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Question

(A) For an intrinsic semiconductor,the electron concentration $n_e$ and hole concentration $n_h$ are equal to the intrinsic carrier concentration $n_i

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$10^6 \ m^{-3}$

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(A) For an intrinsic semiconductor,the electron concentration $n_e$ and hole concentration $n_h$ are equal to the intrinsic carrier concentration $n_i$.Given: $n_e = n_h = 2 	imes 10^8 \ m^{-3}$.The law of mass action states that $n_e 	imes n_h = n_i^2$.Therefore,$n_i^2 = (2 	imes 10^8) 	imes (2 	imes 10^8) = 4 	imes 10^{16} \ m^{-6}$.After doping,the new electron concentration is $n_e' = 4 	imes 10^{10} \ m^{-3}$.Since the product of carrier concentrations remains constant at a given temperature,$n_e' 	imes n_h' = n_i^2$.Substituting the values: $(4 	imes 10^{10}) 	imes n_h' = 4 	imes 10^{16}$.Solving for $n_h'$: $n_h' = \frac{4 	imes 10^{16}}{4 	imes 10^{10}} = 10^6 \ m^{-3}$.

$A$ semiconductor has equal electron and hole concentration of $2 \times 10^8 \ m^{-3}$. On doping with a certain impurity,the electron concentration increases to $4 \times 10^{10} \ m^{-3}$. What is the new hole concentration of the semiconductor?

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