---
title: "Mean free path from drift mobility"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:mean-free-path-43-mobility"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/mean-free-path-43-mobility
site: BCT Engineering
---

# Mean free path from drift mobility

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

- Asked in: 2076 Chaitra, 4 marks; 2076 Asoj, 4 marks; 2073 Chaitra, 4 marks (asked 3 times)
- 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/mean-free-path-43-mobility
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

Drift mobility of conduction electron is 43 cm²V⁻¹s⁻¹ and mean speed is 1.2×10⁶ms⁻¹. Calculate the mean free path of electrons between collisions.

## Short answer

Mean free time τ = μ m_e/e = (43×10⁻⁴ m²/V·s × 9.109×10⁻³¹ kg)/(1.602×10⁻¹⁹ C) = 2.44×10⁻¹⁴ s. Mean free path λ = u τ = 1.2×10⁶ m/s × 2.44×10⁻¹⁴ s = 2.93×10⁻⁸ m ≈ 29.3 nm.

## Answer

In the free-electron (Drude) model the drift mobility is related to the mean free time $\tau$ between collisions by

$$
\mu_d = \frac{e\tau}{m_e} \quad\Rightarrow\quad \tau = \frac{\mu_d m_e}{e}
$$

and the mean free path is the distance travelled at the mean speed $u$ in time $\tau$: $\lambda = u\tau$.

**Given:** $\mu_d = 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 43 \times 10^{-4}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$, $u = 1.2 \times 10^{6}\ \text{m/s}$, $m_e = 9.109 \times 10^{-31}\ \text{kg}$, $e = 1.602 \times 10^{-19}\ \text{C}$.

#### Mean free time

$$
\tau = \frac{43 \times 10^{-4} \times 9.109 \times 10^{-31}}{1.602 \times 10^{-19}} = 2.44 \times 10^{-14}\ \text{s}
$$

#### Mean free path

$$
\lambda = u\tau = 1.2 \times 10^{6} \times 2.44 \times 10^{-14} = 2.93 \times 10^{-8}\ \text{m} \approx 29.3\ \text{nm}
$$

This is roughly 100 interatomic spacings (copper spacing is about 0.26 nm), showing that conduction electrons are not scattered by every ion but only by lattice vibrations and defects.

**Answer:** $\lambda \approx 2.93 \times 10^{-8}\ \text{m} = 29.3\ \text{nm}$ (with $\tau = 2.44 \times 10^{-14}\ \text{s}$).

## Check yourself

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

1. Which relation links drift mobility and mean free time?
   A. mu = e tau / m
   B. mu = m tau / e
   C. mu = e m / tau
   D. mu = tau/(e m)
   Answer: A. mu_d = e tau / m_e, so tau = mu m / e. (Key point: Mobility relation)

2. What is 43 cm^2/Vs in SI?
   A. 43e-4 m^2/Vs
   B. 43e4 m^2/Vs
   C. 43e-2 m^2/Vs
   D. 43 m^2/Vs
   Answer: A. 1 cm^2 = 1e-4 m^2. (Key point: Unit conversion)

3. What is tau?
   A. 2.44e-14 s
   B. 2.44e-8 s
   C. 1.2e-6 s
   D. 4.3e-3 s
   Answer: A. tau = 43e-4 x 9.109e-31 / 1.602e-19 = 2.44e-14 s. (Key point: Mean free time)

4. How is the mean free path obtained, and what is it?
   A. lambda = u tau = 29.3 nm
   B. lambda = u/tau = 5e19 m
   C. lambda = mu tau = 1e-17 m
   D. lambda = u tau = 0.29 nm
   Answer: A. Distance at mean speed during tau: 1.2e6 x 2.44e-14 = 2.93e-8 m. (Key point: Mean free path)

## Close variants set in other exams

- [Relaxation time and mean free path](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Arelaxation-time-mean-free-path) — 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 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
- [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
- [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×
- [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

## 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
- [Drift speed in copper wire, 17 mA](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adrift-speed-copper-seventeen-ma) — Engineering Physics, chapter 8