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(D) Given: Intrinsic carrier concentration $n_i = 10^{16} 	ext{ m}^{-3}$.Donor concentration (phosphorus atoms) $n_D = 10^{21} 	ext{ m}^{-3}$.Since

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$10^{11} 	ext{ per m}^3$

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(D) Given: Intrinsic carrier concentration $n_i = 10^{16} 	ext{ m}^{-3}$.Donor concentration (phosphorus atoms) $n_D = 10^{21} 	ext{ m}^{-3}$.Since phosphorus is a pentavalent impurity, it acts as a donor, making the crystal an $n$-type semiconductor.In an $n$-type semiconductor, the electron concentration $n_e \approx n_D = 10^{21} 	ext{ m}^{-3}$.According to the law of mass action for semiconductors, $n_e \cdot n_h = n_i^2$, where $n_h$ is the hole concentration.Substituting the values: $10^{21} \cdot n_h = (10^{16})^2$.$10^{21} \cdot n_h = 10^{32}$.$n_h = 10^{32} / 10^{21} = 10^{11} 	ext{ m}^{-3}$.Therefore, the new hole concentration is $10^{11} 	ext{ per m}^3$.

$A$ pure silicon crystal at temperature $300 \text{ K}$ has electron and hole concentration $(n_i) = 10^{16} \text{ per m}^3$ each. If $10^{21}$ phosphorus atoms $(n_D)$ are added per cubic metre, what is the new hole concentration in the silicon crystal?

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