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^(4)//m^(3)`
D. `10^(2)//m^(3)`

Answer

A

Explanation

Here `n_(i)=10^(19)m^(3), n_(h)=10^(21) "atoms"//m^(3)`
The intrinsic concentration of electron-hole pair is given by `n_(i)^(2)=n_(e)n_(h)`
`:. n_(e)=n_(i)^(2)/n_(h)=((10^(19))^(2))/10^(21)=10^(38)/10^(21)=10^(17)//m^(3)`