---
title: "Drift mobility of electrons in potassium"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:potassium-drift-mobility-39-5"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/potassium-drift-mobility-39-5
site: BCT Engineering
---

# Drift mobility of electrons in potassium

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

- Asked in: 2079 Bhadra, 4 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/potassium-drift-mobility-39-5
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

If electric conductivity of potassium is 1.39 × 10⁵ Sm/cm, calculate the drift mobility of electron at room temperature. Molar mass and density of potassium are 39.5 and 0.91 gm/cc.

## Short answer

Electron concentration n = (910 kg/m^3 × 6.022e23) / 0.0395 kg/mol = 1.387e28 m^-3. With sigma = 1.39e7 S/m, drift mobility mu_d = sigma / (e n) = 1.39e7 / (1.602e-19 × 1.387e28) = 6.25e-3 m^2/Vs (62.5 cm^2/Vs).

## Answer

Conductivity of a metal is $\sigma = e n \mu_d$, so

$$
\mu_d = \frac{\sigma}{e n}
$$

Potassium is monovalent (one valence electron per atom), so $n$ equals the atomic concentration.

**Given:** $\sigma = 1.39 \times 10^{5}\ \text{S/cm} = 1.39 \times 10^{7}\ \text{S/m}$ (the unit "Sm/cm" is read as S/cm, the standard value for potassium), $M = 39.5\ \text{g/mol} = 0.0395\ \text{kg/mol}$, $\rho = 0.91\ \text{g/cm}^3 = 910\ \text{kg/m}^3$.

#### Electron concentration

$$
n = \frac{\rho N_A}{M} = \frac{910 \times 6.022 \times 10^{23}}{0.0395} = 1.387 \times 10^{28}\ \text{m}^{-3}
$$

#### Drift mobility

$$
\mu_d = \frac{1.39 \times 10^{7}}{1.602 \times 10^{-19} \times 1.387 \times 10^{28}} = 6.25 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}
$$

$$
\mu_d = 62.5\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}
$$

(The corresponding mean free time is $\tau = \mu_d m_e/e = 3.6 \times 10^{-14}$ s.)

**Answer:** $\mu_d \approx 6.25 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 62.5\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$.

## Check yourself

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

1. Drift mobility from conductivity is
   A. μ_d = σ e n
   B. μ_d = σ/(e n)
   C. μ_d = e n/σ
   D. μ_d = σ n/e
   Answer: B. σ = e n μ_d. (Key point: Formula)

2. Why does n equal the atomic concentration for potassium?
   A. It is monovalent: one free electron per atom
   B. It has two valence electrons per atom
   C. All electrons in the atom are free
   D. n is always 1e28 for alkali metals
   Answer: A. With one valence electron per atom, n = ρ N_A / M. (Key point: Electron concentration)

3. The conductivity 1.39e5 S/cm should be converted to
   A. 1.39e3 S/m
   B. 1.39e9 S/m
   C. 1.39e5 S/m
   D. 1.39e7 S/m
   Answer: D. Multiply by 100 cm/m. (Key point: Unit handling)

4. The mobility comes out as about
   A. 6.25 cm^2/V/s
   B. 625 cm^2/V/s
   C. 62.5 cm^2/V/s
   D. 1.39e7 cm^2/V/s
   Answer: C. n = 1.387e28 m^-3 gives μ = 6.25e-3 m^2/V/s = 62.5 cm^2/V/s. (Key point: Result)

## Close variants set in other exams

- [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×