Drude conductivity formula — identifying σ for a metal If a metal has n conduction electrons per m^3, each with charge e and mass m, and the relaxation time is τ, which expression gives the electrical conductivity σ?

Introduction / Context:
The classical Drude model connects microscopic carrier properties to macroscopic conductivity. Recognizing the correct dependence on carrier density, charge, mass, and scattering time is foundational in solid-state physics and electrical engineering.

Given Data / Assumptions:

• Free-electron-like conduction (Drude picture).
• Number density n (m^−3), electron charge e, mass m, relaxation time τ.
• Low-field Ohmic regime where drift velocity is linear in applied field.

Concept / Approach:
In the Drude model, the average drift velocity is v_d = (e * E * τ) / m. Current density is J = n * e * v_d. By definition J = σ * E. Eliminating v_d gives the conductivity formula.

Step-by-Step Solution:
Start: v_d = (e * E * τ) / m.Compute J: J = n * e * v_d = n * e * (e * E * τ / m) = (n * e^2 * τ / m) * E.Identify σ from J = σ E → σ = n * e^2 * τ / m.

Verification / Alternative check:
Dimensional analysis: e^2 has C^2, n in m^−3, τ in s, m in kg; combining yields S/m, the correct unit for conductivity.

Why Other Options Are Wrong:
(a) missing one factor of e; (c) is resistivity, not conductivity; (d) has τ in denominator incorrectly; (e) introduces τ^2 with wrong dependence.

Common Pitfalls:

• Dropping one factor of e when moving from force law to current density.
• Inverting the expression and mistakenly giving resistivity instead of conductivity.

Final Answer:
σ = n * e^2 * τ / m