Transcript E f

제 4 장 Metals I:
The Free Electron Model
Outline
4.1 Introduction
4.2 Conduction electrons
4.3 the free-electron gas
4.4 Electrical conductivity
4.5 Electrical resistivity (T)
4.6 heat capacity of conduction electrons
4.7 Fermi surface
4.8 Electrical conductivity: effects of the Fermi surface
4.9 Thermal conductivity in metals
4.10 Motion in a magnetic field; cyclotron resonance and the Hall effect
4.11 AC conductivity and optical properties
4.12 Thermionic emission
4.13 failure of the free-electron model
4.1 Introduction
Metals in life from ancient to future.
Example: Duralumin (Cu, Mn, and Mg) in Car, Cu in electrical wire, Ag
and Au as jewelry
Common properties:
physical strength, density, electrical and thermal conductivity
Optical (visible) reflectivity
Explanation of many physical properties by assuming
cloud of free-electrons all over sample
Most of optical properties can be explained but with a limit.
4.2 Conduction electrons
•Na gas; a simple collection of free atoms.
Each atom has 11 electrons orbiting around the nucleus.
11Na:
1s2 2s22p6 3s1
Core
electrons
Valence
electron
1 valence el. loosely bound to the rest of the system(=Na1+).
can explain ordinary chemical reaction.
•Na atoms to form a metal
bcc structure (section 1.7)
d = 3.7 Å
d
4.2 Conduction electrons (p138)
•Na atoms to form a metal.
bcc structure (section 1.7)
Fig. 4.1 overlap of the 3s in solid sodium
two atoms overlap slightly  valence el. can hop from one to the
neighboring atom  move around freely all over the crystal.
conduction electrons in a crystal !!
Naming since they carriers electrical current in an electric field.
Cf) core electrons do not give electrical current.
N=concentration M’=atomic weight, Z = atomic valence (1+, 2+,..),
m =density of the substance
N  Zv
m N A
M
.
(4.1)
4.3 Free electron gas (p140)
econd. is completely free,
except for a potential at the surface.
Free motion except occasional reflection
from the surface which confines econd..
Like gas!! So called free electron gas!
How about interaction between 1) econd. and ecore.?
a) Strong coulomb attraction between econd. and core ion is
compensated by repulsive potential due to array of core ion by
quantum effect.
b) econd. spend tiny time near core ion.
E  Ek  E p
c)
Short-range screened coulomb potential rather
than a long-range pure coulomb potential.
interaction between 2) conduction electrons themselves?
Pauli exclusion principle : = 1(a)+2(b)
1) electrons of parallel  or  spins tend to stay away.
2) electron pair having opposite spin  can stay together
Each electron is surrounded by a hole”
The hole with a radius of about 1 Å. The hole move with each electron.
Interaction between two specific electron is always screened by the other
electrons.  very low interaction between the two electrons.
Free electron gas vs ordinary gas
1) charged, i.e. like plasma
2) N is very large N~1023 cm-3 while ordinary gas ~ 1019 cm-3
Jellium model : ions form a uniform jelly into which econd. move around
4.4 electrical conductivity (p142)
I V R
(4.2)
Law of electrical conduction in metal : Ohm’s law
J: current density per area, E=electrical field
1
I
V
ρL
(4.4)
J  , E  , and R  , (4.3) σ 
ρ
A
L
A
Substitute
VA
I
ρL
R
ρL
(4.3)
A
J  σE (4.5)
into (4.2)
I
V

A ρL
Substitute J and E (4.3) and (4.4)
J  E
Electric field accelerates econd. (not ions in the crystal) thus electric current.
4.4 electrical conductivity (p142)
Newton’ equation with friction force with  = collision time.
Collision and friction tends to reduce the velocity to zero.
Final (or steady) state solution.
Terminal velocity (drift velocity) vs. random velocity
dv
v
m *  eE  m * , (4.6)
dt

dv
0
dt
dv
eE vd

 0
dt vd
m* 
e
vd  
E.
m*
(4.7)
vd  
e
E.
m*
(4.7)
4.4 electrical conductivity (p142)
Newton’ equation with friction force with  = collision time.
Collision and friction tends to reduce the velocity to zero.
Final (or steady) state solution.
Terminal velocity (drift velocity) vs. random velocity
dv
v
m *  eE  m * , (4.6)
dt

vd  
dv
 0 long after
dt
dv
eE vd

 0
dt t  ,vvd
m* 
e
vd  
E.
m*
(4.7)
e
E.
m*
(4.7)
Newton’ equation with friction (collision) force with  = collision time.
Collision and friction tends to reduce the (average vector)velocity to zero.
Terminal velocity (drift velocity) vs. random velocity
e
vd  
E.
m*
(4.7)
1019 1014
2
1
vd  
10

10
ms
1030
vter
1
2
 3kT 
6
1
vr  
  10 ms for metal
 m* 
vr /vd  108
J: current density per area is proportional to E field,
 e  Ne 
J  ( Ne)vd  ( Ne) 
E 
E. (4.8)
m*
 m* 
2
A
I
dl  vd dt
I
J
A
I
dQ
dt
dQ  qNdV  qNAdl  qNAvd dt
I dQ  Ne vd dt A
J 

  Ne vd
A A  dt
dt  A
 = collision time, mean free lifetime? ~ 10-14 sec
J  σE (4.5)
Ne 2

,
m*
2
 e  Ne 
J  ( Ne)vd  ( Ne) 
E 
E. (4.8)
m*
 m* 
(4.9)
metal
  5 10 7 m 1
N  1029 m 3  1023 cm 3
semiconductor
  1 m 1
N  10 20 m 3
vr  106 ms 1
vr  10 4 ms 1
Nmetal / Nsemiconductor ~ 109 , vr , metal / vr ,semiconductor ~ 102
 = relaxation time? ~ 10-14 sec
Ne 2

,
m*
(4.9)
vd (t )  vd ,0et  ,
dv
v
m *  m * , (4.6)'
dt

(4.10)
l
  , (4.11)
vr
 vs. mean free path and random velocity
vd  
Ne2l

. (4.12)
m * vr
e
E.
m*
(4.7)
Microscopic expression for Joule heat = power per volume absorbed by the
electron system from the electric field. (i.e. current),
Maybe Skip
 
2
 dl
dW dF  l
 eE  Ne  2
P

 F   NFvd  N (eE ) 
E . (4.13)
 
dt
dt
dt
m*
 m* 
Origin of collision time
l
  , (4.11)
vr
Ne2l

. (4.12)
m * vr
  1014 s; vr  106 ms1
l ≈ 10-8 m ≈ 102 Å
d≈ 4Å
25x unit cell
Quantum mechanics, deBroglie relation:
h

. (4.14)
m * vr
A light wave traveling in a crystal is not scattered at all.
The ions form a perfect lattice, no collision at all, that is l=∞ and hence τ≈∞,
leads to infinite conductivity. In reality, l ≈ 102 Å.
Well trained soldiers at ground, Movie (hero)
SKIP
m* 1
  2 . (4.15)
Ne 
http://sdsu-physics.org/physics180/physics196/Topics/gaussLaw.html
고체 물리 4장 시험
6월 5일 수요일 1,2, 3교시
4.5 Electrical resistivity versus temperature
σ metal
varies with the metal’s temperature, so do resistivity ρ.
σ

1
ρ
(4.4)
m*
,
2
Ne 
Ne 2

,
m*
(4.9)
(4.15)
1
 probability of the electron scatter per
 unit area
Fig. 4.5 The normalized resistivity
(T)/(290 K) versus T for Na at low
Temp. (a), and high Temp. (b). (290
K) ≈ 2.1010-8 m.
  10 14 s  1014 collisions in 1 second .
Electron undergoes a collision only
because the lattice is not perfectly
regular.
Deviation from perfect lattice:
a. Lattice vibration (phonon) of the ions around their equilibrium position due
to thermal excitation of the ions. – Dynamic imperfection!
a. All static imperfections (impurities or defect).
Probability of electron being scattered by phonon and impurities are additive,
since these 2 mechanisms are assumed to act independently.
1


1

1
,
(4.16)
 ph  i
m*
  2 , (4.15)
Ne 
τ ph : collision time due to phonon
τi : collision time due to impurities
Substitute (4.16) into (4.15)
  i   ph (T) 
m* 1 m* 1
 2
.
2
Ne  i Ne  ph
 splits into 2 terms:
(4.17)
1. due to scattering by impurities (independent of T)  residual resistivity
2. due to scattering by phonon (dependent of T)  intrinsic (ideal ?) resistivity.
1


1
 ph

m*
 2 ,
Ne 
1
i
,
(4.16)
(4.15)
τ ph : collision time due to phonon
τi : collision time due to impurities
m* 1 m* 1
  i   ph (T)  2  2
.
Ne  i Ne  ph
(4.17)
 splits into 2 terms:
1. due to scattering by impurities (independent of T)  residual resistivity
2. due to scattering by phonon (dependent of T)  intrinsic resistivity.
@ low T: scattering by phonon is negligible (amplitude of oscillation very small)
τ ph∞ ; ph 0 hence = i constant
As Tscattering by phonons more effective
ph(T) 
@ high T: ph(T) linearly with T
As N ,  i 
Expectation:
(impurity concentration)
For small Ni, i proportional to Ni  ph>>i, except @low T.
Assumption: collision is of the hard sphere
i 
li
,
vr
(4.18)
li 
1
.
N i i
(4.19)
li : mean free path for impurity collision
i: scattering cross section of an impurity
  i   ph (T) 
m* 1 m* 1
 2
.
2
Ne  i Ne  ph
(4.17)
Substitute (4.18) and (4.19) into (4.17)
m*
 i  2 N i i vr
Ne
i proportional to Ni
i: not conductivity but scattering cross section??
WHY??
lph 
1
N ion ion
,
(4.20) 
1
T
lph: mean free path for phonon collision
ion: scattering cross section per ion
Nion : concentration of metallic ion in the lattice
The average scattering cross section
 ion   x 2 ,
(4.21)  T
 ion 
The average of potential energy
1

k x 2  E   kT
 T at high T ,
2
e
1
  i   ph (T) 
m* 1 m* 1
 2
.
2
Ne  i Ne  ph
   1
 ph T   
,
 T
 kM  e  1
(4.23)
lion
,
vr
(4.18) 
1
T
(4.22)
(4.17)
 2 T
 ph 
,
kM 
(4.24)
4.6 Heat capacity of conduction electrons
Drude-Lorentz model  kinetic theory of gasses that a free particle in
equilibrium @T has average energy 3 kT .
2
Average energy per mole:
3  3
E  N A  kT   RT ,
2  2
Ce 
 E
T
(4.25)
3
Ce  R  3 cal mole K.
2
Total specific heat in metal
C  Cph  Ce ,
NA: Avogadro number
R= NA k
(4.27)
(4.26)
Cph: specific heat phonon (motion of
ion core)
Ce: specific heat electrons
3
C  3R  R  4.5R  9 cal mole K.
2
(4.28)
Experiment C in metal ≈3R @high T 도체나 부도체나 실지로는 비슷하다
Contribution Ce<< 1.5 R , WHY the discrepancy (차이)?
Fig. 4.6 (a) occupation of energy levels according to the Pauli exclusion
principle. (b) the distribution function f(E) versus E, at T= 0 K and T > 0 K.
Energy in metal is quantized.
The electrons fill the energy level from the lowest level to the highest.
The highest energy occupied level is called Fermi energy or Fermi level.
Distribution function of electrons:
f(E): Probability that the energy level is occupied by electrons.
f(E)=0 empty level
f(E)=1 filled level
@T=0 K
 1, E  EF
f E   
 0, E  EF
(4.29)
@ T  0 K  electron excites.
f ( E, T  0 K) 
1
e
( E  EF ) kT
1
(4.30)  Fermi Dirac distributi on.
.
Only electron in the range kT near Ef are excited
only fraction kT are affected.
Ef
Total number of electron excited per mole=
Each electron absorb energy kT
N kT 
E  A
Ef
2
Fig. 4.6 (b) the distribution
function f(E) versus E, at
T=0 K and T > 0 K.
Ce 
for E f  5 eV ~ 6000 K
and T  300 K
 E
T
Ce  2 R
kT
.
EF
(4.31)
N A kT
Ef
Tf : Fermi temperature
E f  kTf  Ce  2 R
T
kT
. (4.31)  Ce  2 R
Tf
EF
E f  5 eV ; T f  6 104 K
To excite all electrons, the solid must be heated to Tf, but impossible.
Specific heat of electrons far below its classical value.
Ce 
2
2
R
kT
,
EF
(4.32), more accurate theory
linear function of temperature !!
Unlike specific heat of lattice which is constant at high temperature and
proportional to T3 at low temp..
4.7 The Fermi Surface (FS)
Electrons in metal are in continuous state random of motion.
v:speed of particle
1
2
E
2
m*v
Velocity space (vx, vy, vz)  magnitude and direction of electron.
1
2
E f  m*vf
2
1
2
(4.33)
 2 E f   2  5 1.6 10
  
E f  5eV  v f  
9 x10 31
 m*  
19
1
2

  106 ms 1

Electron in FS move very fast.

2
2
EF 
3

N

2m

2/3
(4.34)
N  1029 m 3  1023 cm 3
Fig. 4.7 The Fermi
surface and the
Fermi sphere
Fig. 4.6 (b) The distribution
function f(E) vs Ef
at T = 0 K and T > 0 K
Chap 3.3과 kittel
pp130-132 설명하기
4.8 Electrical conductivity; effect of the Fermi surface
Fig. 4.8 (a) The Fermi surface at equilibrium. (b) Displacement of the Fermi
sphere due to an electric field.
Absence of an electric field: FS is centered at origin.
Electron moving at high speed and carry currents. Total current system is zero.
Electric field (+) is applied: electron acquired vd. FS is displaced to the left.
e
vd  
E.
m*
(4.7)
Current density:
The fraction electron which remain uncompensated≈
vd
vF
Concentration N  vd  and velocity  vF
v
F 

Current density : (E는 에너지가 아니라 electric field)
vd  
e
E.
m*
(4.7)
 vd 
ne  F


J  e  N     vF   Nevd 
E
m*
 vF 
lF
ne 2lF
F 
 
m * vF
vF
2
1
l F ~  @ high T
T
1
 ~ or  ~ T
T
ne 2 F

m*
(4.35)
Ne 2

,
m*
(4.9)
FS를 고려한 것도 고전적인 결과와 같은 온도 의존성 보임
4.9 Thermal conductivity in metals
Heat current (Q): the amount of thermal energy crossing a unit area per unit
time.
Q  K
dT
dx
K: thermal conductivity
In insulator  heat is carried by phonon.
In metal  heat may be transported by electrons or phonons.
K  K e  K ph
K ph  102 Ke
Heat is transported almost entirely by those
electrons near Fermi level, because those
well below this level cancel each other
contribution.
Fig. 4.9 The physical basis for thermal conductivity. Energetic electrons on the
carry net energy to the right
Evaluate the thermal conductivity
Cv: specific heat
1
K  cv vl
v: speed
3
l: mean free path
2
Ce 
2
R
kT
(4.32); R  Nk
EF
1   2 Nk 2T 
vF lF
K  
3  2 EF 
lF
F
vF
Ef 
1
2
m*vf
2
(4.33)
The thermal conductivity in terms of electron properties of metal:
K
 2 Nk 2T F
Ratio
3m *
K
T
(4.36)
ne 2 F

m*
(4.35)
is known as Lorentz number
K
 2 Nk 2T F m * 1  k 
L
  
2
T
Ne  F 3m * T 3  e 
2
(4.37)
K
 2 Nk 2T F m * 1  k 
L
  
2
T
Ne  F 3m * T 3  e 
2
(4.37)
L only depends on the universal constant k and e should be the same for all
metals. Its numerical value is 5.8x10-9 cal-ohm/s K2.
Electrical and thermal conductivities are related in metal (carried by electrons).
So free electron model is good model to explain conductivity experiment!
4.10 Motion in a magnetic field: cyclotron and the hall effect
Cyclotron resonance
A magnet field applied across a metallic slab causes electrons to move in a
counterclockwise circular fashion in a plane normal to the field.
The frequency of cyclotron
motion:
eB
c 
m*
Fig. 4.10 (a) cyclotron motion.
(b) the absorption coefficient  vs .
vc 
(4.38)
c
eB

 2.8B(in kGauss) GHz
2 2m *
EM signal pass through the slab parallel to B.
The electric field of the signal acts on the electrons, and some of the energy is
absorbed.
  c
(4.39)  the greatest absorption
Cyclotron resonance is used to measure the electron mass in metals and
semiconductors.
The hall effect
Fig. 4.11 Origin of the Hall field and Hall effect
Electric current Jx flowing in x-axis and magnetic field Bz applied normal to the
wire in the Z-axis.
Before B applied:
E flowing in (+) x-direction, the conduction electrons are drifting with a
velocity v in (-) x-direction.
After B applied:
Lorentz force F  ev  B causes the electrons to bend downward. Electrons
accumulate on the lower surface, producing a net (-) charge. Simultaneously a
net (+) charge appears on the upper surface, because of the deficiency of
electrons.
This combination (+) and (-) surface charges create a downward E (called hall
field).
Evaluate the Hall field
The Lorentz force: FL  evx B
E created by the surface charges FH which opposes FL.
The accumulation process continues until the Hall force completely cancels the
FL.
In the steady state FH  FL
 eEH  evx B or eEH  evx B  Hall field
Current density: J x  N  evx
The hall field: EH  
1
JxB
Ne
(4.40)
The proportionality constant:
EH
1
 hall constant  RH  
JxB
Ne
(4.41)
Electrons concentration N can be determined by Hall field.
1
(4.41) depends on charge of electrons.
Ne
Sign of RH indicates the sign of the carriers involved.
RH  
RH positive: current is carried by holes
4.11 The AC conductivity and optical properties
Electrical conductivity in the presence of an alternating current field.
EM wave propagate in x-direction and polarized in y-direction.
E y  E0ei ( qxt )
dv
v
m *  eE  m * , (4.6)
dt

(4.42)
The equation of motion (유도해 보자, 역학강의 참고)
vy  
e
1
Ey
m * 1  i
(4.43)
The current density J y  N  e v y
leads to AC conductivity ~ 
0
1  i
Ne2
1
J y  N  e v y 
E
m * 1  i
(4.44)
Ne2
where  0 
,the conductivity is now
m*
 '
0
1   2 2
 "
 0
1   2 2
~   'i "
(4.45)
’: phase current which produces the resistive joule heating
”: /2 out of phase inductive current (적분을 통해서 왜 0 이 되는지 확인)
0
 '
1   2 2
 0
 "
1   2 2
(4.45)
Low frequency τ « 1, ”« ’ electrons exhibit resistive character
High frequency 1 « τ, ’ « ”  electrons inductive character
Maxwell equation
E
 H  L
J
t
(4.46)
1st term: the displacement current associated with the polarization of the ion core
2nd term: convective current of the conduction electrons
 ~  E
~
Current for AC field: J  E  

  i  t
Rewrite (4.46)
E
  H  ~
t
(4.47)
σ~
~
where total dielectric constant ε  ε L  i
ω
(4.48)
~ from (4.45) into (4.48) yield,
Substitution 

 0
'
"
~

 r   r  i r    L,r 
2 2

1



0



~r  ~r  0

0
  i
2 2


1



0



(4.49)
Complex index refraction of medium:
1
n : usual index refractive
~
~
2
n  εr  n  i (4.50) κ : extinction coefficient
Reflectivity R and absorption coefficient  are related to n and κ.
2

n  1   2
R
n  12   2

(4.51)
2

c
c : velocity of light
(4.52)
~
~  i "


@low frequency τ « 1, r reduces to the imaginary value
r
r , and hence
1
2
    0 
~

n       
 2   2 0 
"
r
Inverse of  is  
1
2
(4.53)
 c 
known as skin depth    0 
 2 0 
2
1

1
2
(4.54)
@high frequency 1 « τ, cover the visible and ultraviolet ranges.
~r reduces to the real value.
 ω 2p 
Ne2
2
(4.56) p: plasma frequency
 r  ε L ,r 1  2  (4.55) where ω p 
 Lm *
 ω 
The frequency region:
1. 1/τ <<  p, r  0  n=0 (순허수라야 제곱이 음의 실수!) and R=1, the
metal exhibits perfect reflectivity.
2. p  , 0  r  κ=0 therefore =0, 0  R 1, metallic medium acts like a
non-absorbing transparent dielectric.
Discontinuous drop in R at  = p (plasma edge reflection).
p proportional to N. In metal densities such that p falls into the high visible or
UV range.
Fig. 4.12 The plasma reflection edge.
전자 밀도가 높으면 모든 가시광을 반사하
므로 은색, 밀도가 낮으면 적색은 완전반
사가 안됨
p can be deduced from Maxwell equation
D   E  0
(4.57)
where D  E is electric displacement.
This equation admits the existence of longitudinal mode for
provide only    0 r  0 (4.58)
r vanishes at  = p  plasma mode
r ’: polarization of the charges induced by the field
r ”: absorption of energy by the system
  E  0,
4.12 Thermionic emission
Thermionic emission: when metal is heated, electrons are emitted from its surface.
@T= 0 K: all levels are filled up to the Fermi level EF,
above EF all levels are empty.
Electrons at EF can not escape from metal because
of the presence of an energy barrier at the surface.
The height of barrier denoted by  (work function).
Fig. 4.13 Thermionic emission
T the level above EF begin to be occupied. Energy higher than (EF +) become
populated to some extent. Electrons have enough energy to overcome the
barrier, and responsible for the emission from the surface.
Current density for the emitted electrons:

 
Velocity electrons in the range v x , v y , v z to v x  dv x , v y  dv y , v z  dvz
3
Concentration:
 m* 
d N  N
 e
 2kT 
3
2

2
 m* v 2 v 2
y vz 
 x

2 kT
dv x dv y dvz
(4.59)
Density of emitted current:
dJ x  evx d 3 N
(4.60)
Total current density:
 m* 
J x   dJ x   eN 

 2kT 
3
2
 v e

 m* v 2x  v 2y  v z2
x
Range for vy and vz is (-∞, ∞). Range vx is such that
Therefore,
 m* 
J x  eN 

 2kT 
Integration leads to
1
2

v e
x
vx 0
dvx
dv x dv y dvz
1
m * vx2  EF   
2
 2EF   
where v x 0  
 m * 
1
2
m * ek 2
2
2
. The numerical value is A  120 amp/cm  K
A
2 2  3
2  / kT
J x  AT e
 m*v 2x 2 kT

(4.61)
4.13 Failure of the free electron model
a. Electrical conductivity is proportional to electron concentration, but divalent
and trivalent metals are consistently less conductive than mono-valent metals.
b. Free electron model always predicts a negative Hall constant but some metals
exhibits positive Hall constant
c. Measurements of the Fermi surface indicate that it is often non-spherical in
shape
E F  12 m vF2 ( 4.33)


2/3
2
2
EF 
3 N
( 4.34)

2m
N e 2 F
  
( 4.35)
m vF
K
 2 N k 2T F
3m

( 4.36)
1  k 

 ( 4.37)
3 e 
eB
c   (4.38)
m
  c (4.39)
2
L
1
J x B(4.40)
Ne
1
RH  
(4.41)
Ne
E y  E0 ei qxt  (4.42)
ΕH  
vy  
e
1
E (4.43)

m 1  i
0
(4.44)
1  i
0
 
 
,
   0 2 2  (4.45)
2 2
1  
1  
~ 
E
 J (4.46)
t
~ E (4.47)
  H 
t
~
~  i  (4.48)

L 
  H L
 0
~  i    

r
r
L,
r

0 1   2 2




0
  i
(4.49)
2 2


1



0



~1/ 2  n  i (4.50)
n~ 
r
2

n  1   2
R
(4.51)
m  12   2

2
 (4.52)
c
  
n~
  r 
 2
1/ 2
1/ 2
 0 
 (4.53)
 
2


 0 
1/ 2
 0 c 2 
 (4.54)
  
2


 0 
  p2 
r L,r 1  2 (4.55)
  


2
Ne
 p2 
(4.56)
L m
  D    E  0(4.57)
0r  0(4.58)
32
 m   m  x2  y2  z2  2 kT
 e
d N  N 
d x d y d z (4.59)
 2kT 
dJ x  e x d 3 N (4.60)
3
J x  AT 2 e  / kT (4.61)