chapter 5 free electron theory - smu
TRANSCRIPT
CHAPTER 5
FREE ELECTRON
THEORY
Free Electron Theory
Many s
olid
s co
nduct
ele
ctrici
ty.
There
are
ele
ctro
ns
that
are
not
bound t
o a
tom
s but
are
able
to m
ove t
hro
ugh t
he
whole
cry
stal.
Conduct
ing
solid
sfa
llin
totw
om
ain
class
es;
meta
lsand
sem
iconduct
ors
.
and
incr
ease
sby
the
additio
nof
small
am
ounts
of
impurity
. The r
esi
stiv
ity n
orm
ally
decr
ease
s m
onoto
nic
ally
with
decr
easi
ng t
em
pera
ture
.
and
can
be
reduce
dby
the
additio
nof
small
am
ounts
of
impurity
.
Sem
iconduct
ors
tend t
o b
eco
me insu
lato
rs a
t lo
w T
.
68
()
;10
10
metals
RT
mρ
−−
−Ω−
()
()
pure
semiconductor
metal
RT
RT
ρρ
−
Why mobile electrons appear in some
solid
s and others?
When
the
inte
ract
ions
betw
een
ele
ctro
ns
are
consi
dere
dth
isbeco
mes
avery
difficu
ltquest
ion
toansw
er.
The
com
mon
physi
calpro
pert
ies
of
meta
ls;
•G
reat
physi
calst
rength
•H
igh
densi
ty•
Hig
hdensi
ty•
Good
ele
ctri
caland
therm
alco
nduct
ivity,
etc
.
This
chapte
rw
illca
lcula
teth
ese
com
mon
pro
pert
ies
of
meta
lsusi
ng
the
ass
um
ption
that
conduct
ion
ele
ctro
ns
exis
tand
consi
stof
all
vale
nce
ele
ctro
ns
from
all
the
meta
ls;
thus
meta
llic
Na,
Mg
and
Al
will
be
ass
um
ed
tohave
1,
2and
3m
obile
ele
ctro
ns
per
ato
mre
spect
ively
.A
sim
ple
theory
of
‘freeelectronmodel’
whic
hw
ork
sre
mark
ably
well
will
be
desc
ribed
toexpla
inth
ese
pro
pert
ies
of
meta
ls.
Why mobile electrons appear in some
solid
s and not others?
Acc
ord
ing
tofr
ee
ele
ctro
nm
odel
(FEM
),th
evala
nce
ele
ctro
ns
are
resp
onsi
ble
for
the
conduct
ion
of
ele
ctrici
ty,
and
for
this
reaso
nth
ese
ele
ctro
ns
are
term
ed
conduct
ion
ele
ctro
ns.
N
a11→
1s2
2s2
2p
63s1
This
vala
nce
ele
ctro
n,
whic
hocc
upie
sth
eth
ird
ato
mic
shell,
isth
eele
ctro
nw
hic
his
resp
onsi
ble
chem
ical
pro
pert
ies
of
Na.
Vala
nce
ele
ctro
n (
loose
ly b
ound)
Core
ele
ctro
ns
W
hen
we
bring
Na
ato
ms
togeth
er
tofo
rma
Na
meta
l,
N
ahas
aBCC
stru
cture
and
the
dis
tance
betw
een
neare
stneig
hbours
is3.7
A˚
Na m
eta
l
neare
stneig
hbours
is3.7
A˚
The r
adiu
s of
the t
hird s
hell
in N
a is
1.9
A˚
Solid
state
of
Na
ato
ms
overl
ap
slig
htly.
From
this
obse
rvation
itfo
llow
sth
at
avala
nce
ele
ctro
nis
no
longer
att
ach
ed
toa
part
icula
rio
n,
but
belo
ngs
toboth
neig
hbouring
ions
at
the
sam
etim
e.
The
rem
oval
of
the
vala
nce
ele
ctro
ns
leaves
aposi
tively
charg
ed
ion.
The
charg
edensi
tyass
oci
ate
dth
eposi
tive
++
+
A
vala
nce
ele
ctro
nre
ally
belo
ngs
toth
ew
hole
cryst
al,
since
itca
nm
ove
readily
from
one
ion
toits
neig
hbour,
and
then
the
neig
hbour’s
neig
hbour,
and
soon.
This
mobile
ele
ctro
nbeco
mes
aco
nduct
ion
ele
ctro
nin
aso
lid.
The
charg
edensi
tyass
oci
ate
dth
eposi
tive
ion
core
sis
spre
ad
uniform
lyth
roughout
the
meta
lso
that
the
ele
ctro
ns
move
ina
const
ant
ele
ctro
static
pote
ntial.
All
the
deta
ilsof
the
cryst
al
stru
cture
islo
stw
hen
this
ass
unption
ism
ade.
++
+
Acc
ord
ing
toFE
Mth
ispote
ntial
ista
ken
as
zero
and
the
repuls
ive
forc
ebetw
een
conduct
ion
ele
ctro
ns
are
als
oig
nore
d.
There
fore
,th
ese
conduct
ion
ele
ctro
ns
can
be
consi
dere
das
movin
gin
dependently
ina
square
well
of
finite
depth
and
the
edges
of
well
corr
esp
onds
toth
eedges
of
the
sam
ple
.
Consi
der
am
eta
lw
ith
ash
ape
of
cube
with
edge
length
of
L,
Ψand
Eca
nbe
found
by
solv
ing
Sch
rödin
ger
equation
V
0L/
2L/
2
22
2E
mψ
ψ−
∇=
h0
V=
Sin
ce,
(,
,)
(,,)
xLy
Lz
Lxyz
ψψ
++
+=
•By m
eans
of periodic
boundary
conditio
ns
Ψ’s
are
runnin
g w
aves.
The s
olu
tions
of Sch
rödin
ger
equations
are
pla
ne w
aves,
w
here
V is
the v
olu
me o
f th
e c
ube,
V=
L3
()
11
(,,)
xy
zikxkykz
ikr
xyz
ee
VV
ψ+
+=
=rr
Norm
aliz
ation c
onst
ant
So t
he w
ave v
ect
or
must
satisf
y
where
p,
q,
r ta
kin
g a
ny inte
ger
valu
es;
+ve,
-ve o
r ze
ro.
Na
pλ
=2
,wherek
π λ
=
2Na
pkπ
=2
2k
pp
Na
L
ππ
==
2xk
pLπ
=2
yk
qLπ
=2
zk
rLπ
=;
;
The w
ave funct
ion Ψ
(x,y
,z)
corr
esp
onds
to a
n
energ
y o
f
th
e m
om
entu
m o
f
22
2
kE
m=h
22
22
()
2x
yz
Ek
kk
m=
++
h
(,,)
pk
kk
=h
Energ
y is
com
ple
tely
kin
etic
(,,)
xy
zp
kk
k=h
22
21 2
2
kmv
m=h
22
22
mv
k=h
pk
=h
W
eknow
that
the
num
ber
of
allo
wed
kvalu
es
insi
de
asp
herica
lsh
ell
of
k-s
pace
of
radiu
sk
of
2 2()
,2Vk
gkdk
dk
π=2π
w
here
g(k
)is
the
densi
tyof
state
sper
unit
magnitude
of
k.
The number of allowed states
per unit energy range?
Each
kst
ate
repre
sents
two
poss
ible
ele
ctro
nst
ate
s,one
for
spin
up,
the
oth
er
issp
indow
n.
()
2()
gEdE
gkdk
=()2()dk
gE
gkdE
=dE
22
2
kE
m=h
2dE
k
dk
m=h
2
2mE
k=
h
()
gE
=2()
gk
dk
dE
222V π
kk
2
2mE
h2m
kh
3/2
1/2
23(2
)2
()
Vm
EgE
π=
h
Ground state of the free electron gas
Ele
ctro
ns
are
ferm
ions
(s=
±1/2
)and
obey
Pauli
excl
usi
on
princi
ple
;each
state
can
acc
om
modate
only
one
ele
ctro
n.
The
low
est
-energ
yst
ate
of
Nfr
ee
ele
ctro
ns
isth
ere
fore
obta
ined
by
filli
ng
the
Nst
ate
sof
low
est
energ
y.
Thus
all
state
sare
fille
dup
toan
energ
yE
F,kn
ow
nas
Ferm
ien
erg
y,
obta
ined
by
inte
gra
ting
densi
tyof
state
sbetw
een
0and
EF,
should
equal
N.
Hence
Rem
em
ber
3/2
1/2
23(2
)2
()
Vm
EgE
π=
h
FF
EE
VV
==
=∫
∫
Solv
e f
or
EF
(Ferm
i energ
y);
2/3
22
3
2F
NE
mVπ
=
h
3/2
1/2
3/2
23
23
00
()
(2)
(2)
23
FF
F
VV
NgEdE
mE
dE
mE
ππ
==
=∫
∫h
h
The o
ccupie
d s
tate
s are
insi
de t
he F
erm
i sp
here
in k
-space
show
n b
elo
w;
radiu
s is
Ferm
i w
ave n
um
ber
kF. 2
2
2
FF
kE
m=h
kz
Ferm
i su
rface
E=
E
2/3
22
3
2F
NE
mVπ
=
h
2F
e
Em
=
ky
kx
E=
EF
kF
From
th
ese
two e
quation k
F
can b
e found a
s,1/3
23
F
Nk
Vπ
=
Th
e s
urfa
ce o
f th
e F
erm
i sp
here r
ep
resen
t th
e
bo
un
dary b
etw
een
occu
pie
d a
nd
un
occu
pie
d
k
sta
tes a
t ab
so
lute
zero
fo
r t
he f
ree e
lectr
on
gas.
Typic
al
valu
es
may
be
obta
ined
by
usi
ng
monovale
nt
pota
ssiu
mm
eta
las
an
exam
ple
;fo
rpota
ssiu
mth
eato
mic
densi
tyand
hence
the
vala
nce
ele
ctro
ndensi
tyN
/Vis
1.4
02x10
28
m-3
soth
at
19
3.4010
2.12
FE
JeV
−=
×=
10.746
kA
−=
°
Fe
rmi (d
egenera
cy)
Tem
pera
ture
TF by
10.746
Fk
A−
=°
FB
FE
kT
=
42.4610
FF
B
ET
Kk
==
×
It
isonly
at
ate
mpera
ture
of
this
ord
er
that
the
part
icle
sin
acl
ass
ical
gas
can
att
ain
(gain
)kin
etic
energ
ies
as
hig
has
EF
.
Only
at
tem
pera
ture
sabove
TF
will
the
free
ele
ctro
ngas
behave
like
acl
ass
icalgas.
Fe
rmim
om
entu
mF
FP
k=h
Fe
FP
mV
=
These
are
the
mom
entu
mand
the
velo
city
valu
es
of
the
ele
ctro
ns
at
the
state
son
the
Ferm
isu
rface
of
the
Ferm
isp
here
.
So,
Ferm
iSphere
pla
ys
import
ant
role
on
the
behavio
ur
of
meta
ls.6
10.8610
FF
e
PV
ms
m
−=
=×
2/3
22
32.12
2F
NE
eVm
Vπ
=
=
h
1/3
21
30.746
F
Nk
AVπ
−
=
=°
Typ
ical
valu
es o
f m
on
ovale
nt
po
tassiu
m m
eta
l;
V
61
0.8610
FF
e
PV
ms
m
−=
=×
42.4610
FF
B
ET
Kk
==
×
The free electron gas at finite temperature
At
ate
mpera
ture
Tth
epro
babili
tyof
occ
upation
of
an
ele
ctro
nst
ate
of
energ
yE
isgiv
en
by
the
Ferm
idis
trib
ution
funct
ion 1
Fe
rmi
dis
trib
ution
funct
ion
dete
rmin
es
the
pro
babili
tyof
findin
gan
ele
ctro
nat
the
energ
yE.
()/
1
1F
BFD
EE
kT
fe
−=
+
f FD(E,T
)
()/
1
1F
BFD
EE
kT
fe
−=
+Fermi Function at T=0
and at a finite temperature
f F
D=
? A
t 0°K
i.E<
EF
11
f=
=
EF
E<
EF
E>
EF
0.5
E
ii.E>
EF
()/
11
FB
FD
EE
kT
fe
−=
=+
()/
10
1F
BFD
EE
kT
fe
−=
=+
Fermi-Dirac distribution function at
various temperatures,
n(E
,T)
g(E
)
n(E
,T)
num
ber
of
free
N
um
ber
of
ele
ctro
ns
per
unit
energ
yra
nge
acc
ord
ing
toth
efr
ee
ele
ctro
nm
odel?
The
shaded
are
ash
ow
sth
ech
ange
indis
trib
ution
betw
een
abso
lute
zero
and
afinite
tem
pera
ture
.
T>
0
T=
0
EE
F
n(E
,T)
num
ber
of
free
ele
ctro
ns
per
unit
energ
yra
nge
isju
stth
eare
aunder
n(E
,T)
gra
ph.
(,)
()
(,)
FD
nET
gEf
ET
=
Fe
rmi-
Dir
ac
dis
trib
ution
funct
ion
isa
sym
metr
icfu
nct
ion;
at
finite
tem
pera
ture
s,th
esa
me
num
ber
of
levels
belo
wE
Fis
em
ptied
and
sam
enum
ber
of
levels
above
EF
are
fille
dby
ele
ctro
ns.
n(E
,T)
g(E
)
T>
0
T=
0
EE
F
Heat capacity of the free electron gas
Fr
om
the
dia
gra
mof
n(E
,T)
the
change
inth
edis
trib
ution
of
ele
ctro
ns
can
be
rese
mble
din
totr
iangle
sof
heig
ht
1/2
g(E
F)and
abase
of
2k
BT
so1/2
g(E
F)k
BT
ele
ctro
ns
incr
ease
dth
eir
energ
yby
kBT.
kBT.
T>
0
T=
0
n(E
,T)
E
g(E
)
EF
The
diffe
rence
inth
erm
al
energ
yfr
om
the
valu
eat
T=
0°K
21
()
(0)
()(
)2
FB
ET
EgE
kT
−
D
iffe
rentiating w
ith r
esp
ect
to T
giv
es
the
heat
capaci
ty a
t co
nst
ant
volu
me,
2()
vF
B
EC
gE
kT
T∂=
=∂
2(
)3
FF
NEgE
=3
33
()2
2F
FB
F
NN
gE
EkT
== 2
23
()
2v
FB
B
BF
NC
gE
kT
kT
kT
==
3 2v
B
FTC
Nk
T
=
H
eat
cap
acit
y o
fFree e
lectr
on
gas
Transport Properties of Conduction Electrons
Fe
rmi-
Dirac
dis
trib
ution
funct
ion
desc
ribes
the
behavio
ur
of
ele
ctro
ns
only
at
equili
brium
.
Ifth
ere
isan
applie
dfield
(Eor
B)
or
ate
mpera
ture
gra
die
nt
the
transp
ort
coeff
icie
nt
of
therm
al
and
ele
ctrica
lco
nduct
ivitie
sm
ust
be
therm
al
and
ele
ctrica
lco
nduct
ivitie
sm
ust
be
consi
dere
d.
Tran
sp
ort
co
eff
icie
nts
σ,E
lectr
ical
co
nd
ucti
vit
yK
,Th
erm
al
co
nd
ucti
vit
y
Tota
l heat
capaci
ty a
t lo
w t
em
pera
ture
s
w
here
γ a
nd
βare
const
ants
and t
hey c
an
3C
TT
γβ
=+
Ele
ctro
nic
Heat
capaci
tyLa
ttic
e H
eat
Capaci
ty
w
here
γ a
nd
βare
const
ants
and t
hey c
an
be f
ound d
raw
ing C
v/T
as
a f
unct
ion o
f T
2
Equation
of
motion
of
an
ele
ctro
nw
ith
an
applie
dele
ctric
and
magnetic
field
.
This
isju
stN
ew
ton’s
law
for
part
icle
sof
mass
me
and
charg
e(-
e).
e
dv
meE
evB
dt=−
−×
rur
rur
This
isju
stN
ew
ton’s
law
for
part
icle
sof
mass
me
and
charg
e(-
e).
The
use
of
the
class
ical
equation
of
motion
of
apart
icle
todesc
ribe
the
behavio
ur
of
ele
ctro
ns
inpla
ne
wave
state
s,w
hic
hexte
nd
thro
ughout
the
cryst
al.
Apart
icle
-lik
eentity
can
be
obta
ined
by
superp
osi
ng
the
pla
ne
wave
state
sto
form
aw
avepack
et.
The v
elo
city
of
the w
avepack
et
is t
he g
roup
velo
city
of
the w
aves.
Thus
So o
ne c
an u
se e
quation o
f m
dv/d
t
1
ee
ddE
kp
vm
mdk
dk
ω=
==
=
rur
rh
rr
h
22
2e
kE
m
pkω
==
=
hh h
So o
ne c
an u
se e
quation o
f m
dv/d
t
e
dv
vm
eE
evB
dt
τ
+=−
−×
rr
urr
ur
τ=
mean fre
e t
ime b
etw
een c
olli
sions.
An e
lect
ron
lose
s all
its
energ
y in t
imeτ
(*)
In
the a
bse
nce
of
a m
agnetic
field
, th
e a
pplie
d E
re
sults
a c
onst
ant
acc
ele
ration b
ut
this
will
not
cause
a c
ontinuous
incr
ease
in c
urr
ent.
Sin
ce
ele
ctro
ns
suffer
colli
sions
with
phonons
ele
ctro
ns
The a
dditio
nal te
rm
cause
the v
elo
city
v t
o
deca
y e
xponentially
with a
tim
e c
onst
ant
when
the a
pplie
d E
is
rem
oved.
e
vm
τ
r
τ
The Electrical Conductivty
In
the p
rese
nce
of
DC fie
ld o
nly
,eq.(
*)
has
the
steady s
tate
solu
tion
ev
Emτ
=−
rur
e
e mτµ=
Mobili
ty f
or
ele
ctro
n
M
obili
ty d
ete
rmin
es
how
fast
the c
harg
e c
arr
iers
m
ove w
ith a
n E
.
em
a c
onst
ant
of
pro
port
ionalit
y(m
obili
ty)
e
em
ele
ctro
n
Ele
ctrica
l cu
rrent
densi
ty,
J
W
here
n is
the e
lect
ron d
ensi
ty a
nd v
is
drift
velo
city
.H
ence
()
Jn
ev
=−
Nn
V=
2 τur
ur2
neτ
e
ev
Emτ
=−
rur
2 e
ne
JE
m
τ=
urur
JE
σ=
urur
2 e
ne m
τσ=
Ele
ctr
ical co
nd
ucti
vit
y
Ohm
’s law
1ρ
σ=
LR
Aρ=
Ele
ctri
cal Resi
stiv
ity a
nd R
esi
stance
Collisions
In
aperf
ect
cryst
al;
the
colli
sions
of
ele
ctro
ns
are
with
therm
ally
exci
ted
latt
ice
vib
rations
(sca
ttering
of
an
ele
ctro
nby
aphonon).
This
ele
ctro
n-p
honon
scatt
ering
giv
es
ate
mpera
ture
dependent
colli
sion
tim
e()
phT
τte
mpera
ture
dependent
colli
sion
tim
ew
hic
hte
nds
toin
finity
as
T0.
In
real
meta
l,th
eele
ctro
ns
als
oco
llide
with
impurity
ato
ms,
vaca
nci
es
and
oth
er
imperf
ect
ions,
this
resu
ltin
afinite
scatt
eri
ng
tim
eeven
at
T=
0.
()
phT
τ
0τ
The t
ota
l sc
att
ering r
ate
for
a s
lightly im
perf
ect
cr
yst
alat
finite t
em
pera
ture
;
So t
he t
ota
l re
sist
ivity ρ
,
0
11
1
()
phT
ττ
τ=
+
Due t
o p
honon
Due t
o im
perf
ect
ions
So t
he t
ota
l re
sist
ivity ρ
,
This
is
know
n a
s M
att
heis
en’s
rule
and illu
stra
ted in
follo
win
g f
igure
for
sodiu
m s
peci
men o
f diffe
rent
purity
.
02
22
0
()
()
ee
e
I
ph
mm
mT
ne
ne
Tne
ρρ
ρτ
ττ
==
+=
+
Ideal re
sist
ivity
Resi
dual re
sist
ivity
Residual resistance ratio
Resi
dual re
sist
ance
ratio =
room
tem
p.
resi
stiv
ity/
resi
dual re
sist
ivity
and it
can b
e a
s hig
h a
s
for
hig
hly
puri
fied s
ingle
cry
stals
.6
10
Tem
pera
ture
pure
impure
Collision tim
e τ
σ
10
15.310(
)pureNa
residual
xm
σ−
=Ω−
71
()
2.010(
)sodium
RT
xm
σ−
=Ω−
em
m=
14
22.610
mx
sneσ
τ−
=
can b
e f
ound b
y
takin
g
at
RT
28
32.710
nx
m−
=ne
11
7.010
xs
−
61.110
/Fv
xm
s=
()29
lRT
nm
=
(0)
77lT
mµ=
=
at
T=
0
Fl
vτ
=Ta
kin
g
;
a
nd
These
mean
free
path
sare
much
longer
than
the
inte
rato
mic
dis
tance
s,co
nfirm
ing
that
the
free
ele
ctro
ns
do
not
colli
de
with
the
ato
ms
them
selv
es.
Thermal conductivity, K
metals
nonmetals
KK
−
Due t
o t
he h
eat
tranport
by t
he c
onduct
ion e
lect
rons
Ele
ctro
ns
com
ing
from
ahott
er
regio
nof
the
meta
lca
rry
more
therm
al
energ
yth
an
those
from
aco
ole
rre
gio
n,
resu
ltin
gin
anet
flow
of
heat.
The
therm
alco
nduct
ivity
1 3V
FK
Cvl
=V
C
net
flow
of
heat.
The
therm
alco
nduct
ivity
l
Fv
BkT
Fε
Fl
vτ
=2
1 2F
eF
mv
ε=
where
is
the s
peci
fic
heat
per
unit v
olu
me
is t
he m
ean f
ree p
ath
;
a
nd
Ferm
i ener
gy
isth
em
ean
speed
of
ele
ctro
ns
resp
onsi
ble
for
therm
al
conduct
ivity
since
only
ele
ctro
nst
ate
sw
ithin
about
of
change
their
occ
upation
as
the
tem
pera
ture
vari
es.
22
22
11
2(
)3
32
3
BV
FB
F
Fe
e
NT
nkT
KCv
kV
Tm
m
ππ
ττ
ετ
==
=2 2
vB
FTC
Nk
T
π
=
where
Wiedem
ann-Franz law
2 e
ne m
τσ=
22
3
B e
nkT
Km
πτ
=
The r
atio o
f th
e e
lect
rica
l and t
herm
al co
nduct
ivitie
s is
independent
of
the
ele
ctro
n g
as
para
mete
rs;
22
82
2.4510
3
Kk
xW
KT
e
πσ
−−
==
Ω
B
ele
ctro
n g
as
para
mete
rs;
82
2.2310
KL
xW
KT
σ−
−=
=Ω
Lore
ntz
num
ber
For
copper
at
0 C