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A metal has $9 \times 10^{28}$ conduction electrons per $\mathrm{m}^3$ and its resistivity is $1 \times 10^{-8} \Omega$. m. If the drift speed of an electron in the metal is $1.6 \times 10^6 \mathrm{~m} / \mathrm{s}$ then its mean free path is (mass of electron $=9 \times 10^{-31} \mathrm{~kg}$ and charge of electron $=1.6 \times 10^{-19} \mathrm{C}$ )
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Verified Answer
The correct answer is:
$62.5 \mathrm{~nm}$
Mean free path is given by
$$
\begin{aligned}
& \lambda=\frac{\mathrm{mv}}{1 \mathrm{Ve}^2 \mathrm{p}} \\
& \lambda=\frac{9 \times 10^{-31} \times 1.6 \times 10^6}{9 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^2 \times 10^{-8}} \\
& \lambda=6.25 \times 10^{-8} \mathrm{~m} \\
& \lambda=62.5 \mathrm{~nm}
\end{aligned}
$$
$$
\begin{aligned}
& \lambda=\frac{\mathrm{mv}}{1 \mathrm{Ve}^2 \mathrm{p}} \\
& \lambda=\frac{9 \times 10^{-31} \times 1.6 \times 10^6}{9 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^2 \times 10^{-8}} \\
& \lambda=6.25 \times 10^{-8} \mathrm{~m} \\
& \lambda=62.5 \mathrm{~nm}
\end{aligned}
$$
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