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

Explanation

For a doped semi-conductor in thermal equilebrium
`n_(e)n_(h)=n_(i)^(2)` (Law of mass action)
`n_(e)=m=n_(i)^(2)/n_(h)=(1.5 times 10^(16))^(2)/(3 times 10^(22))=7.5 times 10^(9) m^(-3)`