Question:
In a conductor, if the number of conduction electrons per unit volume is $8.5 \times 10^{28} \mathrm{~m}^{-3}$ and mean free time is $25 \mathrm{fs}$ (femto second), it's approximate resistivity is: $\left(\mathrm{m}_{\mathrm{e}}=9.1 \times\right.$ $10^{-31} \mathrm{~kg}$ )
Correct Option: , 3
Solution:
(3) $\rho=\frac{m}{n e^{2} \tau}$
$=\frac{9.1 \times 10^{-31}}{8.5 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^{2} \times 25 \times 10^{-15}}$
$=10^{-8} \Omega-\mathrm{m}$