Free electron model of metals

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Free Electron Model Atomic Structure Header Classical Drude Quantum Schrodinger

The free electron model is a simple model for the behavior of valence electrons in a crystal structure of a metallic solid. It was developed principally by Arnold Sommerfeld who combined the classical Drude model with quantum mechanical Fermi-Dirac statistics. Given its simplicity, it is surprisingly successful in explaining many experimental phenomena, especially

• the Wiedemann-Franz law which relates electrical conductivity and thermal conductivity;
• the temperature dependence of the heat capacity;
• the shape of the electronic density of states;
• the range of binding energy values;
• electrical conductivities;
• thermal electron emission and field electron emission from bulk 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.

And from E&M recall that the resistivity of the material is:

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

Also another often quoted value is the mobility, the proportionality constant between the velocity of the electrons and the electric field it is placed in, of the material which is:

$\mu ={\frac {e\tau }{m_{e}}}={\frac {\sigma }{ne}}$

Hall effect

When a metal is placed in a magnetic field $\mathbf {B}$ and a current density $\mathbf {j}$ is passed through it, a transverse electric field $\mathbf {E_{H}}$ is set up given by

$\mathbf {E_{H}} =R_{H}\mathbf {B} \ \times \ \mathbf {j}$

This is the Hall Effect and $R_{H}$ is known as the Hall coefficient.

This effect originates from the Lorentz Force, which makes charges pile up on opposite sides of the conductor. This creates a potential difference, or Hall Voltage, within the conductor. The internal charges will continue to pile up until it is in equilibrium with the Lorentz Force.

This set up is useful because it enables us to measure the electron mobility, $\mu$, from measured values of $R_{H}$ and electronic conductivity, $\sigma$.

$\mu =|R_{H}|\sigma \!$

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.

Pauli Principle and Fermi-Dirac statistics

Fermi Dirac Probability Distribution

$f(E)={\frac {1}{e^{\frac {E-E_{F}}{k_{B}T}}+1}}$

Schrodinger Equation for Free Electrons

Fermi energy and Fermi momentum

Consider a cube of metal with sides $L$ with faces perpendicular to x, y, and z axes. We must solve the time independent Schrodinger equation:

$-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi =\epsilon \psi$

We can do this by imposing periodic boundary conditions such that:

$\psi (x+L,y+L,z+L)=\psi (x,y,z)$

This gives us solutions of Schrodinger's equation are plane waves

$\psi (x,y,z)={\frac {1}{V^{\frac {1}{2}}}}e^{i\mathbf {k} \cdot \mathbf {r} }={\frac {1}{V^{\frac {1}{2}}}}e^{(k_{x}x+k_{y}y+k_{z}z)}$

where $V=L^{3}$ and $V^{-{\frac {1}{2}}}$ is to normalize the wave-function.

Maybe someone who has taken quantum can elaborate on the explanation above?' MatthewHoza 14:35, 29 April 2009 (EDT)

To satisfy the periodic boundary conditions the allowed wave-vectors must be:

$k_{x}={\frac {2\pi p}{L}}\ \ \ k_{y}={\frac {2\pi q}{L}}\ \ \ k_{z}={\frac {2\pi r}{L}}$

where $p,q,$ and $r$ are integer values. This corresponds to an energy

$\epsilon ={\frac {\hbar ^{2}k^{2}}{2m}}={\frac {\hbar ^{2}}{2m}}(k_{x}^{2}+k_{y}^{2}+k_{z}^{2})$

and a momentum

$\mathbf {p} =\hbar \mathbf {k} =\hbar (k_{x},k_{y},k_{z})$