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

When `GE` specimen is doped with `Al`, then concentration of acceptor atoms is also called concentration of holes. Using formula, `n_(i)^(2) =n_(0)p_(0)` where
`n_(1)` = concentration 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) xx n_(0) rArr n_(0)=10^(17) m^(-3)`.