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The resistivity of a metal is $1 \times 10^{-8} \Omega \mathrm{m}$. If it contains $9 \times 10^{28}$ electrons per $\mathrm{m}^3$ then the relaxation time of electrons inside the metal is nearly (electron mass $=9 \times 10^{-31} \mathrm{~kg}$ )
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The correct answer is:
$4 \times 10^{-14} \mathrm{~s}$
We have the expression for electrical resistivity is given by $\rho=\frac{m}{\mathrm{ne}^2 \tau}$
$$
\begin{aligned}
& \tau=\frac{\mathrm{m}}{\mathrm{ne}^2 \rho}=\frac{9.1 \times 10^{-31}}{9 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^2 \times 10^{-8}} \\
& \tau=3.94 \times 10^{-14} \\
& \approx 4 \times 10^{-14} \mathrm{~s}
\end{aligned}
$$
$$
\begin{aligned}
& \tau=\frac{\mathrm{m}}{\mathrm{ne}^2 \rho}=\frac{9.1 \times 10^{-31}}{9 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^2 \times 10^{-8}} \\
& \tau=3.94 \times 10^{-14} \\
& \approx 4 \times 10^{-14} \mathrm{~s}
\end{aligned}
$$
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