The resistivity of a metal is 1 times 10^{-8} | vedclass

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Question

(A) The expression for electrical resistivity $ho$ is given by $ho = \frac{m}{ne^2 	au}$.Rearranging the formula to solve for the relaxation time

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$4 	imes 10^{-14} \ s$

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(A) The expression for electrical resistivity $ho$ is given by $ho = \frac{m}{ne^2 	au}$.Rearranging the formula to solve for the relaxation time $	au$:$	au = \frac{m}{ne^2 ho}$Given values:$m = 9 	imes 10^{-31} \ kg$$n = 9 	imes 10^{28} \ m^{-3}$$e = 1.6 	imes 10^{-19} \ C$$ho = 1 	imes 10^{-8} \ \Omega \cdot m$Substituting these values into the equation:$	au = \frac{9 	imes 10^{-31}}{(9 	imes 10^{28}) 	imes (1.6 	imes 10^{-19})^2 	imes (1 	imes 10^{-8})}$$	au = \frac{9 	imes 10^{-31}}{9 	imes 10^{28} 	imes 2.56 	imes 10^{-38} 	imes 10^{-8}}$$	au = \frac{9 	imes 10^{-31}}{23.04 	imes 10^{-18}}$$	au \approx 0.39 	imes 10^{-13} \ s = 3.9 	imes 10^{-14} \ s$Rounding to the nearest value, we get $	au \approx 4 	imes 10^{-14} \ s$.

The resistivity of a metal is $1 \times 10^{-8} \Omega \cdot m$. If it contains $9 \times 10^{28}$ electrons per $m^3$, then the relaxation time of electrons inside the metal is nearly (electron mass $= 9 \times 10^{-31} \ kg$).

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