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 1.5 xx 10^(16) m^(-3) , n_h = 4.5 xx 10^(22) m^(-3)`
But `n_e n_h= n_(i)^(2)`
` therefore n_e = (n_(i)^(2))/(n_h) = ((1.5 xx 10^(16) )^2)/( 4.5 xx 10^(22))= 5 xx 10^9 m^(-3) `