Question

) State and prove Wiedemann–Franz law (or) Derive expressions for electrical and thermal conductivities on the basis of classical free electron theory and deduce the value of Lorentz number

Answer 1

The ratio of a metal's thermal conductivity (K) to its electrical conductivity () is proportional to the metal's absolute temperature.

$\frac{K}{\sigma} ∝ T \implies \frac{K}{\sigma} =L T$

Using the formulas for electrical and thermal conductivity of metals, we can drive Widemann-Franz's law using classical theory.

The expression for thermal conductivity

$K= \frac{K_B nv \lambda}{2}$

The expression for elecetricl conductivity

$\sigma = \frac{ne^2 \tau}{m}$

$\frac{K}{\sigma}= \frac{1/2 K_B nv \lambda}{ne^2 \tau / m}$

$\frac{K}{\sigma}= \frac{1}{2} \frac{mK_B v^2}{e^2 }$

$\frac{K}{\sigma}= \frac{1}{2} mv^2\frac{K_B}{e^2 }$

We know that kinetic energy s given by

$\frac{1}{2}mV^2= \frac{3}{2} K_BT$

$\frac{K}{\sigma}= \frac{3}{2} K_BT\frac{K_B}{e^2 }$

$\frac{K}{\sigma T}= \frac{3}{2}\frac{K^2_B}{e^2 }$

$\frac{K}{\sigma T}= L$

L is the Lorentz number

As a result, it is demonstrated that the ratio of a metal's thermal and electrical conductivity is proportional to the metal's absolute temperature.

$\implies L = \frac{3}{2} \frac{K_B^2}{e^2}$

$L = \frac{3}{2} \frac{(1.38*10^{-23})^2}{2(1.6*10^{-19})^2} = 1.12 * 10^{-8} W \Omega K^{-2}$