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Conductivity of sodium from electron density

Short answer

Using σ = ne²τ/m, with n = 2.5×10²⁸ m⁻³, τ = 3×10⁻¹⁴ s, e and m constants, conductivity σ ≈ 2.1×10⁷ S/m.

What will be the conductivity of sodium metal having electron density 2.5×10²⁸ m⁻³ and relaxation time 3×10⁻¹⁴ sec?

Conductivity from the free-electron model is σ = ne²τ/m.

Given: n = 2.5×10²⁸ m⁻³, τ = 3×10⁻¹⁴ s, e = 1.602×10⁻¹⁹ C, m = 9.11×10⁻³¹ kg.

σ = n e² τ / m
  = (2.5×10²⁸)(1.602×10⁻¹⁹)²(3×10⁻¹⁴) / (9.11×10⁻³¹)
  = (2.5×10²⁸)(2.566×10⁻³⁸)(3×10⁻¹⁴) / (9.11×10⁻³¹)
  = 1.925×10⁻²³ / 9.11×10⁻³¹
  = 2.11×10⁷ (Ω·m)⁻¹

Answer: σ ≈ 2.1×10⁷ Ω⁻¹m⁻¹ (S/m), so the resistivity is ρ = 1/σ ≈ 4.7×10⁻⁸ Ω·m, which is the right order for a metal.

✓ Answer checked by Claude Opus 5.5 (high effort)

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