---
title: "Conductivity of sodium metal"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:sodium-conductivity-atomic-weight"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/sodium-conductivity-atomic-weight
site: BCT Engineering
---

# Conductivity of sodium metal

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2074 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/sodium-conductivity-atomic-weight
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

What will be the conductivity of sodium metal having atomic weight 22.9 and density 1.013 gm/cm³? The relaxation time of sodium metal is 3×10⁻¹⁴ sec.

## Short answer

Electron density n = dN_A/M = 2.66×10²⁸ m⁻³. Using the Drude model, conductivity σ = ne²τ/m = (2.66×10²⁸)(1.6×10⁻¹⁹)²(3×10⁻¹⁴)/(9.11×10⁻³¹) ≈ 2.2×10⁷ S/m.

## Answer

Use the free-electron (Drude) model: σ = n e² τ / m, assuming one free electron per sodium atom.

Given: atomic weight M = 22.9 g/mol, density d = 1.013 g/cm³, τ = 3 × 10⁻¹⁴ s. Constants: N_A = 6.022 × 10²³ per mol, e = 1.6 × 10⁻¹⁹ C, m = 9.11 × 10⁻³¹ kg.

#### Step 1: electron density
1. Atoms per cm³ = d N_A / M = (1.013 × 6.022 × 10²³)/22.9 = 2.66 × 10²² cm⁻³
2. n = 2.66 × 10²² cm⁻³ = 2.66 × 10²⁸ m⁻³

#### Step 2: conductivity
σ = n e² τ / m
σ = (2.66 × 10²⁸)(1.6 × 10⁻¹⁹)²(3 × 10⁻¹⁴) / (9.11 × 10⁻³¹)

- e² = 2.56 × 10⁻³⁸
- numerator = 2.66 × 10²⁸ × 2.56 × 10⁻³⁸ × 3 × 10⁻¹⁴ = 2.04 × 10⁻²³
- σ = 2.04 × 10⁻²³ / 9.11 × 10⁻³¹ = 2.24 × 10⁷ S/m

(Using e = 1.602 × 10⁻¹⁹ gives 2.25 × 10⁷.)

**Answer: σ ≈ 2.2 × 10⁷ Ω⁻¹m⁻¹ (S/m)**, a resistivity of about 4.4 × 10⁻⁸ Ω·m.

## Check yourself

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

1. Conductivity in the free-electron model is
   A. σ = n e τ/m
   B. σ = n e^2 τ/m
   C. σ = n e^2 m/τ
   D. σ = n e τ^2/m
   Answer: B. σ = n e^2 τ/m. (Key point: Formula)

2. n is obtained from
   A. d N_A/M, one electron per atom
   B. M/(d N_A)
   C. d M/N_A
   D. N_A/d
   Answer: A. Atoms per cm^3 = 1.013 × 6.022e23/22.9 = 2.66e22. (Key point: Electron density)

3. n in m^−3 is
   A. 2.66 × 10^22
   B. 2.66 × 10^25
   C. 2.66 × 10^28
   D. 2.66 × 10^19
   Answer: C. 1 cm^−3 = 10^6 m^−3. (Key point: Unit conversion)

4. The conductivity is about
   A. 2.2 × 10^7 S/m
   B. 2.2 × 10^5 S/m
   C. 4.4 × 10^−8 S/m
   D. 2.2 × 10^9 S/m
   Answer: A. σ ≈ 2.24 × 10^7 S/m, ρ ≈ 4.4 × 10^−8 Ω·m. (Key point: Result)

## Close variants set in other exams

- [Conductivity of sodium from electron density](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asodium-conductivity-electron-density) — Engineering Physics, chapter 8, asked 2×

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