Order 1238142: Condensed matter
tutorthammy
Lect-2-Free-electron.pdf
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Questions A uniform silver wire has a resistivity of 1.54×10–8 Ωm at
room temperature. For an electric field along the wire of 1 volt cm–1, compute the average drift velocity of electron
assuming that there is 5.8 × 1028 conduction electrons /m3. Also calculate the mobility.
Ans: mobility = 6.99x10-3 m2/(V. s); drift velocity = 0.7 m/s
Calculate mobility, then use it to find drift velocity
The Wiedemann Franz law
estimated thermal conductivity (from a classical ideal gas)
or using
average speed
rms value
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Comparison of the Lorenz number to experimental data
at 273 K metal 10-8 Watt Ω K-2
Ag 2.31 Au 2.35 Cd 2.42 Cu 2.23 Mo 2.61 Pb 2.47 Pt 2.51 Sn 2.52 W 3.04 Zn 2.31
Drude prediction: 0.98 – 1.11 Watt ΩK -2
off by a factor of 2, but still very good!
Question
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The thermal conductivity of a metal is 123.92 Wm –1 K –1. Find the electrical conductivity and Lorentz number when the metal posses relaxation time 10–14 sec at 300 K. (Density of
electron = 6×1028 per m3)
Ans: σ=1.686x107 Ω-1m-1 ; L = 2.45x10-8
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Failures of the Drude model
• Despite this correct prediction, there are some serious problems with the Drude model.
• It is fortuitous – since the measured specific heat is not close to
per electron
Due to two mistakes that roughly cancel – specific heat far too large, velocity far too small !
(Due to not taking Fermi statistics of the electron into account)
Failures of the Drude model
The Drude model predicts a roughly 100 times larger value of the Peltier coefficient
The Peltier effect is that running a current through a material also transports heat
thermal current
electrical current
The ratio known as the thermopower or
“Seebeck coefficient”
For most metals value is 100 times smaller
Also
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The electronic properties of Metals: quantum mechanical approach
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Outline
• Basic assumptions: Born-Oppenheimer and one-electron approximations.
• Free electron (Sommerfeld model) • Calculate the quantum mechanical eigenvalues and wave
functions for one electron in some simple potential of all the others and the ions.
• Fill up these eigenvalues using the Fermi-Dirac statistics.
Going to see how Sommerfeld generalized Drude’s model to incorporate Fermi statistics – first digress and look at intuitive picture of how energy levels
in a solid are formed.
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The Born-Oppenheimer principle / the adiabatic approximation
•We discuss the motions of the electrons and the ions separately. We start out by discussion of the electronic states for a fixed lattice of ions in their equilibrium position.
In the Hamiltonian we have: kinetic energy of the electrons (el) , kinetic energy of the ions, the el-el interaction, the el-ion interaction, ion-ion interactions
We can treat the electrons separately from the ions (Born-Oppenheimer approximation) because the electrons are so much faster than the ions, partly due to their lower mass . When the ions move away from the equilibrium position, the electrons follow adiabatically. The electron energy changes but they stay in their ground state. When the ions move back, all the energy is re-gained. So it is sufficient to calculate the electron energies and wave functions for a fixed lattice of ions.
Zv
Z-Zv
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One electron approximation
• Solving the Schrödinger equation for an all-electron wave function in a rigid lattice is still too hard. We assume an effective one-electron potential.
we know for sure that
...but not much more since lattice periodic
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The idea of energy bands: Na consider one atom of Na: 11 electrons
An intuitive picture - think of the solid as built from atoms
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The idea of energy bands: Na
• Focus only on the valence (outer) electrons (3s). •What happens when we move them together?
consider two Na atoms / a Na 2
molecule: 22 electrons
big separation
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The idea of energy bands: Na
• Levels split up in bonding and anti bonding molecular orbitals and are occupied according to the Pauli principle.
• The distance between the atoms must be such that there is an energy gain.
consider two Na atoms / a Na 2
molecule: 2 3s electrons
molecular energy levels
capacity: 2x2=4 electrons upper level anti-bonding
lower level bonding
bonding
anti-bonding
Since each energy level can have 2 electrons in
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The idea of energy bands: Na
• N levels with very similar energies, like in a super-giant molecule. We can speak of a “band” of levels.
• Every band has N levels. We can put 2N electrons into it (but we have only N electrons from N Na atoms).
consider many (N) Na atoms (only 3s level)
Can see why this (Na) is metallic: band half filled
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The idea of energy bands: Si or diamond
4 electrons per atom (8 different states including
spin) 2 atoms per unit cell
sp3: bonding 8 states 8 states per atom 16 states per unit
cell
sp3: anti-bonding 8 states
semiconductor
The idea of energy bands: Si or diamond
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The free electron model (Sommerfeld model) – A&M Ch2
• Completely different approach: consider now free electrons in a box.
• a bit like Drude but quantum: • The depth of the box is the minimum potential energy in the
solid.
We might just as well require that
A constant potential gives just an off-set of the energies we might just as well take the potential to be zero. All we have to do is to solve this with the right boundary conditions
- use periodic boundary conditions
Free electron gas
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Want to calculate the ground-state properties of N electrons confined to a volume V. Find the energy levels of a single electron in the volume V, then filling levels consistent with Pauli exclusion principle (at most one electron in a single electron level)
A single electron can be described by the wave function ψ. If the electron has no interactions the wave function associated with a level of energy E satisfies:
Schrödinger equation (SE)
Confinement of electron to volume V by a boundary condition on above SE: Choose a cube of side L: V=L3 Boundary condition reflecting the electron is confined to the cube “Periodic Boundary Conditions” or “Born-von Karman”
cube of side length L,
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The free electron model
cube of side length L, periodic boundary conditions
solutions
electron density n and total number of electrons N
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solutions
boundary conditions give
Schrödinger equation
energy levels
E(k )= nx2x
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Filling in the electrons
2 electrons per k-point. If the electron density is N/V, we have to distribute N electrons on N/2 k-points.
fill in start at 0,0,0 .... and we should choose those states which have the lowest energy - we will occupy up to a sphere of filled electrons, can work out the radius of this sphere
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Filling in the electrons
N/2 points are within a sphere of radius
highest occupied energy
If have N electrons need only N/2 points because of spin no thermally excited electrons, no entropy aspect. this is only energy and it is only at 0 K.
Equation of a sphere
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The Fermi energy
highest occupied energy
= Fermi energy
is also called
with and the electron density
kF is the size of a sphere in reciprocal space which is characteristic for the metal (and given by the electron density)
Problem
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The free electron density of aluminum is 18.1 × 1028m–3 Calculate its Fermi energy at 0 K.
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The Fermi energy
element EF(eV)
Na 3.22
Cu 7.00
Al 11.63
In the range of typical binding energies of metals
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The Fermi velocity
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The Fermi velocity
• The Fermi velocity does (to first order) not depend on the temperature!
and
what is the speed of the electrons at the Fermi level?
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The Fermi velocity
element EF(eV) vF(ms-1)
Na 3.22 1.07 106
Al 11.63 2.02 106
Cu 7.00 1.57 106
x
x
x
Recall Drude velocity where we had an order of magnitude less (at RT)
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The density of states: free electrons We can now calculate the density of states.
we have
Tells how many states fall in a certain energy interval
