A `Ge` specimen is doped 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 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

`N_(A)=10^(21)//m^(3)`, `n_(i)=10^(19)//m^(3)`
For P-type
Number of holes `=n_(h)=N_(A)=10^(21)//m^(3)`
Now, `n_(e)n_(h)=n_(i)^(2)`
`impliesn_(e)=((10^(19))^(2))/(10^(21))=10^(17)//m^(3)`.