A Ge specimen is doped with Al. The concentration of acceptor atoms is `10^(21)"atoms/m"^(3)`. Given that the intrinsic concentration of electron hole pairs is `10^(19)//m^(3)`, the concentration of electrons in the specimen 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

`10^(17)//m^(3)`
For p-type semiconductor `n_(h)~~ N_(A)`
where `N_(A)` is the concentration of acceptor atoms. Also `n_(h)n_(e )=n_(i)^(2)` where `n_(h)=` concentration of holes `=10^(21) "atom/m"^(3)`
`n_(i)=` concentration of electron-pair `=10^(19)"atom/m"^(3)`
`n_(e )=` concentration of electron,
`therefore n_(e )=(n_(i)^(2))/(n_(h))=((10^(19))^(2))/(10^(21))=10^(17)"atoms/m"^(3)`