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(D) The electrical conductivity $\sigma$ is given by the formula $\sigma = \frac{ne^2	au}{m}$, where $n$ is the charge carrier density, $e$ is the ch

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$2.5 	imes 10^{-14} \,s$

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(D) The electrical conductivity $\sigma$ is given by the formula $\sigma = \frac{ne^2	au}{m}$, where $n$ is the charge carrier density, $e$ is the charge of an electron, $	au$ is the relaxation time (average time between collisions), and $m$ is the mass of an electron.Rearranging the formula to solve for $	au$, we get $	au = \frac{\sigma m}{ne^2}$.Given values: $n = 9.1 	imes 10^{28} \,m^{-3}$, $\sigma = 6.4 	imes 10^7 \,S \,m^{-1}$, $m = 9.1 	imes 10^{-31} \,kg$, and $e = 1.6 	imes 10^{-19} \,C$.Substituting these values into the equation:$	au = \frac{(6.4 	imes 10^7) 	imes (9.1 	imes 10^{-31})}{(9.1 	imes 10^{28}) 	imes (1.6 	imes 10^{-19})^2}$$	au = \frac{6.4 	imes 10^7 	imes 9.1 	imes 10^{-31}}{9.1 	imes 10^{28} 	imes 2.56 	imes 10^{-38}}$$	au = \frac{6.4 	imes 10^{-24}}{2.56 	imes 10^{-10}}$$	au = 2.5 	imes 10^{-14} \,s$.Thus, the average time between two successive collisions is $2.5 	imes 10^{-14} \,s$.

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