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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

Similar Questions

For a pure $Si$ crystal at $300 \ K$,the electron $(n_e)$ and hole $(n_h)$ concentrations are equal,being $1.5 \times 10^{16} \ m^{-3}$. On doping with Indium,the hole concentration increases to $4.5 \times 10^{22} \ m^{-3}$. Calculate the new electron concentration $(n_e)$ in the doped $Si$.

Which of the following statements is not true?

In an extrinsic semiconductor,the number densities of holes and electrons are ${N_p}$ and ${N_e}$ respectively. Then:

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