Order 1238142: Condensed matter
17/09/2018
1
29
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
30
The Fermi-Dirac distribution
• How are the electronic states populated at finite temperature?
17/09/2018
2
31
The Fermi-Dirac distribution
• At T=0 all the states are filled up to the highest occupied state. This state is called the Fermi energy E
F . It is equal to
the chemical potential μ at T=0.
μ kB=1.380×10
−23 J⋅K−1
kB=8.617×10 −5 eV⋅K−1Boltzmann constant
Should be familiar from statistical physics For metals it is a good approximation to say that this is valid for all temperatures For semiconductors it is not!
32
The Fermi-Dirac distribution
17/09/2018
3
33
The Fermi-Dirac distribution
the curved zone in the Fermi Dirac distribution increases with temperature.
If f (E,T) = 0.1 then E-EF=2.2kBT
If f (E,T) = 0.9 then E-EF=-2.2kBT
E
Electrons above EF behave like ideal gas and approximately obey the Maxwell-Boltzmann distribution
34
Occupation of the states at finite T
g(E)
g(E)f(E,T)
17/09/2018
4
Problem:
35
Evaluate the Fermi function for energy k T above the Fermi energy
B
f = 0.2689
36
Heat capacity of the electrons: estimate
Their mean energy is
and the heat capacity
The result of this calculation is quite similar to the correct result which is Adv. See A&M Ch2
For derivation
We say that all the electrons in the yellow zone have energy 3/2 kBT (from ideal gas theory)
Important is that C is prop to g(EF) and to T! The true result is found after straight but lengthy calculation
Excited electrons contribute to the heat capacity - estimated from area of rectangle
17/09/2018
5
37
Heat capacity of the electrons
V
Recall, we had
Rearrange EF to give
substituting into expression for g(E) we get
At the Fermi energy, g(EF) = 3N/2E F
(1)
Substituting for g(EF) into (1), and using EF=kB TF we get:
Gives linear in T, and factor 0.01 smaller than classical result
n=N/V
3/2
,
e
N E
E
so number of electrons excited is g(EF) times kB T =3 kB TN/(2EF), or the fraction excited of total electrons, N, is:
38
Heat capacity of a metal: lattice + electrons
• two contributions: lattice and electrons • electrons unimportant at high T but dominating at
sufficiently low T
17/09/2018
6
2 / 32 23 2.12
2 F
N E eV
m V
1 / 32 13 0 .7 4 6F
N k A
V
6 10.86 10FF e
P V ms
m
42.46 10FF B
E T K
k
Typical values of monovalent potassium metal; Typical values e.g. monovalent potassium metal; for potassium the
atomic density and hence the valance electron density N/V is 1.402x1028 m-3 so that:
40
Free electron quantum version of Wiedemann Franz
We must use the Fermi velocity and the correct heat capacity for the electrons. We know (almost) all this.
we get...
so that L = 2.45 10-8 Watt Ω K-2
so this is almost exactly twice the classical (Drude) value
at RT
17/09/2018
7
41
The Wiedemann Franz law: Drude model
estimated thermal conductivity (from a classical ideal gas)
but now we have Using for the velocity
And Eq (1) for Cv with g = 3N/2E F
42
Comparison of the Lorenz number to experimental data
L = 2.45 10-8 Watt Ω K-2
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
17/09/2018
8
Problems of the Free Electron Model
• Have assumed only 1 electron per atom; but actually many • Cannot explain optical spectra • Specific heat, although better than Drude is still off by up to
factor of 10
• Can’t explain magnetism • Doesn’t take into account electron-electron interactions
43
• Adv read Ch3 A&M
44
17/09/2018
9
45
46