Question

A pure Ge specimen is doped with $Al$. The number density of acceptor atoms is approximately $10^{21} m^{-3}$. If density of electron hole pair in an intrinsic semiconductor is approximately $10^{19} m^{-3}$, the number density of electrons in the specimen is

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

Answer 1

Answer: (C)

Given that, the density of electron hole pair in an intrinsic semiconductor
$(n_{i}) =\frac{10^{19}}{m^3}$
density of holes $(n_{h}) =\frac{10^{21}}{m^3}$
We know that $n_h \cdot n_{e} =n_i^2$
$\therefore 10^{21} \times n_{e} =(10^{19})^2$
$\Rightarrow n_{e} =\frac{10^{19} \times 10^{19}}{10^{21}}$
$=10^{17}m^{-3}$