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,them 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`=pair=10^(19)//m^(3)`
`n_(0)`=concentration of electron
`p_(0)`=concentration of holes=`10^(21) atom//m^(3)`
`implies(10^(19))^(2)=10^(21)xxn_(0)impliesn_(0)=10^(17)//m^(3)`.