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(B) For an intrinsic semiconductor,the law of mass action states that the product of the density of conduction electrons $(n_e)$ and the density of ho

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$1 	imes 10^9 \, m^{-3}$

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(B) For an intrinsic semiconductor,the law of mass action states that the product of the density of conduction electrons $(n_e)$ and the density of holes $(n_h)$ is equal to the square of the intrinsic carrier concentration $(n_i^2)$:$n_e \cdot n_h = n_i^2$Given that for an intrinsic semiconductor,$n_e = n_h = n_i$,we have $n_i = 7.07 	imes 10^{15} \, m^{-3}$.Therefore,$n_i^2 = (7.07 	imes 10^{15})^2 \approx 50 	imes 10^{30} = 5 	imes 10^{31} \, m^{-6}$.When the semiconductor is doped with indium (a trivalent impurity),it becomes a $p$-type semiconductor where the density of holes $n_h = 5 	imes 10^{22} \, m^{-3}$.Using the mass action law for the doped semiconductor:$n_e = \frac{n_i^2}{n_h} = \frac{5 	imes 10^{31}}{5 	imes 10^{22}} = 1 	imes 10^9 \, m^{-3}$.Thus,the density of conduction electrons in the doped semiconductor is $1 	imes 10^9 \, m^{-3}$.

In an intrinsic semiconductor,the density of conduction electrons is $7.07 \times 10^{15} \, m^{-3}$. When it is doped with indium,the density of holes becomes $5 \times 10^{22} \, m^{-3}$. Find the density of conduction electrons in the doped semiconductor.

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