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Microscopic resistivity and temperature

Short answer

In the free-electron model, conductivity σ = ne²τ/m, so resistivity ρ = m/(ne²τ). With increasing temperature, intensified lattice vibrations shorten the mean collision time τ, thereby raising resistivity.

Derive a relation for resistivity of a conductor using microscopic view. From your result, explain why resistivity of a conductor increases with increasing temperature.

Microscopic view (free-electron / Drude model)

In a metal, free electrons move randomly. On average they collide with lattice ions after a mean time τ (relaxation or mean collision time). Between collisions, an applied field E gives each electron an acceleration a = eE/m.

  1. Velocity gained between collisions: v_d = a τ = eEτ/m. This is the average drift velocity (the electron loses it on collision).
  2. Current density: J = n e v_d (n = free electrons per unit volume)

J = (n e² τ / m) E

  1. Compare with Ohm's law, J = σE, with conductivity σ = 1/ρ:

σ = n e² τ / m

ρ = m / (n e² τ)

  1. Equivalent forms: with mean free path λ and mean speed v, τ = λ/v, so ρ = m v/(n e² λ).

Why resistivity increases with temperature

  1. In a metal, n (free electrons) is practically constant with temperature, as are m and e.
  2. When the temperature rises, the lattice ions vibrate with larger amplitude, so electrons collide with them more often and the mean collision time τ decreases (also v increases slightly).
  3. Since ρ ∝ 1/τ, resistivity increases.

Experimentally, ρ = ρ₀[1 + α(T − T₀)], approximately linear for metals.

(In semiconductors, n rises rapidly with temperature, which outweighs the fall of τ, so their resistivity decreases.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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