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Mean free path for 53 cm²/V·s mobility

Short answer

Mean free time τ = μ_d m_e/e = 3.01×10⁻¹⁴ s. Mean free path λ = u τ = 6.63×10⁻⁸ m (66.3 nm).

Conduction electrons with drift mobility of 53cm²V⁻¹s⁻¹ and mean speed of 2.2 × 10⁶ms⁻¹ collides. Calculate the mean free path of electrons between collision.

The drift mobility is related to the mean free time τ\tau by μd=eτ/me\mu_d = e\tau/m_e, and the mean free path is λ=uτ\lambda = u\tau, where uu is the mean speed.

Given: μd=53 cm2V−1s−1=53×10−4 m2V−1s−1\mu_d = 53\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 53 \times 10^{-4}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}, u=2.2×106 m/su = 2.2 \times 10^{6}\ \text{m/s}.

Mean free time

τ=μdmee=53×10−4×9.109×10−311.602×10−19=3.01×10−14 s\tau = \frac{\mu_d m_e}{e} = \frac{53 \times 10^{-4} \times 9.109 \times 10^{-31}}{1.602 \times 10^{-19}} = 3.01 \times 10^{-14}\ \text{s}

Mean free path

λ=uτ=2.2×106×3.01×10−14=6.63×10−8 m≈66.3 nm\lambda = u\tau = 2.2 \times 10^{6} \times 3.01 \times 10^{-14} = 6.63 \times 10^{-8}\ \text{m} \approx 66.3\ \text{nm}

This is a few hundred atomic spacings, so electrons pass many ions before being scattered.

Answer: λ≈6.63×10−8 m=66.3 nm\lambda \approx 6.63 \times 10^{-8}\ \text{m} = 66.3\ \text{nm} (τ=3.01×10−14\tau = 3.01 \times 10^{-14} s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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