---
title: "Microscopic resistivity and temperature"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:resistivity-microscopic-temperature"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/resistivity-microscopic-temperature
site: BCT Engineering
---

# Microscopic resistivity and temperature

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2073 Shrawan, 5 marks
- Video: [Variation of Resistivity with Temperature | Current & Resistance](https://www.youtube.com/watch?v=h69CQv--9NQ) — JIWAN THAPA PHYSICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/resistivity-microscopic-temperature
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

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

## Answer

#### 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

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

σ = n e² τ / m

**ρ = m / (n e² τ)**

4. 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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Drift velocity is
   A. eEτ/m
   B. eE/(mτ)
   C. mEτ/e
   D. e τ/(mE)
   Answer: A. v_d = aτ with a = eE/m. (Key point: Drift velocity)

2. The resistivity is
   A. n e^2 τ/m
   B. m/(n e^2 τ)
   C. m n/(e^2 τ)
   D. τ/(n e m)
   Answer: B. σ = ne^2τ/m and ρ = 1/σ. (Key point: Resistivity)

3. Why does ρ rise with T in a metal?
   A. n falls
   B. Ion vibrations shorten τ
   C. e increases
   D. m decreases
   Answer: B. More collisions reduce τ, and ρ ∝ 1/τ. (Key point: Temperature)

4. Semiconductor resistivity falls with T because
   A. τ rises
   B. n rises rapidly, outweighing the fall in τ
   C. Collisions stop
   D. m falls
   Answer: B. Carrier density increases quickly with temperature. (Key point: Semiconductors)

## Related in Engineering Physics

- [Atomic view of resistivity, ρ = m/ne²τ](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acurrent-density-mobility-atomic-resistivity) — Engineering Physics, chapter 8
- [J = σE, resistivity and superconductors](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aresistivity-j-sigma-e-superconductor) — Engineering Physics, chapter 8

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