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

Answer

A

Explanation

`n_(h) = 10^(21) "atoms"//m^(3)`
`n_(i) = 10^(19) "atoms"//m^(3) , n_(e) = ?`
`n_(e) n_(b) = n_(i)^(2)`
`n_(e)= (n_(i)^(2))/(n_(b)) = ((10^(19))^(2))/(10^(21)) = (10^(38))/(10^(21)) = 10^(17) "atoms"//m^(3)`.