---
title: "Drift mobility in potassium (molar mass 39.95)"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:potassium-drift-mobility-39-95"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/potassium-drift-mobility-39-95
site: BCT Engineering
---

# Drift mobility in potassium (molar mass 39.95)

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

- Asked in: 2072 Kartik, 4 marks
- Video: [Carrier Drift Current: Mobility](https://www.youtube.com/watch?v=LZ6hcaieFMU) — Jordan Edmunds Chetty
- 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-39-95
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

If electrical 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 39.95 and 0.91 gm/cc.

## Short answer

Drift mobility mu_d = sigma/(e n). Electron concentration n = (rho N_A)/M = 1.37×10^28 m^-3, giving mu_d = 6.33×10^-3 m^2/(V·s) (or 63.3 cm^2/(V·s)).

## Answer

Conductivity $\sigma = e n \mu_d$, so $\mu_d = \sigma/(en)$. Potassium is monovalent, so the free electron concentration 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), $M = 39.95\ \text{g/mol}$, $\rho = 0.91\ \text{g/cm}^3 = 910\ \text{kg/m}^3$.

#### Electron concentration

$$
n = \frac{\rho N_A}{M} = \frac{910 \times 6.022 \times 10^{23}}{0.03995} = 1.372 \times 10^{28}\ \text{m}^{-3}
$$

#### Drift mobility

$$
\mu_d = \frac{\sigma}{e n} = \frac{1.39 \times 10^{7}}{1.602 \times 10^{-19} \times 1.372 \times 10^{28}} = 6.33 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}
$$

$$
\mu_d = 63.3\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}
$$

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

**Answer:** $\mu_d \approx 6.33 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 63.3\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$.

## Check yourself

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

1. Drift mobility is found from
   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. n for potassium is
   A. ρ N_A/M = 1.372e28 m^-3
   B. ρ M/N_A
   C. 2 ρ N_A/M
   D. M N_A/ρ
   Answer: A. Monovalent, one electron per atom. (Key point: Concentration)

3. ρ = 0.91 g/cm^3 equals
   A. 0.91 kg/m^3
   B. 9.1e5 kg/m^3
   C. 91 kg/m^3
   D. 910 kg/m^3
   Answer: D. 1 g/cm^3 = 1000 kg/m^3. (Key point: Unit conversion)

4. The mobility is about
   A. 6.33 cm^2/V/s
   B. 633 cm^2/V/s
   C. 63.3 cm^2/V/s
   D. 1.37e28 cm^2/V/s
   Answer: C. 6.33e-3 m^2/V/s = 63.3 cm^2/V/s. (Key point: Result)

## Close variants set in other exams

- [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×

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