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

Answer

A

Explanation

When Ge specimen is doped with Al, then concentration of acceptor atoms is alos called concentration of holes. Using formula `n_(i)^(2)=n_(0)p_(0)` where
`n_(1)`= concentratio of electron hole pair `=10^(19)m^(-3)`
`n_(0)`= concentration of electron
`p_(0)`= concentration of holes `=10^(21) "atom" m^(-3)`
`rArr (10^(-19))^(2)10^(21)xxn_(0) rArr n_(0)=10^(-17) m^(-3)`