The length of a germanium rod is 0.928 cm an | vedclass

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(A) Given data: $l = 0.928 \ cm = 0.928 	imes 10^{-2} \ m$,$A = 1 \ mm^2 = 1 	imes 10^{-6} \ m^2$,$n_i = 2.5 	imes 10^{19} \ m^{-3}$,$\mu_h = 0.15

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$50 \ \Omega \ cm$

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(A) Given data: $l = 0.928 \ cm = 0.928 	imes 10^{-2} \ m$,$A = 1 \ mm^2 = 1 	imes 10^{-6} \ m^2$,$n_i = 2.5 	imes 10^{19} \ m^{-3}$,$\mu_h = 0.15 \ m^2 V^{-1} s^{-1}$,and $\mu_e = 0.35 \ m^2 V^{-1} s^{-1}$.Conductivity $\sigma$ for an intrinsic semiconductor is given by $\sigma = n_i e (\mu_e + \mu_h)$.Substituting the values: $\sigma = (2.5 	imes 10^{19}) 	imes (1.6 	imes 10^{-19}) 	imes (0.35 + 0.15)$.$\sigma = 2.5 	imes 1.6 	imes 0.5 = 2 \ \Omega^{-1} m^{-1}$.Resistivity $ho$ is the reciprocal of conductivity: $ho = \frac{1}{\sigma} = \frac{1}{2} \ \Omega m = 0.5 \ \Omega m$.Converting to $\Omega \ cm$: $0.5 \ \Omega m = 0.5 	imes 100 \ \Omega \ cm = 50 \ \Omega \ cm$.

The length of a germanium rod is $0.928 \ cm$ and its area of cross-section is $1 \ mm^2$. If for germanium $n_i = 2.5 \times 10^{19} \ m^{-3}$,$\mu_h = 0.15 \ m^2 V^{-1} s^{-1}$,and $\mu_e = 0.35 \ m^2 V^{-1} s^{-1}$,then the resistivity is:

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