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(D) The conductivity $\sigma$ of an $n$-type semiconductor is given by $\sigma = n_e e \mu_e$,where $n_e$ is the number density of donor atoms,$e$ is

Vedclass · Accepted Answer

$10^{22} \ m^{-3}$

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(D) The conductivity $\sigma$ of an $n$-type semiconductor is given by $\sigma = n_e e \mu_e$,where $n_e$ is the number density of donor atoms,$e$ is the electronic charge,and $\mu_e$ is the electron mobility.Given: $\sigma = 6.4 \ \Omega^{-1} \ cm^{-1} = 6.4 	imes 10^2 \ \Omega^{-1} \ m^{-1}$.Electron mobility $\mu_e = 4 	imes 10^3 \ cm^2 \ V^{-1} \ s^{-1} = 4 	imes 10^3 	imes 10^{-4} \ m^2 \ V^{-1} \ s^{-1} = 0.4 \ m^2 \ V^{-1} \ s^{-1}$.Electronic charge $e = 1.6 	imes 10^{-19} \ C$.Substituting the values: $6.4 	imes 10^2 = n_e 	imes (1.6 	imes 10^{-19}) 	imes 0.4$.$n_e = \frac{6.4 	imes 10^2}{1.6 	imes 10^{-19} 	imes 0.4} = \frac{6.4 	imes 10^2}{0.64 	imes 10^{-19}} = 10 	imes 10^{21} = 10^{22} \ m^{-3}$.

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