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Conductivity of sodium metal

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.

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.

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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