Question

In a pure Silicon $\left(n_{i} = \left(10\right)^{16} m^{- 3}\right)$ crystal at $300 \, K, \, 10^{21}$ atoms of phosphorus are added per cubic meter. The new hole concentration will be

• A. $10^{21}\text{per}m^{3}$
• B. $10^{19}\text{per}m^{3}$
• C. $10^{11}\text{per}m^{3}$
• D. $10^{5}\text{per}m^{3}$

Answer 1

Answer: (C)

By using mass action law $n_{i}^{2}=n_{e}n_{h}$
$\Rightarrow n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{\left(\left(10\right)^{16}\right)^{2}}{\left(10\right)^{21}}=\left(10\right)^{11} \, per \, m^{3}$