Pure Si at 300 K has equal electron `(n_(e))` and hole `(n_(h))` concentrations of `1.5xx10^(16)m^(-3)` doping by indium increases `n_(h)` to `4.5xx10^(22)m^(-3)`. Caculate `n_(e)` in the doped Si-

Explanation

Here `n_(i)=1.5 xx 10^(16)m^(-3)`, `n_(h)=4.5 xx 10^(22)m^(-3)`, Now `n_(h)n_(e)` =`n_(i)^(2)` or `h_(e)` =`n_(i)^(2)/n_(h)=(1.5xx10^(16))^(2)/(4.5xx10^(22))=5xx10^(9)m^(-3)`