Difficulty: Easy
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:
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:
Final Answer:
10^−14 s
Discussion & Comments