A metal has 9 times 10^{28} conduction electr | vedclass

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(D) The resistivity $ho$ is given by $ho = \frac{m}{ne^2	au}$, where $	au$ is the relaxation time. Also, the mean free path $\lambda$ is given b

Vedclass · Accepted Answer

$62.5$

Vedclass · Answer

(D) The resistivity $ho$ is given by $ho = \frac{m}{ne^2	au}$, where $	au$ is the relaxation time. Also, the mean free path $\lambda$ is given by $\lambda = v_d 	au$, where $v_d$ is the drift speed. From the resistivity formula, $	au = \frac{m}{ne^2ho}$. Substituting this into the mean free path formula: $\lambda = v_d \left( \frac{m}{ne^2ho} ight) = \frac{m v_d}{ne^2ho}$. Given: $m = 9 	imes 10^{-31} \ kg$, $v_d = 1.6 	imes 10^6 \ m/s$, $n = 9 	imes 10^{28} \ m^{-3}$, $e = 1.6 	imes 10^{-19} \ C$, and $ho = 1 	imes 10^{-8} \ \Omega \cdot m$. $\lambda = \frac{(9 	imes 10^{-31}) 	imes (1.6 	imes 10^6)}{(9 	imes 10^{28}) 	imes (1.6 	imes 10^{-19})^2 	imes (1 	imes 10^{-8})}$. $\lambda = \frac{14.4 	imes 10^{-25}}{9 	imes 10^{28} 	imes 2.56 	imes 10^{-38} 	imes 10^{-8}} = \frac{14.4 	imes 10^{-25}}{23.04 	imes 10^{-18}} = 0.625 	imes 10^{-7} \ m = 62.5 \ nm$.

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