Consider two conductors A and B such that the number density of free electrons in conductor A is half of that in conductor B, all other factors (including the relaxation time) being the same. What will be the resistivity of conductor A with respect to conductor B?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The resistivity of conductor A will be double that of conductor B.
From the drift of electrons picture, the conductivity is \( \sigma=ne^2\tau/m \), and the resistivity is its reciprocal, \( \rho=1/\sigma \).
\( \displaystyle \rho=\frac{m}{ne^2\tau} \)
Here \( e \), \( m \) and the relaxation time \( \tau \) are the same for both conductors. So the resistivity is inversely proportional to the number density \( n \) of free electrons.
With \( n_A=n_B/2 \):
\( \begin{aligned} \frac{\rho_A}{\rho_B} &= \frac{n_B}{n_A} \\ &= \frac{n_B}{n_B/2} \\ &= 2 \end{aligned} \)
Therefore, \( \rho_A=2\rho_B \). Conductor A has fewer free electrons to carry the current, so its resistivity is twice as large.
NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… if we identify the conductivity σ as”
\( \displaystyle \sigma=\frac{ne^2}{m}\,\tau \)\( \qquad (3.23) \)