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Question

(A) The conductivity $\sigma$ is the reciprocal of resistivity $\rho$,so $\sigma = \frac{1}{\rho} = \frac{1}{0.5} = 2 \ \Omega^{-1} m^{-1}$.For an int

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$2.16 	imes 10^{19} / m^3$

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(A) The conductivity $\sigma$ is the reciprocal of resistivity $ho$,so $\sigma = \frac{1}{ho} = \frac{1}{0.5} = 2 \ \Omega^{-1} m^{-1}$.For an intrinsic semiconductor,the electron concentration $n_e$ equals the hole concentration $n_h$,which is equal to the intrinsic carrier concentration $n_i$.The conductivity formula is $\sigma = n_i e (\mu_e + \mu_h)$.Substituting the given values: $2 = n_i 	imes (1.6 	imes 10^{-19}) 	imes (0.39 + 0.19)$.$2 = n_i 	imes (1.6 	imes 10^{-19}) 	imes (0.58)$.$2 = n_i 	imes (0.928 	imes 10^{-19})$.$n_i = \frac{2}{0.928 	imes 10^{-19}} \approx 2.155 	imes 10^{19} / m^3$.Rounding to two decimal places,we get $2.16 	imes 10^{19} / m^3$.

The resistivity of a pure semiconductor is $0.5 \ \Omega m$. If the electron and hole mobility are $0.39 \ m^2 / V-s$ and $0.19 \ m^2 / V-s$ respectively,then calculate the intrinsic carrier concentration.

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