According to standard school models, the specific resistance (or conductivity) depends on the movability of the charges. In a metal lattice, some electrons can freely move and are only hindered by lattice impurities and lattice vibrations. These vibrations and thus the specific resistance increase with increasing temperature. When the metal melts, there is no more lattice left, so this argument can't be continued.

Does the resistance increase or decrease with the phase transition and what is the thermal behaviour in the liquid phase? And of course, why?

Possibility 1: With the phase transition, the resistance increases because the atomic rests move and thus disturb the electron flow even more.

Possibility 2: With the phase transition, the resistance jumps to nearly $\infty$, because the lattice collapses and the electrons get bound to the atoms.

Possibility 3: With the phase transition, the resistance decreases, because the atomic rests get movable as well.

In each case, I'd assume the resistance to decrease with increasing temperature in the liquid phase, because movement increases.


1 Answer 1


Consider the fermi-dirac state density distribution:

$ f(E) = \frac{1}{1 + \exp^{\frac{E}{k_{B}T}}} $

Fermi-Dirac Distribution

This shows that as the temperature of a system of spin 1/2 particles increases, the density of states lowers itself over more states.

To model the solid-state of a metal, we take the k-space of dimension:

$ V = (\frac{\pi}{L})^3 $

and the fermi-sphere, of radius equal to the fermi wave-vector:

$V = \frac{4}{24} \pi r^{3} $

because $ r = k_{F} $ and add factor of two for each spin +-1/2 of N, number of electrons in volume, then you can get:

$ k_{F} = (\frac{2\pi^{2}V}{N})^{1/3} $

Where N is the number of electrons, generally $ N_{A} \times N_{e} $, V is the volume.

This enables us to find the fermi-energy:

$ E_{F} = \frac{\hbar^{2}k_{F}^{2}}{2m_{e}} $

In the first formula and graph, this energy defines what energy the states reach at 0K.

This distribution of states relates to electricity due to the Band Structure of metals, of the valency and conduction bands. As you can see in the image the fermi-energy is flotted on the axes for different types of metal and DOS along the bottom stands for density of states which is the first formula.


$ V = IR $

$ R = \frac{\rho L}{A} $ - $ \rho $ is resistivity

$ \rho = \frac{1}{\sigma} $ $ \sigma $ is conductivity

$ \sigma = \frac{ nq^{2}a}{\sqrt{3k_{B}Tm}} $

$a$ = mean free path ~ lattice spacing

$n$ = number of electrons, $q$ = charge of electron, $T$ = Temperature, $m$ = mass of particle.


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