Question

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`. Calculate `n_(e_(2))` in the doped Si.

Answer 1

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