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(C) For an intrinsic semiconductor, the law of mass action states that $n_i^2 = n_e n_h$, where $n_i$ is the intrinsic carrier concentration, $n_e$ is

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$9 	imes 10^8 \,m^{-3}$

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(C) For an intrinsic semiconductor, the law of mass action states that $n_i^2 = n_e n_h$, where $n_i$ is the intrinsic carrier concentration, $n_e$ is the electron concentration, and $n_h$ is the hole concentration.Given $n_i = 6 	imes 10^{15} \,m^{-3}$ and the doped electron concentration $n_e = 4 	imes 10^{22} \,m^{-3}$.Substituting these values into the formula:$(6 	imes 10^{15})^2 = (4 	imes 10^{22}) 	imes n_h$$36 	imes 10^{30} = 4 	imes 10^{22} 	imes n_h$$n_h = \frac{36 	imes 10^{30}}{4 	imes 10^{22}}$$n_h = 9 	imes 10^8 \,m^{-3}$.

The concentration of electrons in an intrinsic semiconductor is $6 \times 10^{15} \,m^{-3}$. On doping with an impurity, the electron concentration increases to $4 \times 10^{22} \,m^{-3}$. In thermal equilibrium, the concentration of the holes in the doped semiconductor is:

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