Question

In a conductor, if the number of conduction electrons per unit volume is $8.5\times {10}^{28}{\text{m}}^3$ and mean free time is $25\text{fs}$ (femto second), it's approximate resistivity is :
$(m_e=9.1\times {10}^{-31}\text{kg})$

• A. ${10}^{-8}\Omega \text{m}$
• B. ${10}^{-7}\Omega \text{m}$
• C. ${10}^{-5}\Omega \text{m}$
• D. ${10}^{-6}\Omega \text{m}$

Answer 1

The correct option is A ${10}^{-8}\Omega \text{m}$

$\because J=\sigma E\Rightarrow \rho =\frac{E}{J}$

The resistivity, $\rho =\frac{m}{n{e}^2\tau }$

$=\frac{9.1\times {10}^{-31}}{8.5\times {10}^{28}\times (1.6\times {10}^{-19}{)}^2\times 25\times {10}^{-15}}$

$=1.6\times {10}^{-8}\Omega \text{m}\approx {10}^{-8}\Omega \text{m}$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.