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Pure Si at 300 K has equal electron `(n_(e))` and hole `(n_(h))` concentrastions of `1.5xx10^(16)m^(-3)` doping by indium increases `n_(h)` to `4.5xx10^(22)m^(-3)`. Caculate `n_(theta)` in the doped Si- |
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Answer» `n_(e)n_(h)=n_(i)^(2)` `n_(h)=4.5xx10^(22)m^(-3)` so, `n_(e)=5.0xx10^(9)m^(-3)` |
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