free electron in_metal
TRANSCRIPT
UEEP2024 Solid State Physics
Topic 3 Free Electron in Metals
Drude’s classical theory of Electrical Conduction
• Drude assumed that a metal is composed of ions, which are stationary, and valence electrons, which are free to move.
• If no voltage applied to the metal then at each collision the electron is deflected in a different direction so that the overall motion is quite random.
• The valence electrons appears to be similar to the molecules in an idea gas.
• The velocity of the electrons with mass m at temperature T is given by the equation
where KB is the Boltzmann’s constant.
TKmv B23
21 2
Drude’s classical theory of Electrical Conduction
• The mean free path is the average distance that an electron travels between collisions.
• The relaxation time is the average time duration between collisions.
• The mean speed of the electron • The speed of electron at room temperature is
about 105 ms-1.• The mean free path is about 1 nm and the
relaxation time about 10-14 s.
.
v
Drude’s classical theory of Electrical Conduction
• When we apply an electric field to a sample, the electrons are attracted towards the positive end of the sample, a net flow of electrons in this direction.
• If the electric field is E, then the force on each electron is eE.
• Acceleration on each electron is .• Change in velocity • This quantity is called the drift velocity.
emeEa
.em
eEav
Drude’s classical theory of Electrical Conduction
• Electron mobility µ =e/m.
• The current density
where n is the number of valence electrons per unit volume,• Conductivity of the metal is given by • Drube’s model is a) consistent with ohm’s law. b) Explain the phenomenon of electrical resistance. c) Gives good values of conductivity.
.eEnAIJ
.en
Failures of Classical model• Conductivity should be directly proportional to the valence electron
concentration, not in good agreement with experimental data.• Electrical properties of alloys should be intermediate between the values of
corresponding pure materials. However many alloys have resistivity which are considerably larger than those of either of the pure constituents.
• The dependence of resistivity on temperature: Drube’s model predicted resistivity should be proportional to T1/2,
experimental measurement show that resistivity is actually proportional to T over a wide range of temperature.
• Drude ‘s model predict the molar specific heat capacity for monovalent metal of 9R/2, 6R for divalent metal and 15R/2 for trivalent metal. However, experiment results show that the molar specific heat capacity at room temperature is approximately 3R, regardless of the valency of the metal.
Example
Estimate the typical conductivity of a metal at 295 K assuming that the mean free path is about 1 nm and the number of valency electrons is about 1029 m-3.
Solution
• Thermal velocity
• Relaxation time
• Conductivity
1531
23
1016.11011.9
)295)(1038.1(33
ms
mTkv B
sv
155
9
1062.81016.1
101
117
31
15219292
1042.21011.9
1062.8106.110
mm
ne
Free Electron Fermi Gas• Atoms bounded by “free electrons”. Good
example is alkali metals (Li, K, Na, etc.)
Electron in a metal can move freely in a straight path over many atomic distances, undeflected by collisions with other conduction electron or by collisions with the atom cores
Free Electron Fermi Gas1. A conduction electron is not deflected by ion cores
arranged on a periodic lattice
2. A conduction electron is scattered only infrequently by other conduction electrons
Consequence of the Pauli exclusion principle
Free electron Fermi gas – A gas of free electrons subject to the Pauli principle
Free Electron Fermi Gas
),(ˆ),( trHtrt
i
For a general quantum system
(1)
For a single particle in three dimensions
),()(),(2
),( 22
trrVtrm
trt
i
(2)
Wave functionHamiltonian operator
Free Electron Fermi Gas(1-D)
nnn E
dxd
m
2
22
2
Consider a free electron gas in 1-D (electron of mass m is confined to a length L by infinite barriers), (2) can be expressed as
With boundary conditions
0)(0)0( Lnn
(3)Energy of the electron
Free Electron Fermi Gas(1-D)
LnwherexA nn
n
212sin
Lnkwhere
mkE nn
n
2
22
The solution of (3)
Energy En is given bywavevector
0 /L 2/L
K-space
k /L
Free Electron Fermi Gas3-D
)()(2
22
rErm kkk
Consider a free electron gas in 3-D (electron of mass m is confined to a cube of edge L), (2) can be expressed as
With boundary conditions
0)(0)0(0)(0)0(0)(0)0(
LzzLyyLxx
kk
kk
kk
(4)
Free Electron Fermi Gas(3-D)
LnLnLn zzyyxx 21,
21,
21
zyxA
zyxk
2sin2sin2sin
The solution of (4)
where
(5)
Free Electron Fermi Gas(3-D, periodic boundary conditions)
• 3-D system• periodic boundary
conditions
,.....8,6,4,2,0,,LLLL
kkk zyx
),,(),,(),,(),,(),,(),,(
zyxLzyxzyxzLyxzyxzyLx
kk
kk
kk
The solution of (4)
rikArk exp)( the components of k satisfy
One allowed value of k per volume (2/L)3
Free Electron Fermi Gas(3-D, periodic boundary conditions)
2222
22
22 zyxk kkkm
km
E
Energy Ek is given by
K space
(6)
Free Electron Fermi Gas(density of state)
2/3
22
3
3
3 23/2
3/42
mEL
LkN
total number of allowed energy state
Density of states – number of allowed energy state per unit energy range
2/12/3
22
3 22
)( EmLdEdNED
(7)
(8)
Free Electron Fermi Gas(Fermi energy)
3/2
3
22 32
LN
mEF
The Fermi energy is defined as the energy of the topmost filled level in the ground state of the N electron system.
The state of the N electron system at absolute zero
From (7), we get
For N electron system
Free Electron Fermi Gas(Fermi-Dirac distribution)
• The Fermi-Dirac distribution gives the probability that an energy state with energy E is occupied by an electron
1exp
1)(
kTE
Ef
The electron density is given by
dEEfEDn )()(
Fermi level – the energy at which the probability of occupation is 1/2
Example
Show that the probability for an electron state at the Fermi energy is equal to 0.5 for all finite temperature.
Solution
• When E = EF,
.21
111
11
11
11)(
0
/)(/)(
e
eeEf TKEETKEE BFFBF
Example
Using the Fermi-Dirac distribution, determine the values of energy corresponding to f = 0.9 and f = 0.1 at a temperature of 300 K.
Solution
TKEETKEE
TKEEe
e
eEf
BF
BF
BF
TKEE
TKEE
TKEE
BF
BF
BF
20.220.2
)111.0ln(/)(111.01111.1
9.011
9.01
1)(
/)(
/)(
/)(
Fermi-Dirac Distribution
1.0
0.0
EF
Probability of occupationf(E)
Energy
Solution
TKEETKEE
TKEEe
e
eEf
BF
BF
BF
TKEE
TKEE
TKEE
BF
BF
BF
20.220.2
)9ln(/)(9110
1.011
1.01
1)(
/)(
/)(
/)(
Ohm’s Law
BvqF
qFmvmomentum
In an electric field and magnetic field B, the force on an electron of charge –q is
If B = 0,
mqv
thus
collision time
Ohm’s Law
mnqqnvJ
2
mnq
2
The electric current density is
If we define the electrical conductivity as
J
then
Ohm’s Law
Hall Effect
JBRBqnJ
qvBqF
H
0B
Hall coefficient
JBRBqnJ
qvBqF
H
Quantum StatisticsA single quantum mechanical system consists of N particles constrained to some volume V
assumption
1. weakly interacting particles
2. the particle density is low enough so that the energy of the system can be considered as the sum of the individual particle energies
1
ˆ
iii
sii
si
EnE
EHwavefunction of particle s
number of particles with energy Ei
Quantum Statistics• How to determine the most probable distribution of
finding n1 particles out of a total of N with energy E1, n2 with energy E2, and so on?
The most probable distribution (n1, n2 …,ns,….) is the one associated with the largest number of microscopically distinguishable arrangements
Thus, the plan of attack would be to derive an expression for P, the total number of microscopically distinct arrangement corresponding to a given arbitraty sequence (n1, n2 …,ns,….) and then find the particular sequence that maximizes P.
Three types of quantum particles
1. Identical but distinguishable particles• harmonic oscillators
2. Identical indistinguishable particles of half-odd integral spin (Fermions)• electrons• protons
3. Identical indistinguishable particles of integral spin (Bosons)• photons• phonons
The particles of concern to us fall into one of three categories
obey Pauli exclusion principle
Pauli exclusion principle not apply
N particle system
,.......,......,, 21 sEEE
Our model system is taken to contain N particles, each with a spectrum of allowed energy levels
We divide the single-particle energy spectrum into energy “bins”. Bin s, as an example, represents all the elementary quantum states whose energies lie within some arbitrarily chosen interval Es centered about Es. The number of quantum states in bin s is denoted by gs.
Identical but distinguishable particles
1
11
111
11 !!
!nN
nn
CgornNn
NgP
Determine P of N particles such that bin 1 contains n1 particles, bin 2 n2 particles, and so on.
For bin 1, the total number of distinguishable choices is
For bin 2, the total number of distinguishable choices is
21
22
2212
122 !!
!nnN
nn
CgornnNn
nNgP
Identical but distinguishable particles
the total number of distinguishable choices in which bin 1 contains n1 particles, bin 2 n2 particles, and so on, is
1
2121
!!
........,....,....,,
s s
ns
ss
ngN
PPPnnnPs
Identical indistinguishable particles of half-odd integral spin (Fermions)
11!!!
111
11 ng Cor
ngngP
Determine P of N particles such that bin 1 contains n1 particles, bin 2 n2 particles, and so on.
For bin 1, the total number of distinguishable choices is
For bin 2, the total number of distinguishable choices is
22!!!
222
22 ng Cor
ngngP
Identical indistinguishable particles of half-odd integral spin (Fermions)
the total number of distinguishable choices in which bin 1 contains n1 particles, bin 2 n2 particles, and so on, is
1
2121
!!!
........,....,....,,
s sss
s
ss
nngg
PPPnnnP
Identical indistinguishable particles of integral spin (Bosons)
!1!
!1
11
111
gngnP
Determine P of N particles such that bin 1 contains n1 particles, bin 2 n2 particles, and so on.
For bin 1, the total number of distinguishable choices is
For bin 2, the total number of distinguishable choices is
!1!
!1
22
222
gngnP
Identical indistinguishable particles of integral spin (Bosons)
the total number of distinguishable choices in which bin 1 contains n1 particles, bin 2 n2 particles, and so on, is
1
2121
!!1!1
........,....,....,,
s ss
ss
ss
nggn
PPPnnnP
Lagrange method
constEnE
constNn
sss
ss
1
1
we wish to find the set of ns for which P is maximized subjected to conditions :
We constrain our solution using Lagrange multipliers forming the function:
1121 )ln(,.......,,
sss
sss nEEnNPnnnF
Lagrange method
sallfornF
s
0
0lnln sss Eng
For Identical but distinguishable particles
ss
s Egn
expMaxwell-Boltzmann
Lagrange method
0lnln ssss Enng
For Identical indistinguishable particles of half-odd integral spin (Fermions)
1exp
s
ss E
gn
Fermi-Dirac
Lagrange method
0lnln ssss Enng
For Identical indistinguishable particles of half-odd integral spin (Fermions)
1exp
s
ss E
gn
Bose-Einstein
Energy Band
Electron in a metal can move freely in a straight path over many atomic distances, undeflected by collisions with other conduction electron or by collisions with the atom cores
Energy Band
),(ˆ),( trHtrt
i
For a general quantum system
(1)
For a single particle in three dimensions
),()(),(2
),( 22
trrVtrm
trt
i
(2)
Wave functionHamiltonian operator
Energy Band
nnn E
dxd
m
2
22
2
free electron gas
(3)Energy of the electron
V(r)=0
Free Electron Model
Energy Band0)( rVif
Considering a periodic potential, V(r+T)=V(r)
)()()(2
)( 22
rrVrm
rE
(2) can be expressed as (4)
Energy Band
G
riGGeUrV )(
G
iGxGeUxV )(
The periodic potential V(r) may be expanded as a Fourier series in the reciprocal lattice vectors G
or
(5)
For 1-D system
Thus, (4) can be rewritten as
)()(2
)( 2
22
xeUdx
xdm
xEG
iGxG
(6)
Energy Band
G k
xGkikG
k
ikxk
eCUxxV
eCkmdx
xdm
)(
22
2
22
)()(
2)(
2
k
ikxkeCx)(
The wavefunction Ψ(x) may be expressed as a Fourier series summed over all values of the wavevector permitted by the boundary conditions
then
(7)
Energy Band
k
ikxk
G k
xGkikG
ikxk
k
eCEeCUeCkm
)(22
2
From (6), we get
mkwhere
CUCE
k
GGkGkk
2
0
22
Each Fourier component must have the same coefficient on both sides of the equation
(8)
central equation
Energy Band
G
xGkiGk
ikxkk eCeCx )()(
Once we determine The Ck from central equation, the Ψ(x) is given as
(9)
Rearrange (9), we get
)(
)()(
xueCwhere
exueeCx
kG
iGxGk
ikxk
ikx
G
iGxGkk
(10)
Energy Band (Bloch Theorem)
)()()exp()()(
Trururikrur
kk
kk
The solutions of the Schrodinger equation for a periodic potential must be of a special form
The eigenfunctions of the wave equation for a periodic potential are the product of a plane wave exp(ik•r) times a function uk(r) with the periodicity of the crystal lattice
(11)Translation vector
Bloch theorem
Energy Band (Kronig-Penney Model)
00
)(xbDeCe
axBeAex
QxQx
iKxiKx
a0-b
Uo
If V(x) is
The solutions for (2) are
wheremKEandmQEUo 2
222
22
Energy Band (Kronig-Penney Model)
)()(00
00
DCQBAikdxd
dxd
DCBA
xx
xx
The constants A,B,C,D are chosen so that Ψ and dΨ/dx are continuous at x=0 and x=a
)()0()( baikexbbaxa Bloch theorem
(12)
(13)
Energy Band (Kronig-Penney Model)
)(
)(
)(
)(
baikQbQbiKaiKa
baik
bxax
baikQbQbiKaiKa
baikbxax
eDeCeQBeAeiK
edxd
dxd
eDeCeBeAe
e
(14)
(15)
Energy Band (Kronig-Penney Model)
)(coscoscoshsinsinh2
22
bakKaQbKaQbQKKQ
oUband limlim
0
From (12)-(15), we can get
If the potential is a periodic delta function
kaKaKaKaP coscossin)/(
Then, (16) can reduce to
(16)
(17)P=Q2ba/2
Energy Band (Kronig-Penney Model)
-6 -4 -2 0 2 4 6
-2
-1
0
1
2
3
4
5
6
P=3/2
(P/K
a)si
n(K
a)+c
os(K
a)
Ka ()
Not allowed
Energy Band (Kronig-Penney Model)
Energy gap
(Ka that is not allowed)
Band Structure
Actual band structures are usually exhibited as plots of energy versus wavevector in the first Brillouin zone. This is helpful in visualization and economical of graph paper.
2/a
Band Structure
/a 2/a 3/ak
E
If the band structure of A is :
/ak
E
Band Structure
k
E
Eg
k
E
Eg
k
E
Eg
Direct Indirect Overlap
(negative bandgap)
Bandgap – the difference in energy between the lowest point of the conduction band the highest point of the valence band.
valence band
conduction band
Band Structure
band structure of Si
Energy Band(conductor and insulator)
A band is filled from low energy to high energy
Energy Band (conductor and insulator)
A full band cannot conduct electricity. A full band is always a full band no matter what the external field are
An empty band cannot conduct electricity because it does not have charge carrier.
Only a partially filled band can conduct electricity. The occupied states are not “balance” under an external field. This unbalance causes the current flow.
Energy Band(Equation of motion and effective mass)
2
2
2
11dkEd
me
The equation of motion of an electron in an energy band is :
dtdkEdm
dkd
dtdm
dtdvm
dtdvm
dtdkF
eee
e
2
From (1), we can get
(1)
effective mass
)( BvEqdtdkF
Energy Band(Equation of motion and effective mass)
)(2
22
0 AmkEE
Approximate solution near a zone boundary
(2)
From (1), we get
Ammthus
mA
dkEd
e 2
22
2
2
11 (3)
An electron of mass m when put into a crystal respond to applied fields as if the mass were me (effective mass)
Electron rest mass
Example
By determine the number of grid points contained beneath the surface of radius nmax, show that the Fermi energy (i.e. the energy of the highest occupied state at T = 0K ) is given by
.32
3/222
VN
mEF
Each allowed quantum state can be represented by a grid point (nx, ny, nz).Grid points with the same energy are connected by a
surface of constant radius. In order to determine the Fermi energy, find the values of the radius nmax, just contains sufficient states to accommodate all the valence electrons in the crystal
ny
nx
nz
nmax
Solution
• The volume contained beneath the surface corresponds to one-eight of a sphere of radius nmax, the volume is
• Each grid point corresponding to a cube of unit volume, this expression also gives the number of grid points beneath the surface.
• As each grid point corresponding to a state which can accommodate two electrons for a crystal containing N electrons we require N/2 states.
•
634
81 3
max3max
nn
3/1
max
3max 362
NnnN
Solution
• By writing
• Energy
2/1222max zyx nnnn
3/2223/2
3
32
3/2222
max
22
2222222
32
32
322
22
VN
mh
LN
mh
NLm
hnLm
h
nnnLm
hmkhE zyx
Thermal Conductivity in Metals
• Thermal conductivity of a Fermi gas is
where n is the electron concentration k is the Boltzmann constant T is the temperature m is them mass of electron and is the collision time.
mTnklv
mvTnkK FF
el 33
22
2
22
Thermal Conductivity in Metals
• Do the electrons or phonons carry the greater part of the heat current in a metal?
• In pure metals the electronic contribution is dominant at all temperatures.
• In impure metals or in disordered alloys, the electron mean free path is reduced by collisions with impurities, and phonon contribution may be comparable with the electronic contribution.
Wiedemann-Franz law
• The Wisdemann-Franz law states that for metals at not too low temperatures the ratio of the thermal conductivity to the electrical conductivity is directly proportional to temperature, with the value of the constant of proportionality independent of the particular metal.
Tek
mne
mTnkK 22
2
22
33
Lorenz number L
• The Lorenz number L is defined as
• The value of
• This remarkable result involves neither n nor m.• Experimental values of L at 0oC and 100oC are in
good agreement.
TKL
.ohm/deg- watt1045.23
2822
ekL
Experimental Lorenz numbersL ×108 watt-ohm/deg2
Metal 0oC 100oC
Ag 2.31 2.37
Au 2.35 2.40
Cd 2.42 2.43
Cu 2.23 2.33
Mo 2.61 2.79
Pb 2.47 2.56
Pt 2.51 2.60
Su 2.52 2.49
W 3.04 3.20
Zn 2.31 2.33