Pure `Si` at 300 K has equal electron `(n_(e))` and hole `(n_(h))` concentrations of `1.5 xx 10^(18) m^(-3)`. Doping by indium increases `n_(h)` to `4.5 xx 10^(22) m^(-3)`. Calculate `n_(e)` in the dipoed `Si`.

Explanation

At thermal equilibrium `n_(e)n_(h)=n_(1)^(2)=n_(1)^(2) rArr n_(e)=(n_(i)^(2))/(n_(e))=((1.5 xx10^(16)))/(4.5 xx 10^(22))=5 xx 10^(9)m^(-3)`.