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Mean free path from drift mobility

Short answer

Mean free time τ = μ m_e/e = (43×10⁻⁴ m²/V·s × 9.109×10⁻³¹ kg)/(1.602×10⁻¹⁹ C) = 2.44×10⁻¹⁴ s. Mean free path λ = u τ = 1.2×10⁶ m/s × 2.44×10⁻¹⁴ s = 2.93×10⁻⁸ m ≈ 29.3 nm.

Drift mobility of conduction electron is 43 cm²V⁻¹s⁻¹ and mean speed is 1.2×10⁶ms⁻¹. Calculate the mean free path of electrons between collisions.

In the free-electron (Drude) model the drift mobility is related to the mean free time τ\tau between collisions by

μd=eτme⇒τ=μdmee\mu_d = \frac{e\tau}{m_e} \quad\Rightarrow\quad \tau = \frac{\mu_d m_e}{e}

and the mean free path is the distance travelled at the mean speed uu in time τ\tau: λ=uτ\lambda = u\tau.

Given: μd=43 cm2V−1s−1=43×10−4 m2V−1s−1\mu_d = 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 43 \times 10^{-4}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}, u=1.2×106 m/su = 1.2 \times 10^{6}\ \text{m/s}, me=9.109×10−31 kgm_e = 9.109 \times 10^{-31}\ \text{kg}, e=1.602×10−19 Ce = 1.602 \times 10^{-19}\ \text{C}.

Mean free time

τ=43×10−4×9.109×10−311.602×10−19=2.44×10−14 s\tau = \frac{43 \times 10^{-4} \times 9.109 \times 10^{-31}}{1.602 \times 10^{-19}} = 2.44 \times 10^{-14}\ \text{s}

Mean free path

λ=uτ=1.2×106×2.44×10−14=2.93×10−8 m≈29.3 nm\lambda = u\tau = 1.2 \times 10^{6} \times 2.44 \times 10^{-14} = 2.93 \times 10^{-8}\ \text{m} \approx 29.3\ \text{nm}

This is roughly 100 interatomic spacings (copper spacing is about 0.26 nm), showing that conduction electrons are not scattered by every ion but only by lattice vibrations and defects.

Answer: λ≈2.93×10−8 m=29.3 nm\lambda \approx 2.93 \times 10^{-8}\ \text{m} = 29.3\ \text{nm} (with τ=2.44×10−14 s\tau = 2.44 \times 10^{-14}\ \text{s}).

✓ Answer checked by Claude Opus 5.5 (high effort)

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