In a semiconductor,the number densities of el | vedclass

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(A) Given: $n_e = 8 	imes 10^{18} \ m^{-3}$,$n_h = 5 	imes 10^{18} \ m^{-3}$,$\mu_e = 2.3 \ m^2/V \cdot s$,$\mu_h = 0.01 \ m^2/V \cdot s$.Since $n_e

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

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(A) Given: $n_e = 8 	imes 10^{18} \ m^{-3}$,$n_h = 5 	imes 10^{18} \ m^{-3}$,$\mu_e = 2.3 \ m^2/V \cdot s$,$\mu_h = 0.01 \ m^2/V \cdot s$.Since $n_e > n_h$,the semiconductor is $N$-type.The conductivity $\sigma$ is given by $\sigma = e(n_e \mu_e + n_h \mu_h)$.The resistivity $ho = \frac{1}{\sigma} = \frac{1}{e(n_e \mu_e + n_h \mu_h)}$.Substituting the values:$ho = \frac{1}{1.6 	imes 10^{-19} 	imes [(8 	imes 10^{18} 	imes 2.3) + (5 	imes 10^{18} 	imes 0.01)]}$$ho = \frac{1}{1.6 	imes 10^{-19} 	imes [18.4 	imes 10^{18} + 0.05 	imes 10^{18}]}$$ho = \frac{1}{1.6 	imes 10^{-1} 	imes 18.45} = \frac{1}{0.16 	imes 18.45} = \frac{1}{2.952} \approx 0.3387 \ \Omega \cdot m \approx 0.34 \ \Omega \cdot m$.Thus,it is $N$-type with resistivity $0.34 \ \Omega \cdot m$.

In a semiconductor,the number densities 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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