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7/29/2019 EngineeriEngineering Physics 2 Unit-1.pdfng Physics 2 Unit-1
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Unit
Unit
Unit
Unit1111
1
M
aterials
M
aterials
M
aterials
M
aterials
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2
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Band
Structures
Plentyoffreee-s
Veryfew(or)no
freee-s
Fewfree
e-s
OverlappingVB&CB
La
rgeE
g(7eV)
NarrowEg(1eV)
Verysmallelectricfield
forconduction.
Verylargeelectric
field
forconduction.
Smallelectricfieldfor
conduction.
+TCR(Temp
Co-effiof
Resistan
ce)
-TCR
-TCR
3
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Classicalfreeelectronth
eory(or)Mac
roscopicTheory
1900-DrudeandLorentz-
freeelectronsob
eyingthelawsofclassical
mechanics-freeelectronsassumedtomoveina
constantpotentia
l.
Quantumfreeelectrontheory(or)MicroscopicTheory
1928-Somm
erfeld-freeelectronsobeythequantumlaws-freeelectrons
ElectronTheoryofSolids
ElectronTheoryofSolids
ElectronTheoryofSolids
ElectronTheoryofSolids
areassumed
tomoveina
constantpotentialandthefermilevel
electronsareresponsiblefor
thepropertiesofmetals.
Zonetheory(or)Bandth
eory(or)Brillouintheory
1928-Bloch-electronsmoveinaperiodicfield
providedbythelattice.
Conceptofh
oles,originof
Bandgapandef
fectivemassof
electrons,
mechanismof
semiconductivity
basedonband.
4
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ClassicalFree
e-
Theory
Qu
antumFree
e-Theory
Macroscopictheory
Mic
roscopictheory
Alle-sin
cludingcoreand
-
5
valencee
-sareresponsible
forconduction.
responsibleforconduction.
Alle-splayamajorrolein
co
nduction.
Ferm
ilevele-splaya
majorroleinconduction.
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Mobility&Conduct
ivityinMetalConductors
Mobility&Conduct
ivityinMetalConductors
Mobility&Conduct
ivityinMetalConductors
Mobility&Conduct
ivityinMetalConductors
Inmetalstheelectricalconductivitydepends
onthenum
berofchargecarriers(free
electrons)presentinthatma
terial.
Letusconsiderasolidmaterial(S)(metal)
oflengthlandareaofcros
s-sectionA.
6
Letnnum
berofchargecarriers(freeelectrons)bepresentinit.
Totalnumb
erofelectronsin
thesolid(metal)
N=nAl.......(1)
TotalchargeQ=Totalnumberofelectronsx
Chargeofoneelectron
Q=N(e)
.......(2)
Negative
signindicatestha
tthechargeofth
eelectronisnegative.
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Substitutingeqn.(1)i
n(2)weget
Totalchargepresentinthe
solid
Q=nA
l(e)
.......(3)
Whenvoltage(V)isappliedtothemetal,
theelectr
onsstartsmovingwithan
averagev
elocitycalleddriftvelocity(vd)
fromone
endtotheotheri
.e.,alonga
lengthl
(or)throughadistancel,giving
7
.
Therefore
currentI
=
To
ta
lch
eintheso
lid
Time
ta
ken
for
themovemen
to
fch
es
arg
arg
I
=
.......(4)
Substitutingeqn.(3)in(4)weget
Current
I
=
.......(5)
Q tn
Al
e
t
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Currentd
ensity(J)i.e.,the
current,flowing
throughthesolid
per
unitareaisgivenbyJ
=
.......(6)
Substitutin
geqn.(5)ineqn.
(6)weget
J
=
.......(7)
I A nAl
e)
tA((((
Sincedriftvelocity
vd=
averagedis
cetravelledbytheel
ectron
tan
8
vd=
Equation(7
)canbewrittena
s
J
=
nvd(e)
.......(8)
l t
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Currentdensit
yfromOhmsLaw
Currentdensit
yfromOhmsLaw
Currentdensit
yfromOhmsLaw
Currentdensit
yfromOhmsLaw
FromOhmsl
aw,
I
=
.......(9)
Electricalres
istance(R)-oppositionofferedbythesolid(metal)forthe
movementof
electrons,given
by
R
=
.......(10)
V R
lA
9
whereistheresistivityofthesolid(metal).
Substitutinge
qn.(10)in(9)we
get
Current
I
=
.......(11)
Fromeqn(6)
J
=
=
=
VA l
I A
VA l
A
Vl
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J
=
.......(12)
Conductivity()istherecipro
calofresistivity
()
=
(a)
Electricfield
(E)=
(b)
Subabovetwovaluesineqn.(12),weget
V l1 V l
10
J
=
E
.......(13)
Comparingeqn.(
8)&(13),we
get
E
=nvd(e)
=
nv
e
Ed
=
n
(e)
.......(
14)
dv E
=
Mobility
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Thedriftvelocity(vd)acquiredb
ytheelectronperunitelectricfield(E)
appliedtoit.
dv E
=
(i.e.,)
Mobility
mV1S1
Mobility()
Theave
ragevelocityac
quiredbythefr
eeelectronina
particular
direction,duetotheapplication
ofelectricfieldiscalleddriftvelocity.
11
-
Average
distan
ce
trave
lle
db
yan
e
Time
tak
en
d
l
v
t
=
=
ms1
d
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Itisthetimetakenby
thefreeelectronto
reachitsequilibriumpositionfromitsdisturbed
position,in
thepresenceofa
ppliedfield.
Itistheaveragetimetaken
byafree
electronbetweentwosuccessivec
ollision.
c
dv
=
1 dv
=
Relaxatio
ntime()
Collisiontime(c)
Foraisotropicsolidlikemetals
=c
ismeanfree
path.
Theaveragedistancetravelledbetweentwo
successivecollisioniscalledm
eanfreepath.
where
istherootmean
squarethevelocityoftheelectron.
c
c
=
12
Meanfreepath()
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(Classical
FreeElectronTheory)
1sttheorytoex
plaintheelectricalconduction
inconductingm
aterials.
DrudeandLo
rentzin1900.
Freeelectron
sarefullyresponsibleforelectricalconduction.
Dr
ude-Lore
ntzTheory
13
om-centra
nuceuswt+
vecarge
surroundedby
electronsofvecharges.
Drudeassumedthattheelectro
nsin
ametalarefre
etomoveinall
directionsand
formanelectrongas.
Thesefreeelectronsmoverandomlyinallpos
sibledirections
justlikethe
gasmolecules
inacontainer.
Nucleus
aen
ceeecrons
Coreelectrons
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Postulates-Classicalfreeelec
trontheory
1.
Inanatomelectronrevolvearoundthenucleus.
2.Thevalenc
eelectronsofato
msarefreetomoveaboutthewho
levolume
ofmetalsli
kethemolecules
ofaperfectgasin
acontainer.
3.Intheabsenceofelectricfie
ld,t
hesefreeelec
trons
moveinrandomdirections
andcollidewitheach
otherandallthecollisionsareperfectlyelastic.
4.Sincetheelectronsareassumedtobeerfect
astheo
beth
elawsof
14
ClassicalK
ineticTheoryofGases.
5.AlsotheelectronvelocitiesinametalobeytheClassicalMaxw
ell
Boltzmann
DistributionofV
elocities.
6.Whenanelectricfield(E)is
appliedto
themetal,thefreeelectronsare
acceleratedinthedirectionopposite
tothedirec
tionofappliedelectricfield.
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the
quantityofelectricchargeflow
sinunittimeper
unitarea
ofcrosssectionoftheconductorperunitpoten
tialgradient.
QAtE
=
ohm1m1
ElectricalCondu
ctivity()
ThermalConductivity(K)
15
theamountofheatflowingthroughan
unitareaofamaterialper
unittemperatu
regradient.
K=
()
W/
m/K.
Thenegativesignindic
atesthatheatflowsfromhotendtocoldend.
Q dT
dx
F HG
I KJ
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Quantity
ofelectricchargeflowinunittimeperunitareaof
cross
sectionoftheconductorperu
nitpotentialgrad
ient.
QA
tE
=
ohm1m1
Expression
forElectricalConductivity
ElectricalCo
nductivity()
16
enaneecrc
e
sappe
oaconucor,
ereee
ecronsare
accelerate
dandgiveriseto
current(I)whichflowsinthedirectionof
electricfield.
2)
Theflowofchargesisgivenintermsofcurrentdensity(J).
3)
Theelectr
onsmovewithavelocitycalleddr
iftvelocity(vd)andthedrift
velocitydirectionisoppositetothefielddirection.
4)
Inanordinaryconductor,th
ecurrentdensity
isproportionalto
theapplied
electricfield.
J
E
J=E
...(1)
(-proportionalityconstant-electricalcondu
ctivityofaconduc
tor).
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Duetoth
eappliedelectricfield(E),theele
ctronsacquirean
acceleration
acanbeg
ivenby
Acceleration(a)=
...(2)
Letusconsiderthat
Ebetheelectricfieldintensityappliedtoaconductor,
Dr
iftveloc
ity
v
laxationtime
d
b
gb
g
Re
accelerationisthechangeinvelocityovertime
17
,
mbe
themassoftheelectron,
vbet
hevelocityofelectronand
Abe
theareaofcrosssection.
Whenanelectricfieldofstrength(E)isappliedtotheconducto
r,theforce
experienced
bythefreeelectronsisgivenby
F=eE
...(3)
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Dueto
thisforce,thefreeelectronswillacquireanaccelerationa.
FromN
ewtons2ndlawo
fmotion,t
heforceacquiredbyth
eelectrons
canbewrittenas
F=ma
...(4)
Equating(3)and(4)
e
E
=
ma
a
=
...(5)
eE
18
F HG
I KJ
e m
E
Now,substitutingthevalueofafromeqn(2)in
eqn.(
5),weget
=
vd
=
...(6)
eE
mm
vd
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Substitutingeq
n.(6)ineqn.(7),weget
J=
n(e)
J=
...(8)
Curren
tdensity(currentp
erunitareaperuni
ttime)isdetermine
dbythe
numberofcharge
carriersanditsdriftvelocity
J
=
n(e)vd
...(7)
FFFF HHHHGGGG
IIII KKKKJJJJ
e m
E
ne
E
2
19
Comparingeqn.(8)andeqn.(1),weget
E=
=
.....(9)
Thisistheexpressionforelectricalconductivity.
ne m
E
2
ne m
2
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i.
Thustheelectricalconductivityisdirectlypro
portionaltoelec
tron
density
andrelaxationtim
eoftheelectrons.
ne m
2
=
Conclusion
20
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ThermalConductivity(K)
Amou
ntofheatflowin
gthroughanunitareaofamaterialperunit
temperaturegradient.
K=
()
W/m/K.
Thenegativesignindicatesthat
heatflowsfromh
otendtocoldend
Q dT
dx
F HG
I KJ
21
wereK-Coefficientofthermalcon
ductivityofmateria
l,
Q-Amoun
tofheatflowingpe
runittimethrough
anunitcross-sectio
nalarea
dT/dx-Temperaturegradient.
Ingen
eral,t
hethermalconductivityofamaterialisduetothe
presence
oflatticevibrations(i.e.,phonons)andelectrons.Hencethetotaltherma
l
conductioncanbewrittenas,
Ktotal
=K
electron
+
Kphon
ons
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A
phono
n
isaquasiparticlecha
racterized
by
the
quantizationofthemodesoflatticevibrationsofperio
dic,
elasticcry
stalstructureso
fsolids.
Thestudy
ofphononsis
animportantp
artofsolidstate
physics,becausephononsplayamajor
roleinmany
of
thephysicalproperties
ofsolids,includingamaterial's
22
termaaneectrcaconuctvtes.
Aphonon
isaquantumm
echanicaldesc
riptionofaspecial
typeofvibrationalmotion,
known
asnormalmodes
in
classic
almechanics,
in
whicha
latticeuniform
ly
oscillatesatthesamefrequency.
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Letuscon
siderauniformrodABwithtemperaturesT
1(hot)atendA
andT
2(cold)atendB.
HeatflowsfromhotendA
tothecoldend
B.Letuscons
ideracross
sectionalareaCwhichisatadistanceequa
ltothemeanfreepath()of
theelectronbetweentheend
sAandBoftherodasshowninfigure.
Derivation
23
Theamoun
tofheat(Q)conductedbytherod
fromtheendAto
Bof
length2
isgivenby
Q
Q
=
...(1)
T1T2=
Temperaturedifferencebetweentheen
dsAandB(Kelvin)
AT
T
t
(
)
1
2
2
K=C
oefficientofthermal
conductivity.
A=
Areaofcrosssection
oftherod.
2=
Lengthofrod.
t=Timeforconduction.
KA
T
T
t
(
)
1
2
2
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Co-efficientoftherm
alconductivityperunitareaperunittime
K
=
...(
2)
Letus
assumethatthereisequalproba
bilityfortheele
ctronsto
move
inallthesixdirections.
Since
eachelectrontravelswiththermal
velocityvandifnisthefreeelectrondensity,
thenonanave
rage1/6nvelectronswilltravelin
Q
T
T
(
)
1
2
2
24
Sixprobable
directionsof
electron
movement
anyonedirection.
Thenumberelectrons
crossingunitarea
per
unittimeatC
=
...(3)
1 6nv
AccordingtoKineticTh
eoryofGas,free
electronsareassumedtobe
gasmoleculeswhicharefreelymoving.
Theaveragekineticen
ergyofanelectron
=
athotendAoftem
perature(T
1)
3 2
1
K
TB
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Avera
gekineticenergyofanelectron
atcoldendBoftemperature(T
2)
=
3 2
2
K
TB
Numberofelectrons
AverageK
E
of
electron
movingfrom
AtoB
x
.
.
=
1 6
3 2
1
nv
K
TB
.
=
1
Heatenergytransferre
dperunit
area
perunittimefromend
A
toBacrossC
25
=
4
1
B
...
Similarly,t
heheatenergytransferred
perunitareaperunit
timefrom
endBtoAacrossC
1 6
3 2
2
nv
K
TB
.
=
1 4
2
nv
K
TB
=
...(
5)
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Thenetheatenergy
transferredfrom
endAtoBperunitareaper
unittimeacrossCcan
begotbysubtrac
tingeqn.(
5)from
eqn.(
4)
Q
=
Q
=
...(
6)
1 4
1 4
1
2
nvK
T
nvK
T
B
B
1 4
1
2
nvK
T
T
B(
)
Substi
tutine
n.6in
en.2weet
K
=
K
=
...(7)
1 4
2
1
2
1
2
nv
K
T
T
T
T
B(
).
(
)
nv
KB
2
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Substitutingeqn.(8)ineq
n.(7)weget
K
=
nvK
v
B
Relaxationtime()=
Collisiontim
e(c)
v=
...(8)
Wekn
owformetals,
v
(
c=
)
K=
nvK
B
2
27
K
=
...(9)
nv
KB
22
Equatio
n(9)istheclassicalexpressionfo
rthermalcondu
ctivity.
Conclusion
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theratiobetweenthethermalconductivityandtheelectrical
conductivityofametalisdirectlyproportiona
ltotheabsolute
temperatureofthemetal.
K
T
W
iedemann
-FranzL
aw
28
where
Lisaconstantc
alledLorentznum
berwhosevalueis
equalto2.4
4x108W
k2at293K(Quantu
mmechanicalvalue).
K
=
LT
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(a)
ByC
lassicaltheor
y
Theexpressionforelectricalc
onductivity()=
Theexpressionforthermalco
nductivity(K)=
ne m2
nvK
B
22
Thermalconductivit
y
nv
KB
2
29
Electricalconductivity
ne m
2
=
Weknowthat,
Kineticenergyofanele
ctron
mv2=
KBT
3 2
K
=1 2
2 2
mv
Ke
B
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=
K
3 2
2
K
T
K
eB
B
.
=
32
2
KeB
F HG
I KJ
T
K T
=
3 2
2
KeB
F HG
I KJ
K T
=
L
=
L
3 2
2
KeB
F HG
I KJ
K
=LT
3 2
138
10
1602
10
23 1
9
2
. .
L M M
O P P
e
j
30
a.
Itisfoun
dthattheclassicalvalueofLorentznumber,isonlyhalf
oftheexperimental(i.e.,)
2.44x108W
k
2.
b.
ThisdiscrepancyofLv
alueisOneofthefailureofclassical
theo
ry(experimenta
landtheoretical).T
hisdrawbackcanbe
rectifiedbyQuant
umTheory.
L
=1.112x108W
k2
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Theexpressionforelectricalc
onductivity()=
(b)
ByQuantumtheory
Massofthe
electron(m)is
replacedbyth
eeffectivema
ss-m
*
ne m
2
*
L NM M
O QP P
Rearrangingtheexpressionfor
thermalconductivityandsubstitutingthe
electronicspe
cificheat,thethermalconductivity
canbewrittenas
31
Theexpressionforthermalcon
ductivity(K)=
Therm
alconductivity
Electricalconductivity
=
2
2
3
nK m
T
B *
L NM
O QP
K
=
2
22
3
nK m
T
ne m
B * *
L NM
O QP
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K T
=
L
K
=
2
2
3
K
e
T
B
F HG
I KJ
K T
=2
2
3
K
eB
FFFF HHHHGGGG
IIII KKKKJJJJ
=
L
2
2
3
KeB
F HG
I KJ
(3.
)
..
14 3
1
38
10
1602
10
2
23 1
9
2
L M
O P
L M M
O P P
d
i
32
ThusquantumtheoryverifiesWiedem
ann-Franzlawan
dithas
goodagreem
entwiththeexperimentalvalueof
Lorentznumber.
L
=2.44x10
8W
k2
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SuccessofClassica
lFreeElect
ronTheory
1.
ItverifiesOhmslaw
.
2.
Itis
usedtoderiveW
iedemann-Franzlaw.
3.
Itex
plainstheelectricalandthermalconductivitiesofm
etals.
4.
Itex
plainsopticalpropertiesofmetals
.
33
DrawbacksofClassicalFreeElectronTheory
1.
Itisamacroscopic
theory.
2.
Th
istheorycannotexplaintheelectro
nconductivityof
semiconductorsandinsulators.
3.
Fe
rromagnetismca
nnotbeexplaine
dbythistheory.
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4.
Thisth
eorycannotexpla
inthePhotoelectriceffect,Compton
effectandBlackbodyradiation.
5.
Thecalculatedvalueofspecificheatofm
etalsisnotmatching
wi
ththeexperimentalvalue.
6.
Atlow
temperature,Lorentznumberisnotaconstant.Butby
classicaltheoryitisaco
nstant.
34
7.
Dualnaturecannotbeexplained.
8.
Atom
icfinespectraca
nnotbeexplained.
9.
Classic
altheorystatesth
atallthefreeelectronswillabsor
benergy,
butqua
ntumtheorystate
sthatonlyfewelectronswillabso
rbenergy.
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1928-Somm
erfeld-freeelectronsobeythequantumlaws-freeelectrons
areassumed
tomoveina
constantpotentialandthefermilevel
electronsareresponsiblefor
thepropertiesofmetals.
Quantum
Theory
Quantum
Theory
Quantum
Theory
Quantum
Theory
35
ommere
-reane
va
eauresote
assca
reee-teory
included(i)Pa
ulisExclusio
nPrinciple.
(ii)FermiDiracStatistics.
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QuantumTheory
Inclassicaltheorythepropertiesofmetalssuch
aselectricaland
thermal
conductivitiesarewellexplained
ontheassumptionthattheelectronsinthe
metalfreelym
oveslikethep
articlesofagasandhencecalledfree-
electrongas.
Accordingto
classicaltheory,
theparticles(electrons)ofagas
atzero
kelvinwillha
vezerokineticenergy(3/2K
BT)andhenc
eallthe
36
.
Butaccordingtoclassicaltheorywhenalltheparticlesareatrest,a
llofthem
shouldbefille
donlyinthegroundstateenergylevel,whichisi
mpossible
andiscontroversialtothePaulisexclusionprin
ciple.
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"Inanatom
notwoelectr
onscanhavethesameseto
ffour
quantumnu
mbers."
PaulisEx
clusionPrinciple
electron
n
l
m
s
e1
1
0
0
+1/2
e2
1
0
0
-1/2
37
Inaclo
sedsystem,NO
twoelectrons
canOCCUPYtheSAME
STATE.
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Thusinorderto
filltheelectronsinagivenenergylevel,weshouldknowthe
following. (
i)E
nergydistributionofelectrons
(ii)N
umberofavailableenergystates
(iii)N
umberoffilledenergystates
38
(iv)P
robabilityoffillinganelectronin
agivenenergystate,etc.
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TheprobabilityF(E)ofanelectronoccupyingagiven
energylevelisgivenbyFe
rmi-Diracdistrib
utionfunction
F(E)
=
.......(1)
whereEis
calledFermiene
rgy
Fermidistributionfunction
1
1+
F HG
I KJ
exp
E
E
K
TF
B
39
a.
Inmetals,thee-saredistribute
damongthedifferentpossibleenergystates.
b.
EnergyofthehighestfilledstateatOKiscalled
theFermiEnerg
y(EF).
c.
ThemagnitudeofE
Fdependsonhowmanyfr
eeelectronsthere.
d.
AtOKallstatesuptoE
Fareo
ccupiedandstatesaboveE
Fareempty.
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AtT=0K
and
EEF
F(E)
=
=
=
=0
%
Itmeansthat0%
chancetofindtheparticle[electron]
At0Kalle
nergystatesabov
eEFareempty.
Case-2
1
1+
exp
(
)
11+
1
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AtT>
0K
and
E=E
F
F(E)=
=
=
=
50%
#Itmeansth
at50%
chanceto
findtheparticle
[electron].
#At0KenergystatesaboveE
Fareemptyandb
elowE
Farefilled
.
1
1
0
+ex
p(
)
11
1+
1
Case-3
41
Fermidistributionfunctionatdifferenttemperatures
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DensityofEnergyStates
Aparameterofinterestinthestudyofconduc
tivityofmetalsand
semicond
uctorsisthedensityofstates.
TheFerm
ifunctionF(E)givesonlytheprobabilityoffillingup
ofelectrons
inagiven
energystate,itdo
esnotgivesthe
informationabout
thenumber
ofelectronsthatcanbefille
dinagivenenergystate.
Toknowthatweshouldknowthenumberofavailableenergystates,so
calledden
sityofstates.
42
Definition
Thenu
mberofenergystatespresentintheenergyrangefromE
toE+dEperunitvolumeofthe
material.
Z(E)dE= Z
(E)dE=
Noo
fstates
between
E
an
dE
dEinameta
lp
iece
Vo
lumeo
fthemeta
lp
iece
.
+
N
dE
V
olume
(E)
.......(
1)
N
dE
a(E) 3
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Thee
nergyofthefreeelectronisthesameastheenergy
ofa
particleinabox.i.e.,
n2=n
x2+ny2+nz2,
nx,nyandnzarequantumnumbers
correspondingtothree
perpendicularaxesx,y
andz.
m=M
assoftheelectro
n
a=Sideofcubicallyshapedmetal.
E
=h m
a
n
n
n
x
y
z
22
2
2
2
8
+
+
E=
hn
ma
2
2 2
8
....
(2)
....
(3)
43
NumberofenergystateswithaparticularvalueofEdependson
thepossiblecombinationsofquantum
numbershavin
gthe
same
valueofn.
Tocalculatethenumber
ofenergystatesw
ithallpossible
energie
s,withnasradius,constructaspherein3Dimspace
andeve
rypointwithinthespacerepresen
tsanenergystate.
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Thesphe
reisfurtherdividedintomany
shellsandeachofthisshellreresentsa
Alsoeveryintegerrepresentsoneene
rgy
sta
te,unitvolumeo
fthisspacecontains
exactlyonestate.
Hencethe
numberofstates
inanyvolumeis
equaltothevolumeexpressedinunitsof
cubesoflatticeparameters.
44
particula
rcombinationo
fquantum
number
s(nx,ny
andnz)a
ndtherefore
represen
tsaparticularene
rgyvalue.
Therefore,thenumberofenergystateswithinasphereofradius
(n)=
.
Sincencan
haveonlypositiv
eintegervalues,wehaveto
consideronlyoneoctantofthesphere.
4 3
3
n
In3Dview,w
hatevertheportionstakenintheco-ordinateplanes-Octant
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In3Dview,
whatevertheportionstakenintheco-ordinateplanes-Octant
Consideracube-perpendicularcutand
bisectit-Octant
AnoctantisoneoftheeightdivisionsofaEuc
lidean3Dcoordinate
system
45
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Hence
availableenergy
states=
.
Inorderto
calculatethenum
berofstateswith
inasmall
energyintervalEandE+dE
,wehavetocons
tructtwo
sphereswithradiinand(n+
dn)andcalculatethe
spaceoccu
piedwithinthese
twospheres.
Similarly,t
henumberofav
ailableenergysta
teswithinthesph
ere
1 8
4 3
3
n
F HG
KJ
46
oraus
n+dn
=
Thereforenumberofavailableenergystates
betweentheshellsof
radiusn
andn+dn(or)betweentheenergy
levelsEandE+
dEis
givenby
N(E)dE=
1 8
4 3
3
(
)
n
dn
+
L NM
O QP
1 8
4 3
1 8
4 3
3
3
n
dn
n
+
F HG
I KJ
F HG
I KJ
b
g
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Omittinghighe
rpowersofdn
N(E)dE
=
.......(
4)
From
equa
tion
(3),
2
2n
dn
n
h2
2
1 8
4 3
3
3
3
3
2
2
3
F HG
I KJ
+
+
+
n
dn
ndn
ndn
n
=
47
E
=
ma
2
8
n2
=
.......
(5)
n
=
.......
(6)
8
2
2
ma
h
E
8
22
1
2
1
2
ma
h
E
F HG
I KJ
/
/
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n
dn
=
.......
(7)
Su
bs
tituting
eqns.
(6)&(7)in
eqn.
(4)
4
22
ma h
dE
n
ndn
2n
dn
=
8
22
ma
h
dE
Also
differen
tia
tingeqn
.(5)wege
t,
n2
=
....
(5)
8
2
2
ma
h
E8
22
1
2
1
2
ma
h
E
F HG
KJ
/
/
n
=
48
=
N(E
)dE
=
.......(
8)
28
4
22
1
2
1
2
22
ma
h
E
ma
h
dE
F HG
I KJ
F HG
KJ
/
/
4
8
22
3
2
12
ma
h
E
dE
F HG
I KJ
/
/
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Densityofsta
tesisgivenasnumberofenergystatesperunitvo
lume,
i.e.,Densityo
fstates,
Z(E)dE=
4
8
2
2
3
2
1
2
3
ma
h
E
dE
a
F HG
I KJ
/
/
Z(E)dE=
.......(9)
82
3
2
1
2
m
E
d
E
F G
I J
/
/
49
AccordingtoPaulisexclusionprinciple,
eac
henergystatec
anaccommodatetwoelectrons(one
spinupandanotherspindown).
Hencethenumberofenergystatesavailable
forelectron
occupancyisgivenby
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Z(E
)dE
=
Z(E
)dE
=
.......
(10)
4
8
2
2
3
2
1
2
mh
E
dE
F HG
I KJ
/
/
2
82
3
2
1
2
mh
E
dE
F HG
I KJ
/
/
50
The
aboveequatio
nsrepresentsth
edensityofstat
es
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CarrierConcen
trationinmetals
LetN(E)dErepresentsthe
numberoffilledenergystatesbetweenthe
intervalof
energydE.
Normally
alltheenergystateswillnotbefilled.
Theproba
bilityoffillingofelectronsinagivenenergystateisgivenby
Fermifunc
tionF(E).
51
N
(E)dE
=
Z(E)dE.
F(E)
.......(
12)
Z
(E)dE=
2
82
3
2
12
mh
E
dE
F HG
I KJ
/
/
WeknowdensityofstatesZ(E)dEas
.......(
11)
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Number
offilledenergystatesperunitvolu
me
N(E
)dE
=
.......(
13)
N(E)isknownasCarrierd
istributionfunction(or)
Car
rierconcentratio
ninmetals.
2
82
32
12
mh
E
dE
F
E
L NM
O QP
/
/
.
(
)
Inthecaseofamateri
alatabsolutezero,thehighestoccupiedlevel
52
F
F,
[Sinc
eat0K,themaximumenergylev
elthatcanbeoccupiedby
theelectronis
calledfermien
ergylevel(EF)]
Ther
efore,integrating
eqn.(
13)withinthelimits0toEF,
wecanget
thedensityofelectronswithin
thefermienerg
ylevel.
dn
z
2
82
3
2
1
2
0
mh
E
dE
EF
F HG
I KJ
z
/
/
=
(SinceF
(E)=1)
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= =
2
82
3
2
1
2
0
mh
E
dE
EF
F HG
I KJ
z
/
/
2
8
2 3
2
3
2
3
2
0
mh
E
EF
F HG
I KJ
/
/
n=
.......(14)
3
82
3
2
3
2
mh
EF
F HG
I KJ
/
/
b
g
53
Equation(14)g
ivesthecarrie
rconcentration(or)
densityofcharge
carriersat0K.
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Theabov
eequationcanbe
rewritteninterm
sofE
Fas
EF
3/2
=
EF
=
.......(15)
E
=
3.65x101
9.n
2/3eV
3
82
3
n
h m
F HG
I KJ
/2
h m
n
2
23
8
3
F HG
I KJ
F HG
I KJ
/
54
Hence
theFermienerg
yofametaldepe
ndsonlyonthed
ensityof
electronsofthatmetal.
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To
estimatetheaverageelectronenerg
yatabsolutezero
,let
uscalculate
thetotalenergyE
Tat0K.
Averageenergyofanelectron
(Eav
e)
=
...(16)
=
Tota
lener
gyo
fthee
lectronsat
K
E
Num
bero
fenergystates
at
K
n
T
0 0
(
)
(
)
Herethetota
lenergyo
fthe
,
Num
bero
fene
rgy
Energyof
F G
J
F G
I J
x
Meanenergyofe
lectronatabs
olutezero
55
ET
=
Substutingeqn.(
13)inabovee
quationweget
ET
=
N
E
dE
xE
EF
b
g
0z
2
82
3
2
12
0
mh
E
F
dE
E
F
F HG
I KJ
z
/
/
.
(E).
.
b
g
E
N(E)dE
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Sincethetemperatureis
0K,
F(E)=1
ET
= = =
2
82
3
2
1
2
0
mh
E
E
dE
EF
F HG
I KJ
z
/
/
.
.
b
g
2
82
3
2
3
2
0
mh
E
dE
EF
F HG
I KJ
z
/
/
b
g
2
8
2 5
2
3
2
5
2
0
mh
E
EF
F HG
I KJ
/
56
=
ET
=
.......(17)
2
5
2
5
2
mh
EF
HG
K
/
b
g
5
82
3
2
5
2
mh
EF
F HG
I KJ
/
/
b
g
Tota
lenergyofanelectronat0K
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Substitutingeqns.(
17)and
(14)ineqn.(
16)weget
Meanenergy
at0Kis,
Eave
=
Theaverage
energyofanelectronat0Kis
E n
mh
E
h
m
E
T
FF
=
F HG
I KJ
5
882
3
5
3
3
3
/2
/2
/2
/2
b
g
b
g
b
g
3
82
3
2
3
2
mh
EF
F HG
KJ
/
/
b
g
57
Eave
=
......(18)
5
F
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58
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Par
ticlein1dimensionalb
ox
Considera
particle(e-)of
massmmovingalongx-axisin1dim
potential
boxofwidthlandofinfiniteheight.
Theparticleisbouncingba
ckandforthbetweenthewallsofthe
box.
(i.e.,)particleismo
ving
toandfrobetweenthetw
owalls
atx=0andx=l.
59
en
co
esw
ewa,
en
eresnoossoenergyo
e
particle.
Hence,t
he
collisionsarepe
rfectlyelastic.
Sothereisnochangeinp
otentialenergyV.
TheP.EV
oftheparticle
insidetheboxisconstantandcanbe
takenaszeroforsimplicity.
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Particlein1dimensionalbox
Sincethe
wallsareofinfinitepotential,theparticledoesnot
penetrateoutfromthe
box.
Wemay
expressthefactbysayingth
atoutsidethe
box,
the
potentialenergyisfiniteasshowninfig
ure.
Inotherwordswe
canwrite
60
V(x)=0
when
0