---
title: "Copper drift mobility and scattering time"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:copper-mobility-scattering-time"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/copper-mobility-scattering-time
site: BCT Engineering
---

# Copper drift mobility and scattering time

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

- Asked in: 2069 Asar, 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/copper-mobility-scattering-time
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

Calculate the drift mobility and mean scattering time of conduction electrons in copper at room temperature, given the conductivity of the copper is 5.9×10⁵ Ω⁻¹ cm⁻¹, the density of copper is 8.96gcm⁻³ and mass is 63.5g/mol.

## Short answer

Electron concentration n = (rho N_A)/M = 8.50×10^28 m^-3. Drift mobility mu_d = sigma/(e n) = 4.33×10^-3 m^2/(V·s) (43.3 cm^2/(V·s)). Mean scattering time tau = mu_d m_e/e = 2.46×10^-14 s.

## Answer

Use $\sigma = e n \mu_d$ for the mobility and $\mu_d = e\tau/m_e$ for the mean scattering time. Copper gives one free electron per atom.

**Given:** $\sigma = 5.9 \times 10^{5}\ \Omega^{-1}\text{cm}^{-1} = 5.9 \times 10^{7}\ \Omega^{-1}\text{m}^{-1}$, $\rho = 8960\ \text{kg/m}^3$, $M = 63.5\ \text{g/mol}$.

#### Electron concentration

$$
n = \frac{\rho N_A}{M} = \frac{8960 \times 6.022 \times 10^{23}}{0.0635} = 8.50 \times 10^{28}\ \text{m}^{-3}
$$

#### Drift mobility

$$
\mu_d = \frac{\sigma}{e n} = \frac{5.9 \times 10^{7}}{1.602 \times 10^{-19} \times 8.50 \times 10^{28}} = 4.33 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 43.3\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}
$$

#### Mean scattering time

$$
\tau = \frac{\mu_d m_e}{e} = \frac{4.33 \times 10^{-3} \times 9.109 \times 10^{-31}}{1.602 \times 10^{-19}} = 2.46 \times 10^{-14}\ \text{s}
$$

**Answer:** $\mu_d \approx 43.3\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $\tau \approx 2.46 \times 10^{-14}\ \text{s}$.

## Check yourself

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

1. Mobility of copper from the given data is
   A. μ = σ e n
   B. μ = σ/(e n)
   C. μ = e n/σ
   D. μ = σ/n
   Answer: B. Rearranging σ = e n μ. (Key point: Mobility)

2. 5.9e5 Ω^-1 cm^-1 equals
   A. 5.9e7 Ω^-1 m^-1
   B. 5.9e3 Ω^-1 m^-1
   C. 5.9e9 Ω^-1 m^-1
   D. 5.9e5 Ω^-1 m^-1
   Answer: A. Multiply by 100 cm per m. (Key point: Unit conversion)

3. Mean scattering time is
   A. τ = μ e/m_e
   B. τ = e/(μ m_e)
   C. τ = μ/(e m_e)
   D. τ = μ m_e/e
   Answer: D. From μ = eτ/m_e. (Key point: Scattering time)

4. The values are about
   A. μ = 4.33 cm^2/V/s, τ = 2.46e-12 s
   B. μ = 433 cm^2/V/s, τ = 2.46e-14 s
   C. μ = 43.3 cm^2/V/s, τ = 2.46e-14 s
   D. μ = 43.3 cm^2/V/s, τ = 2.46e-10 s
   Answer: C. n = 8.50e28 m^-3 gives these. (Key point: Result)

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