The electrical conductivity of a metal is \(3.2\times10^{7}\ \mathrm{S\,m^{-1}}\). The relaxation time of its free electrons is \(2.25\times10^{-14}\ \mathrm{s}\). The number of free electrons per cubic metre of the metal is (Take \(e=1.6\times10^{-19}\ \mathrm{C}\) and \(m=9\times10^{-31}\ \mathrm{kg}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(5\times10^{28}\ \mathrm{m^{-3}}\)
The drift picture of conduction gives the conductivity of a metal as \( \sigma=ne^{2}\tau/m \). Here \( n \) is the number of free electrons per unit volume and \( \tau \) is the relaxation time. So \( n=\sigma m/(e^{2}\tau) \).
\( \begin{aligned} n &= \frac{\sigma m}{e^{2}\tau} \\ &= \frac{3.2\times10^{7}\times9\times10^{-31}}{(1.6\times10^{-19})^{2}\times2.25\times10^{-14}} \\ &= \frac{2.88\times10^{-23}}{5.76\times10^{-52}} \\ &= 5\times10^{28}\ \mathrm{m^{-3}} \end{aligned} \)
NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… Ohm’s law, if we identify the conductivity σ as”
\( \sigma=\frac{ne^{2}}{m}\tau \)\( \qquad (3.23) \)