Problem Statement
State the Wiedemann-Franz law relating thermal and electrical conductivity of metals, and verify it for copper at $T=300\ \text{K}$. ($\kappa=400\ \text{W/m·K}$, $\sigma_{el}=6\times10^7\ \Omega^{-1}\text{m}^{-1}$)
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Wiedemann-Franz law: $\kappa/(\sigma_{el}T) = L_0$ (Lorenz number).
Step 2 — Apply the relevant physical law or equation: Theoretically from free electron model: $L_0 = \frac{\pi^2}{3}\left(\frac{k_B}{e}\right)^2 = 2.44\times10^{-8}\ \text{W·}\Omega\text{/K}^2$.
Step 3 — Solve algebraically for the unknown: For copper:
Step 4 — Substitute numerical values with units: $$L = \frac{\kappa}{\sigma_{el}T} = \frac{400}{6\times10^7\times300} = \frac{400}{1.8\times10^{10}} = 2.22\times10^{-8}\ \text{W·}\Omega\text{/K}^2$$
Step 5 — Compute and check the result: This is within 10% of the theoretical Lorenz number — good agreement. The law reflects that both heat and charge are carried by conduction electrons.
Step 6: Result: $L\approx2.2\times10^{-8}\ \text{W·}\Omega/\text{K}^2$ vs theoretical $2.44\times10^{-8}$.
Worked Calculation
$$L = \frac{\kappa}{\sigma_{el}T} = \frac{400}{6\times10^7\times300} = \frac{400}{1.8\times10^{10}} = 2.22\times10^{-8}\ \text{W·}\Omega\text{/K}^2$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{L = \frac{\kappa}{\sigma_{el}T} = \frac{400}{6\times10^7\times300} = \frac{400}{1.8\times10^{10}} = 2.22\times10^{-8}\ \text{W·}\Omega\text{/K}^2}$$
Wiedemann-Franz law: $\kappa/(\sigma_{el}T) = L_0$ (Lorenz number).
Theoretically from free electron model: $L_0 = \frac{\pi^2}{3}\left(\frac{k_B}{e}\right)^2 = 2.44\times10^{-8}\ \text{W·}\Omega\text{/K}^2$.
For copper:
$$L = \frac{\kappa}{\sigma_{el}T} = \frac{400}{6\times10^7\times300} = \frac{400}{1.8\times10^{10}} = 2.22\times10^{-8}\ \text{W·}\Omega\text{/K}^2$$
This is within 10% of the theoretical Lorenz number — good agreement. The law reflects that both heat and charge are carried by conduction electrons.
Result: $L\approx2.2\times10^{-8}\ \text{W·}\Omega/\text{K}^2$ vs theoretical $2.44\times10^{-8}$.
Answer
$$\boxed{L = \frac{\kappa}{\sigma_{el}T} = \frac{400}{6\times10^7\times300} = \frac{400}{1.8\times10^{10}} = 2.22\times10^{-8}\ \text{W·}\Omega\text{/K}^2}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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