Question

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

• 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 ${10}^{17}/m^3$

From law of mass action:
$n_i^2=n_e\times n_h$
$\Rightarrow n_e=\frac{{10}^{38}}{{10}^{21}}={10}^{17}/m^3$