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^(15)//m^(3)`
B. `10^(17)//m^(3)`
C. `10^(4)//m^(3)`
D. `10^(2)//m^(3)`

Answer

B

Explanation

From law of mass-action
`n_(i)^(2)=n_(e)xxn_(h)`
Where `n_(i)` is concentration of electron-hole pair and `n_(h)` is concentration of acceptor or holes.
Given, `n_(i)=10^(19) per m^(3),n_(h)=10^(21) per m^(3)`
`(10^(19))^(2)=n_(e)xx10^(21)`
`implies n_(e)=(10^(38))/(10^(21))=10^(17) per m^(3)`