Question #206333

) State and prove Wiedemannâ€“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

Expert's answer

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

\frac{K}{\sigma} âť 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}

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