Question

A Ge specimen is doped with Al. The concentration of acceptor atoms is ~10²¹ atoms/m³. 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

Answer: (A)

$10^{17}/m^{3}$

Answer 2

$n_{i}^{2} = n_{h}n_{e}$ ⇒ $(10^{19})^{2} = 10^{21} \times n_{e}$ ⇒ $n_{e} = 10^{17}/m^{3}.$