Free electron model of metals

Classical Electron Model

Drude transport theory

The Drude theory assumes that movement of electrons can be described classically. The model resembles a pinball machine where the electrons accelerate, hit a scattering surface (positive ion), and then begin accelerating once more.

During the acceleration phase the velocity gained by the electron is described as:

$\Delta v=at=-{\frac {eE}{m}}t$

This gives the electron a total velocity

$v=v_{0}+\Delta v\!$

Then the electron scatters off a surface. The distance between scattering surfaces is called the mean free path, $l$.

Note that on average $v_{0}$ is equal to zero since the movement is in completely random directions. So now we can calculate the average total velocity to be

$<v>\ =\ <v_{0}+\Delta v>\ =\ <v_{0}>+<\Delta v>\ =\ <\Delta v>\!$

$<\Delta v>\ \sim {\overrightarrow {E}}$

Now using the equation for current

${\overrightarrow {j}}=-ne<v>=-ne<\Delta v>\ =-{\frac {ne^{2}}{m}}<t>$

Here $<t>$ is the average time between collisions, which is called the scatter time, $\tau$.

This gives us the Drude formula ,

$\sigma =-{\frac {ne^{2}}{m}}\tau$, where $\tau ={\frac {l}{v_{0}}}$, and $\sigma$ is conductivity.

Limitations of the classical electron model

Looking at the Drude model again and taking into account the Equipartition Theory:

${\frac {1}{2}}mv_{0}^{2}=k_{B}T$

we can show that the scattering time should be proportional to $T$

$v_{0}={\sqrt {\frac {2k_{B}T}{m}}}\sim {\sqrt {T}}$

$\Rightarrow \tau \sim {\frac {1}{\sqrt {T}}}$

However, this is wrong since velocity is not proportional to $T$. When $T=0$ there is still quantum uncertainty.