A pure semiconductor has equal electron and hole concentration of `10^(16) m^(-3)`. Doping by indium increases number of hole concentration `n_(h)` to `5 xx 10^(22) m^(-3)`. Then, the value of number of electron concentration `n_(e)` in the doped semiconductor is

Options

A. `10^(6)//m^(3)`
B. `10^(22)//m^(3)`
C. `2xx10^(9)//m^(3)`
D. `10^(19)//m^(3)`

Answer

C

Explanation

Here, `n_(i)=10^(16)m^(-3), n_(h)=5xx10^(22)m^(-3)`
As `n_(e)n_(h)=n_(i)^(2) therefore" "n_(e)=(n_(i)^(2))/(n_(h))=((10^(16)m^(-3))^(2))/(5xx10^(22)m^(-3))=2xx10^(9)m^(-3)`