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(A) Given: Electron concentration $n_e = 8 	imes 10^{18} \, m^{-3}$,Hole concentration $n_h = 5 	imes 10^{18} \, m^{-3}$.Since $n_e > n_h$,the semic

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$N-$type and its resistivity is $0.34 \, \Omega \cdot m$

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(A) Given: Electron concentration $n_e = 8 	imes 10^{18} \, m^{-3}$,Hole concentration $n_h = 5 	imes 10^{18} \, m^{-3}$.Since $n_e > n_h$,the semiconductor is $N-$type.Mobilities are $\mu_e = 2.3 \, m^2/V \cdot s$ and $\mu_h = 0.01 \, m^2/V \cdot s$.The conductivity $\sigma$ is given by $\sigma = e(n_e \mu_e + n_h \mu_h)$,where $e = 1.6 	imes 10^{-19} \, C$.$\sigma = 1.6 	imes 10^{-19} 	imes [8 	imes 10^{18} 	imes 2.3 + 5 	imes 10^{18} 	imes 0.01]$$\sigma = 1.6 	imes 10^{-19} 	imes 10^{18} 	imes [18.4 + 0.05]$$\sigma = 0.16 	imes 18.45 = 2.952 \, \Omega^{-1} \cdot m^{-1}$.Resistivity $ho = 1/\sigma = 1 / 2.952 \approx 0.3387 \, \Omega \cdot m \approx 0.34 \, \Omega \cdot m$.

In a semiconductor,the concentrations of electrons and holes are $8 \times 10^{18} \, m^{-3}$ and $5 \times 10^{18} \, m^{-3}$ respectively. If the mobilities of electrons and holes are $2.3 \, m^2/V \cdot s$ and $0.01 \, m^2/V \cdot s$ respectively,then the semiconductor is:

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