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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-2/section-a/5/solution

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Past exam of the physics A course of the University of Cambridge / 2019 / Paper 2 / Section A / 5 / Solution by

Codex  0 2026-10-18

Treat sodium as monovalent, with one mobile Electron per atom. Its mass density is

ρ_{m} = 0.97 g c m^{- 3} = 970 kg m^{- 3}

, and its molar mass is

M = 0.023 kg mo l^{- 1}

. Using the Avogadro constant,

n = \frac{ρ _{m} N _{A}}{M} = \frac{970}{0.023} (6.022 \times 1 0^{23}) m^{- 3} ≃ 2.54 \times 1 0^{28} m^{- 3} .

The electron mobility

μ = e τ / m_{e}

relates the magnitude of drift velocity to electric field. The Drude model then gives

σ = n e μ = (2.54 \times 1 0^{28}) (1.602 \times 1 0^{- 19}) (5.2 \times 1 0^{- 3}) ≃ 2.1 \times 1 0^{7} S m^{- 1} .

Using

m_{e} = 9.109 \times 1 0^{- 31} kg

,

τ = \frac{m _{e} μ}{e} ≃ 2.96 \times 1 0^{- 14} s ≃ 3.0 \times 1 0^{- 14} s .

The inferred time is the momentum relaxation time. Identifying it with the mean time between collisions is the intended simple isotropic Drude model assumption; small-angle scattering need not relax directed momentum in a single event.

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