In a pure silicon crystal,the electron-hole c | vedclass

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(B) Given: Intrinsic carrier concentration $n_i = 10^{16} \ m^{-3}$.Phosphorus is a pentavalent impurity,so it acts as a donor. The concentration of d

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$10^{11}$

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(B) Given: Intrinsic carrier concentration $n_i = 10^{16} \ m^{-3}$.Phosphorus is a pentavalent impurity,so it acts as a donor. The concentration of donor atoms added is $N_D = 10^{21} \ m^{-3}$.Since $N_D \gg n_i$,the electron concentration in the n-type semiconductor is approximately equal to the donor concentration: $n_e \approx N_D = 10^{21} \ m^{-3}$.According to the law of mass action for semiconductors,$n_e \cdot n_h = n_i^2$,where $n_h$ is the hole concentration.Substituting the values: $10^{21} \cdot n_h = (10^{16})^2$.$10^{21} \cdot n_h = 10^{32}$.$n_h = \frac{10^{32}}{10^{21}} = 10^{11} \ m^{-3}$.Therefore,the new hole concentration is $10^{11} \ m^{-3}$.

In a pure silicon crystal,the electron-hole concentration is $10^{16} \ m^{-3}$ at $301 \ K$. Now,$10^{21}$ atoms of phosphorus are added per cubic metre. The new hole concentration in silicon is (in per $m^3$):

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