The free electrons in a metal have a number density of \(6.25\times10^{28}\ \mathrm{m^{-3}}\) and a relaxation time of \(4.5\times10^{-14}\ \mathrm{s}\). The resistivity of the metal is (Take \(e=1.6\times10^{-19}\ \mathrm{C}\) and \(m=9\times10^{-31}\ \mathrm{kg}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.25\times10^{-8}\ \Omega\,\mathrm{m}\)
In the drift picture of conduction, the resistivity of a metal is \( \rho=m/(ne^{2}\tau) \). Here \( n \) is the number of free electrons per unit volume and \( \tau \) is their relaxation time. More free electrons, or a longer time between collisions, give a smaller resistivity.
\( \begin{aligned} \rho &= \frac{m}{ne^{2}\tau} \\ &= \frac{9\times10^{-31}}{6.25\times10^{28}\times(1.6\times10^{-19})^{2}\times4.5\times10^{-14}} \\ &= \frac{9\times10^{-31}}{7.2\times10^{-23}} \\ &= 1.25\times10^{-8}\ \Omega\,\mathrm{m} \end{aligned} \)
NCERT (2026-27):
Temperature Dependence of Resistivity – “… From this equation, resistivity of a material is given by”
\( \rho=\frac{m}{ne^{2}\tau} \)\( \qquad (3.27) \)