---
title: "Relaxation time and mean free path"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:relaxation-time-mean-free-path"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/relaxation-time-mean-free-path
site: BCT Engineering
---

# Relaxation time and mean free path

Electrical Engineering Material · Chapter 2: Free Electron Theory of Conduction in Metal

- Asked in: 2080 Bhadra, 4 marks
- Video: [Important terms in Free electron theory (Drift velocity, Mean free path, Relaxation time) PHYSICS](https://www.youtube.com/watch?v=CI1R3r4sfDY) — Gautam Varde
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/relaxation-time-mean-free-path
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

The drift mobility of electron is 43 cm² V⁻¹ s⁻¹ and the mean speed is 2×10⁷ m/s. Calculate the relaxation time and mean free path of electrons between collisions.

## Short answer

Relaxation time tau = mu_d m_e / e = (4.3e-3 m^2/Vs × 9.109e-31 kg) / 1.602e-19 C = 2.44e-14 s. Mean free path lambda = u tau = 2e7 m/s × 2.44e-14 s = 4.89e-7 m (489 nm).

## Answer

The drift mobility and the mean free (relaxation) time $\tau$ are related by

$$
\mu_d = \frac{e\tau}{m_e} \quad\Rightarrow\quad \tau = \frac{\mu_d m_e}{e}, \qquad \lambda = u\tau
$$

where $u$ is the mean speed and $\lambda$ the mean free path.

**Given:** $\mu_d = 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 4.3 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$, $u = 2 \times 10^{7}\ \text{m/s}$, $m_e = 9.109 \times 10^{-31}\ \text{kg}$, $e = 1.602 \times 10^{-19}\ \text{C}$.

#### Relaxation time

$$
\tau = \frac{4.3 \times 10^{-3} \times 9.109 \times 10^{-31}}{1.602 \times 10^{-19}} = 2.44 \times 10^{-14}\ \text{s}
$$

#### Mean free path

$$
\lambda = u\tau = 2 \times 10^{7} \times 2.44 \times 10^{-14} = 4.89 \times 10^{-7}\ \text{m} \approx 489\ \text{nm}
$$

(Using the speed as given. The usual Fermi speed in copper is about $1.6 \times 10^{6}$ m/s, which would give $\lambda \approx 40$ nm; the mean free path scales directly with the speed used.)

**Answer:** $\tau = 2.44 \times 10^{-14}\ \text{s}$, $\lambda = 4.89 \times 10^{-7}\ \text{m}$.

## Check yourself

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

1. Relaxation time is obtained from mobility using
   A. τ = μ_d e / m_e
   B. τ = μ_d m_e / e
   C. τ = e/(μ_d m_e)
   D. τ = μ_d m_e e
   Answer: B. From μ_d = eτ/m_e. (Key point: Relation for τ)

2. 43 cm^2/V/s equals
   A. 4.3e-3 m^2/V/s
   B. 4.3e-1 m^2/V/s
   C. 43e-2 m^2/V/s
   D. 4.3e-5 m^2/V/s
   Answer: A. 1 cm^2 = 1e-4 m^2, so 43 cm^2 = 4.3e-3 m^2. (Key point: Unit conversion)

3. Mean free path is found from
   A. λ = τ/u
   B. λ = u/τ
   C. λ = u τ^2
   D. λ = u τ, with u the mean speed
   Answer: D. Distance travelled between collisions equals speed times time between them. (Key point: Mean free path)

4. Using the given data, τ is about
   A. 2.44e-12 s
   B. 4.89e-7 s
   C. 2.44e-14 s
   D. 2.44e-16 s
   Answer: C. τ = 4.3e-3 x 9.109e-31 / 1.602e-19 = 2.44e-14 s, giving λ = 489 nm with u = 2e7 m/s. (Key point: Result)

## Close variants set in other exams

- [Mean free path from drift mobility](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Amean-free-path-43-mobility) — Electrical Engineering Material, chapter 2, asked 4×

## Related in Electrical Engineering Material

- [Mean free path for 53 cm²/V·s mobility](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Amean-free-path-53-mobility) — Electrical Engineering Material, chapter 2
- [Drift mobility of electrons in potassium](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apotassium-drift-mobility-39-5) — Electrical Engineering Material, chapter 2, asked 2×
- [Copper drift mobility and scattering time](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Acopper-mobility-scattering-time) — Electrical Engineering Material, chapter 2
- [Mean free time of electrons in Si](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asilicon-mean-free-time) — Electrical Engineering Material, chapter 2
- [Copper drift mobility from scattering cross-section](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Acopper-mobility-from-cross-section) — Electrical Engineering Material, chapter 2
- [Drift mobility in potassium (molar mass 39.95)](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apotassium-drift-mobility-39-95) — Electrical Engineering Material, chapter 2, asked 2×

## Related in other subjects

- [Drift speed of electrons, 10 A](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adrift-speed-copper-ten-amp) — Engineering Physics, chapter 8, asked 2×
- [Drift speed of electrons, 20 A](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adrift-speed-copper-twenty-amp) — Engineering Physics, chapter 8, asked 2×
- [Drift velocity of electrons in copper](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adrift-velocity-copper-two-hundred-amp) — Engineering Physics, chapter 8
- [Free space path loss: 6 GHz, 10,500 km](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afspl-derivation-6ghz-10500km) — Propagation and Antenna, chapter 5