=4.5xx10^(22)m^(-3)` But `n_(e)n_(h)=n_(i)^(2)` `therefore n_(e)=(n_(i)^(2))/(n_(h))=((1.5xx10^(16))^(2))/(4.5xx10^(22))=5xx10^(9)m^(-3)`

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Pure Si at 300 K has equal electron (n_(e)) and hole (n_(h)) concentration of 1.5xx10^(16)m6(-3). Doping by indium increases into 4.5xx10^(22)m^(-3). Calculate n_(e) in the doped silicon .

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