In a conductor,if the number of conduction el | vedclass

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(D) The resistivity $ho$ of a conductor is given by the formula: $ho = \frac{m_e}{n e^2 	au}$.Given values:$n = 8.5 	imes 10^{28} \ m^{-3}$$	au

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$10^{-8} \ \Omega m$

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(D) The resistivity $ho$ of a conductor is given by the formula: $ho = \frac{m_e}{n e^2 	au}$.Given values:$n = 8.5 	imes 10^{28} \ m^{-3}$$	au = 25 \ fs = 25 	imes 10^{-15} \ s$$m_e = 9.1 	imes 10^{-31} \ kg$$e = 1.6 	imes 10^{-19} \ C$Substituting these values into the formula:$ho = \frac{9.1 	imes 10^{-31}}{(8.5 	imes 10^{28}) 	imes (1.6 	imes 10^{-19})^2 	imes (25 	imes 10^{-15})}$$ho = \frac{9.1 	imes 10^{-31}}{8.5 	imes 10^{28} 	imes 2.56 	imes 10^{-38} 	imes 25 	imes 10^{-15}}$$ho = \frac{9.1 	imes 10^{-31}}{544 	imes 10^{-25}}$$ho \approx 0.0167 	imes 10^{-6} \ \Omega m \approx 1.67 	imes 10^{-8} \ \Omega m$.Comparing this with the given options,the approximate value is $10^{-8} \ \Omega m$.

In a conductor,if the number of conduction electrons per unit volume is $8.5 \times 10^{28} \ m^{-3}$ and the mean free time is $25 \ fs$ (femtosecond),what is its approximate resistivity? (Given: $m_e = 9.1 \times 10^{-31} \ kg$,$e = 1.6 \times 10^{-19} \ C$)

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