# In a conductor , if the number of conduction electron per unit volume is $8.5\times 10^{28}m^{-3}$  and mean free time is $25\:fs$ ( femto second ) , it's approximate resistivity is : $(m_{e}=9.1\times10^{-31}kg)$ Option 1) $10^{-6}\:\:\Omega\:m$ Option 2) $10^{-7}\:\:\Omega\:m$ Option 3) $10^{-8}\:\:\Omega\:m$ Option 4) $10^{-5}\:\:\Omega\:m$

Answers (1)

Resistivity -

$\rho=\frac{m}{ne^{2}\tau}$

- wherein

It is independent of shape and size of the body i.e. $l$ and $A$

we know that ,

$\rho=\frac{m}{ne^2\tau}$

$\rho=\frac{9.1\times10^{-31}}{(8.5\times10^{28})(1.6\times10^{-19})^{2}(25\times10^{-15})}$

$\rho=\frac{9.1}{8.5\times(1.6)^{2}\times25}\times 10^{-6}\simeq 1.6\times10^{-8}$

Option 1)

$10^{-6}\:\:\Omega\:m$

Option 2)

$10^{-7}\:\:\Omega\:m$

Option 3)

$10^{-8}\:\:\Omega\:m$

Option 4)

$10^{-5}\:\:\Omega\:m$

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