Mobility of electrons in a semiconductor is d | vedclass

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(C) The conductivity $\sigma$ of an $n$-type semiconductor is given by $\sigma = ne\mu_e$,where $n$ is the electron density,$e$ is the elementary char

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(C) The conductivity $\sigma$ of an $n$-type semiconductor is given by $\sigma = ne\mu_e$,where $n$ is the electron density,$e$ is the elementary charge $(1.6 	imes 10^{-19} \ C)$,and $\mu_e$ is the electron mobility.Given: $n = 10^{19} \ m^{-3}$,$\mu_e = 1.6 \ m^2/(V \cdot s)$,and $e = 1.6 	imes 10^{-19} \ C$.Substituting these values: $\sigma = (10^{19}) 	imes (1.6 	imes 10^{-19}) 	imes (1.6) = 1.6 	imes 1.6 = 2.56 \ \Omega^{-1} \cdot m^{-1}$.The resistivity $ho$ is the reciprocal of conductivity: $ho = \frac{1}{\sigma} = \frac{1}{2.56} \approx 0.39 \ \Omega \cdot m$.Rounding to the nearest option,the value is $0.4 \ \Omega \cdot m$.

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