Q. In a pure silicon $(n_{i} = 10^{16}/m^{3})$ crystal at $300\ K$, $10^{21}$ atoms of phosphorus are added per cubic meter. The new hole concentration will be
A) $10^{21}\ per\ m^{3}$
B) $10^{19}\ per\ m^{3}$
C) $10^{11}\ per\ m^{3}$
D) $10^{5}\ per\ m^{3}$
Explanation
(c) By using mass action law $n_{i}^{2} = n_{e}n_{h}$
$\Rightarrow n_{h} = \frac{n_{i}^{2}}{n_{e}} = \frac{\left( 10^{16} \right)^{2}}{10^{21}} = 10^{11}per\ m^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.