= Solution
For degenerate carriers with approximately elastic scattering and a common transport <relaxation time>, $\sigma=ne^2\tau/m^*$ and $\kappa_e=(C_e/V)v_F^2\tau/3$. Using $C_e/V=\pi^2nk_B^2T/(2E_F)$ and $v_F^2=2E_F/m^*$ gives
$$
\boxed{\frac{\kappa_e}{\sigma T}=L_0,\qquad
L_0=\frac{\pi^2k_B^2}{3e^2}=2.44\times10^{-8}\ \mathrm{W\,\Omega\,K^{-2}}}.
$$
This is the Wiedemann–Franz law. The same mobile <electrons> carry charge and heat, but Pauli statistics set the <electronic heat capacity> and hence the <Lorenz number>. The relation applies to electronic <thermal conductivity>, not automatically to total <electrical conductivity> including <phonons>. Inelastic <electron-phonon scattering> can relax <energy> and charge currents differently; nondegeneracy and strong <energy> dependence of scattering also limit the simple form.
A meaningful test combines a four-terminal electrical measurement with a steady thermal-gradient measurement on the same sample and <temperature> range. Contact heat resistance, parasitic heat leaks, <phonon> conduction and sample geometry must be controlled before interpreting a measured Lorenz ratio as an electronic departure from the law.
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