A Ge specimen is doped with Al. The concentration of acceptor atoms is `10^(21)"atoms/m"^(3)`. Given that the intrinsic concentration of electron hole pairs is `10^(19)//m^(3)`, the concentration of electrons in the specimen is …….

Explanation

In pure semiconductor electron-hole `n_(1)=10^(19) m^(-3)` acceptor impurity `N_(A) = 10^(21) m^(-3)`
Holes concentration `n_(h) = 10^(21) m^(-3)`
Electrons concentration `= n_(e) = (n_(1)^(2))/(n_(h)) = ((10^(19))^(2))/(10^(21))=10^(17)m^(-3)`.