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(B) The conductivity of an intrinsic semiconductor is given by $\sigma = n_i e (\mu_e + \mu_h)$.Given $n_i = 2.5 	imes 10^{19} \ m^{-3}$,$e = 1.6 	i

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$4.0 \ k\Omega$

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(B) The conductivity of an intrinsic semiconductor is given by $\sigma = n_i e (\mu_e + \mu_h)$.Given $n_i = 2.5 	imes 10^{19} \ m^{-3}$,$e = 1.6 	imes 10^{-19} \ C$,$\mu_e = 0.39 \ m^2/V\cdot s$,and $\mu_h = 0.19 \ m^2/V\cdot s$.$\sigma = (2.5 	imes 10^{19}) 	imes (1.6 	imes 10^{-19}) 	imes (0.39 + 0.19) = 4 	imes 0.58 = 2.32 \ \Omega^{-1}m^{-1}$.The resistivity $ho = \frac{1}{\sigma} = \frac{1}{2.32} \ \Omega\cdot m$.The resistance $R = ho \frac{L}{A}$.Given $L = 0.925 \ cm = 9.25 	imes 10^{-3} \ m$ and $A = 1 \ mm^2 = 10^{-6} \ m^2$.$R = \frac{1}{2.32} 	imes \frac{9.25 	imes 10^{-3}}{10^{-6}} = \frac{9250}{2.32} \approx 3987 \ \Omega \approx 4.0 \ k\Omega$.

The length of a germanium rod is $0.925 \ cm$ and its area of cross-section is $1 \ mm^2$. If for germanium $n_i = 2.5 \times 10^{19} \ m^{-3}$,$\mu_h = 0.19 \ m^2/V\cdot s$,and $\mu_e = 0.39 \ m^2/V\cdot s$,then the resistance of the rod is:

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