In a semiconductor,the intrinsic carrier conc | vedclass

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(B) Given,the intrinsic carrier concentration $n_i = 6 	imes 10^8 \ m^{-3}$.After doping,the electron concentration $n_e = 9 	imes 10^{12} \ m^{-3}$

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$4 	imes 10^4 \ m^{-3}$

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(B) Given,the intrinsic carrier concentration $n_i = 6 	imes 10^8 \ m^{-3}$.After doping,the electron concentration $n_e = 9 	imes 10^{12} \ m^{-3}$.According to the law of mass action for semiconductors,the product of electron and hole concentrations is equal to the square of the intrinsic carrier concentration:$n_e 	imes n_h = n_i^2$Substituting the given values:$n_h = \frac{n_i^2}{n_e} = \frac{(6 	imes 10^8)^2}{9 	imes 10^{12}}$$n_h = \frac{36 	imes 10^{16}}{9 	imes 10^{12}}$$n_h = 4 	imes 10^4 \ m^{-3}$.

In a semiconductor,the intrinsic carrier concentration of electrons and holes is $6 \times 10^8 \ m^{-3}$. After doping with some impurity,the electron concentration increases to $9 \times 10^{12} \ m^{-3}$. Find the new hole concentration.

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