A `Ge` specimen is dopped with `Al`. The concentration of acceptor atoms is `~10^(21) at oms//m^(3)`. Given that the intrinsic concentration of electron hole pairs is `~10^(19)//m^(3)`, the concentration of electron in the speciman is

Options

A. `10^(17)//m^(3)`
B. `10^(15)//m^(3)`
C. `10^(12)//m^(3)`
D. `10^(16)//m^(3)`

Explanation

Given `n_(1) = 10 ^(19) //m^(9)`
`n_(e) = (n_(1)^(2))/( n_(h)) = ((10^(19))^(2))/( 10^(22))`
`=10^(16)//m^(2)`