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

Explanation

Here `n_(i)=1.5xx10^(16)m^(-3),n_(h)=4.5xx10^(22)m^(-3)`
But `n_(e)n_(h)=n_(1)^(2)`
`therefore n_(e)=(n_(1)^(2))/(n_(h))=((1.5xx10^(16))^(2))/(4.5xx10^(22))=5xx10^(9)m^(-3)`