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

# Copper drift mobility from scattering cross-section

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

- Asked in: 2068 Baisakh, 6 marks
- 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-from-cross-section
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

The mean speed of conduction electrons in copper is 1.5×10⁶ m/s. The cross sectional area of scattering is 3.9×10⁻²²m². Estimate the drift mobility of electrons and conductivity of copper. Given density of copper is 8.96g/cm³ and the atomic mass is 63.56 g/mole.

## Short answer

Electron density n = ρN_A/M = 8.49×10²⁸ m⁻³. Mean free path λ = 1/(S n) ≈ 3.02×10⁻⁸ m, giving τ = λ/u = 2.01×10⁻¹⁴ s. Drift mobility μ_d = eτ/m_e ≈ 35.4 cm²/V/s, conductivity σ = e n μ_d ≈ 4.82×10⁷ S/m.

## Answer

In the free electron (Drude) model, an electron moving at mean speed $u$ is scattered when it comes within the cross-sectional area $S$ of a scattering centre. With $N_s$ scatterers per unit volume:

$$
\lambda = \frac{1}{S N_s}, \qquad \tau = \frac{\lambda}{u} = \frac{1}{S N_s u}, \qquad \mu_d = \frac{e\tau}{m_e}, \qquad \sigma = e n \mu_d
$$

Each copper atom (ion) is taken as a scatterer, and each gives one conduction electron, so $N_s = n$.

**Given:** $u = 1.5 \times 10^{6}\ \text{m/s}$, $S = 3.9 \times 10^{-22}\ \text{m}^2$, $\rho = 8960\ \text{kg/m}^3$, $M = 63.56\ \text{g/mol}$.

#### Atomic (and electron) concentration

$$
n = N_s = \frac{\rho N_A}{M} = \frac{8960 \times 6.022 \times 10^{23}}{0.06356} = 8.49 \times 10^{28}\ \text{m}^{-3}
$$

#### Mean free path and mean free time

$$
\begin{aligned}
\lambda &= \frac{1}{S N_s} = \frac{1}{3.9 \times 10^{-22} \times 8.49 \times 10^{28}} = 3.02 \times 10^{-8}\ \text{m} \\
\tau &= \frac{\lambda}{u} = \frac{3.02 \times 10^{-8}}{1.5 \times 10^{6}} = 2.01 \times 10^{-14}\ \text{s}
\end{aligned}
$$

#### Drift mobility

$$
\mu_d = \frac{e\tau}{m_e} = \frac{1.602 \times 10^{-19} \times 2.01 \times 10^{-14}}{9.109 \times 10^{-31}} = 3.54 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 35.4\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}
$$

#### Conductivity

$$
\sigma = e n \mu_d = 1.602 \times 10^{-19} \times 8.49 \times 10^{28} \times 3.54 \times 10^{-3} = 4.82 \times 10^{7}\ \text{S/m}
$$

This is close to the measured value for copper ($5.9 \times 10^{7}$ S/m), showing that the simple model gives the right order of magnitude.

**Answer:** $\mu_d \approx 35.4\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $\sigma \approx 4.82 \times 10^{7}\ \text{S/m}$ (with $\lambda = 30.2$ nm, $\tau = 2.01 \times 10^{-14}$ s).

## Check yourself

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

1. With scattering cross-section S and N_s scatterers per volume, λ equals
   A. S N_s
   B. 1/(S N_s)
   C. S/N_s
   D. N_s/S
   Answer: B. An electron sweeps volume S λ containing one scatterer, so λ = 1/(S N_s). (Key point: Mean free path)

2. N_s for copper is taken as
   A. Atomic concentration ρ N_A/M = 8.49e28 m^-3
   B. Electron concentration divided by 2
   C. ρ M/N_A
   D. 1e19 m^-3
   Answer: A. Each atom is a scatterer and gives one free electron. (Key point: Scatterer density)

3. τ is obtained from
   A. τ = λ u
   B. τ = u/λ
   C. τ = λ u^2
   D. τ = λ/u
   Answer: D. Time = path / speed: 3.02e-8 / 1.5e6 = 2.01e-14 s. (Key point: Mean free time)

4. After τ, mobility and conductivity are
   A. μ = m_e/(eτ); σ = μ/(e n)
   B. μ = 354 cm^2/V/s; σ = 4.8e9 S/m
   C. μ = eτ/m_e = 35.4 cm^2/V/s; σ = e n μ = 4.82e7 S/m
   D. μ = 3.54 cm^2/V/s; σ = 4.8e6 S/m
   Answer: C. Apply μ = eτ/m_e then σ = e n μ. (Key point: Mobility and conductivity)

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