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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
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Calculate the mean free path in $Cu$ at room temperature $300K$, if the number density of free electrons is $8.5 \times 10^{28}/m^3$ and resistivity $r = 1.7 \times 10^{-8} mho-m$. Given $k = 1.38 \times 10^{-23} J/K \: and\: 1A = 10^{-10} m$

$\begin {array} {1 1} (A)\;25A & \quad (B)\;20A \\ (C)\;5A & \quad (D)\;30A \end {array}$


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 The value of $\tau$ can be calcutated from the formula
$ne^2\tau=m\sigma^2$ by substituting the value of mass of electron,charge of electron, and value of conductivity
$V_{rms} = \sqrt{ \large\frac{3KT}{m}}$
Mean free path $ = \lambda = V_{rms} \times \tau$
Ans : (A)


answered Feb 10, 2014 by thanvigandhi_1
edited Sep 10, 2014 by thagee.vedartham

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