The conductivity of a metal in terms of the number density n of free…

The correct answer is Option (2) → \( \displaystyle\frac{ne^{2}\tau}{m} \)

In an electric field \( E \) the free electrons gain a drift velocity \( v_d=eE\tau/m \), and the current density is \( j=nev_d \). Comparing \( j=ne^{2}\tau E/m \) with Ohm’s law \( j=\sigma E \) gives the conductivity.

Option (1) \( \displaystyle\frac{m}{ne^{2}\tau} \) — Incorrect. This is the reciprocal of the conductivity, that is, the resistivity \( \rho=1/\sigma \) of the metal.

Option (2) \( \displaystyle\frac{ne^{2}\tau}{m} \) — Correct. The conductivity increases with the number of free electrons and with the time between collisions. It decreases as the electron mass increases. One factor \( e \) comes from the force \( eE \). The other comes from the charge carried by each electron.

Option (3) \( \displaystyle\frac{m}{ne\tau} \) — Incorrect. It has \( m \) in the numerator, so a heavier carrier would conduct better. It also has only one factor of \( e \), so it does not have the units of conductivity.

Option (4) \( \displaystyle\frac{ne\tau}{m} \) — Incorrect. This leaves out one factor of \( e \). Here \( e\tau/m \) is the drift velocity per unit field, called the mobility. Multiplying it by the charge density \( ne \) gives \( ne^{2}\tau/m \).

NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… if we identify the conductivity σ as”

    \( \displaystyle \sigma=\frac{ne^2}{m}\,\tau \)\( \qquad (3.23) \)