JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2019
- Medium
- MCQ
In a conductor, if the number of conduction electrons per unit volume is \(8.5 \times 10^{28}\, m^{-3}\) and mean free time is \(25\,fs\) (femto second), its approximate resistivity is \(\left( {{m_e} = 9.1 \times {{10}^{ - 31}}\,kg} \right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \({10^{ - 8}}\,\Omega m\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\rho = \frac{{2m}}{{n{e^2}\tau }}\\
= 3.34 \times {10^{ - 8}}\,\Omega \,m
\end{array}\)
\rho = \frac{{2m}}{{n{e^2}\tau }}\\
= 3.34 \times {10^{ - 8}}\,\Omega \,m
\end{array}\)
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