In a sample of pure silicon,10^{13} text{ ato | vedclass

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(D) The conductivity of an $n$-type semiconductor is given by $\sigma = \frac{1}{ho} = e n_e \mu_e$.Here,$e = 1.6 	imes 10^{-19} 	ext{ C}$ is the

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$520.9$

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(D) The conductivity of an $n$-type semiconductor is given by $\sigma = \frac{1}{ho} = e n_e \mu_e$.Here,$e = 1.6 	imes 10^{-19} 	ext{ C}$ is the elementary charge.$n_e$ is the concentration of donor electrons,which is equal to the concentration of phosphorus atoms,$n_e = 10^{13} 	ext{ cm}^{-3}$.$\mu_e = 1200 	ext{ cm}^2/	ext{V} \cdot 	ext{s}$ is the mobility of electrons.Substituting these values into the formula for resistivity $ho$:$ho = \frac{1}{e n_e \mu_e} = \frac{1}{1.6 	imes 10^{-19} 	imes 10^{13} 	imes 1200}$.$ho = \frac{1}{1.6 	imes 1200 	imes 10^{-6}} = \frac{1}{1920 	imes 10^{-6}} = \frac{1}{1.92 	imes 10^{-3}} = \frac{1000}{1.92} \approx 520.83 \, \Omega \cdot 	ext{cm}$.Thus,the resistivity is approximately $520.9 \, \Omega \cdot 	ext{cm}$.

In a sample of pure silicon,$10^{13} \text{ atoms/cm}^3$ of phosphorus is added. If all donor atoms are active,what will be the resistivity at $20 ^oC$ if the mobility of electrons is $1200 \text{ cm}^2/\text{V} \cdot \text{s}$? (in $\Omega \cdot \text{cm}$)

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