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

For a doped semi-conductor in thermal equilibrium `n_(e)n_(h) = n_(1)^(2)` (Law of mass action)
`n_(e)=(n_(1)^(2))/(h_(h))=((1.5xx10^(16))^(2))/(3xx10^(22))=7.5 xx10^(9)m^(-3)`