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

Options

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

Answer

D

Explanation

`(n_(i))^(2) = n_(e)n_(h)`
`(10^(16))^(2) = n_(e) xx 5 xx 10^(22)`
`therefore n_(e) = (10^(16) xx 10^(16))/(5 xx 10^(22))`
`n_(e) = 2 xx 10^(9) //m^(3)`