Question

A Ge specimen is doped with Al. The concentration of acceptor atom is $\sim {10}^{21}$ atoms$/m^3$. Given that the intrinsic concentration of electron hole pairs is $\sim {10}^{19}/m^3$. The concentration of electrons in the specimen is$?$

• A. ${10}^{17}/m^3$
• B. ${10}^{15}/m^3$
• C. ${10}^4/m^3$
• D. ${10}^2/m^3$

Answer 1

The correct option is A ${10}^{17}/m^3$

$\therefore n_1^2=n_e{n}_n$
$({10}^{19}{)}^2=n_e\times {10}^{21}$
[given, $n_l=\sim {10}^{19}/m^3,n_h=\sim {10}^{21}atom/m^3$]
$n_e={10}^{17}/m^3$.