Charge transport — typical order of electron relaxation time in metals In the classical Drude model (room temperature, common metals), the mean relaxation time τ of conduction electrons is typically of the order of:
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A10^−6 s
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B10^−10 s
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C10^−14 s
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D10^−20 s
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E10^−2 s
Answer
Correct Answer: 10^−14 s
Explanation
Introduction / Context:The relaxation time τ characterizes the average time between momentum-randomizing collisions for conduction electrons. It is a key parameter in Ohmic conduction and appears in mobility, conductivity, and dielectric response models.
Given Data / Assumptions:
- Room temperature metallic conductor.
- Classical Drude response with τ roughly constant over low frequencies.
- No strong impurity or phonon anomalies.
Concept / Approach:In the Drude picture, conductivity σ is σ = n e^2 τ / m. Using typical σ ≈ 10^7 S/m and n ≈ 10^29 m^−3 gives τ on the order of 10^−14 s. More refined models (Drude–Sommerfeld) adjust numbers slightly but keep the same order of magnitude.
Step-by-Step Solution:Assume n ≈ 8 × 10^28 m^−3 and σ ≈ 5.8 × 10^7 S/m (copper).Solve τ = m σ / (n e^2).Insert constants → τ ~ few × 10^−14 s, confirming the order 10^−14 s.
Verification / Alternative check:From mean free path l = v_F τ with Fermi velocity v_F ~ 10^6 m/s and l ~ 10^−8 m, τ ~ 10^−14 s again.
Why Other Options Are Wrong:10^−6 s and 10^−2 s are macroscopic timescales, far too long. 10^−10 s is still orders too large. 10^−20 s is unphysically short for electron transport in metals.
Common Pitfalls:
- Confusing scattering time with dielectric relaxation time of materials.
- Using carrier densities for semiconductors instead of metals.
Final Answer:10^−14 s