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(D) The intrinsic carrier concentration is given as $n_i = 1.41 	imes 10^{16} \, m^{-3}$.Phosphorus is a pentavalent impurity,so it acts as a donor d

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$2 	imes 10^{11} \, m^{-3}$

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(D) The intrinsic carrier concentration is given as $n_i = 1.41 	imes 10^{16} \, m^{-3}$.Phosphorus is a pentavalent impurity,so it acts as a donor dopant. The concentration of donor electrons is $N_D = 10^{21} \, m^{-3}$.Since $N_D \gg n_i$,the concentration of electrons $n_e \approx N_D = 10^{21} \, m^{-3}$.According to the law of mass action for semiconductors,$n_e \cdot n_h = n_i^2$,where $n_h$ is the concentration of holes.Therefore,$n_h = \frac{n_i^2}{n_e} = \frac{(1.41 	imes 10^{16})^2}{10^{21}}$.$n_h = \frac{1.9881 	imes 10^{32}}{10^{21}} \approx 1.9881 	imes 10^{11} \, m^{-3} \approx 2 	imes 10^{11} \, m^{-3}$.

If a semiconductor has an intrinsic carrier concentration of $1.41 \times 10^{16} \, m^{-3}$,when doped with $10^{21} \, m^{-3}$ phosphorus,then the concentration of holes at room temperature will be

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