EngineeriEngineering Physics 2 Unit-1.pdfng Physics 2 Unit-1

7/29/2019 EngineeriEngineering Physics 2 Unit-1.pdfng Physics 2 Unit-1

1/65

Unit

Unit

Unit

Unit1111

1

M

aterials

M

aterials

M

aterials

M

aterials


2/65

2


3/65

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


4/65

Classicalfreeelectronth

eory(or)Mac

roscopicTheory

1900-DrudeandLorentz-

freeelectronsob

eyingthelawsofclassical

mechanics-freeelectronsassumedtomoveina

constantpotentia

l.

Quantumfreeelectrontheory(or)MicroscopicTheory

1928-Somm

erfeld-freeelectronsobeythequantumlaws-freeelectrons

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


5/65

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.


6/65

Mobility&Conduct

ivityinMetalConductors

Mobility&Conduct


Mobility&Conduct


Mobility&Conduct


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.


7/65

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


8/65

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


9/65

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


10/65

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


11/65

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


12/65

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()


13/65

(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


14/65

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.


15/65

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


16/65

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).


17/65

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)


18/65

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


19/65

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


20/65

i.

Thustheelectricalconductivityisdirectlypro

portionaltoelec

tron

density

andrelaxationtim

eoftheelectrons.

ne m

2

=

Conclusion

20


21/65

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


22/65

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.


23/65

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


24/65

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


25/65

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)


26/65

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


27/65

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


28/65

theratiobetweenthethermalconductivityandtheelectrical

conductivityofametalisdirectlyproportiona

ltotheabsolute

temperatureofthemetal.

K

T

W

iedemann

-FranzL

aw

28

where

Lisaconstantc

alledLorentznum

berwhosevalueis

equalto2.4

4x108W

k2at293K(Quantu

mmechanicalvalue).

K

=

LT


29/65

(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


30/65

=

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


31/65

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


32/65

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


33/65

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.


34/65

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.


35/65

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.


36/65

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.


37/65

"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.


38/65

Thusinorderto

filltheelectronsinagivenenergylevel,weshouldknowthe

following. (

i)E

nergydistributionofelectrons

(ii)N

umberofavailableenergystates

(iii)N

umberoffilledenergystates

38

(iv)P

robabilityoffillinganelectronin

agivenenergystate,etc.


39/65

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.


40/65

AtT=0K

and

EEF

F(E)

=

=

=

=0

%

Itmeansthat0%

chancetofindtheparticle[electron]

At0Kalle

nergystatesabov

eEFareempty.

Case-2

1

1+

exp

(

)

11+

1


41/65

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


42/65

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


43/65

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.


44/65

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


45/65

In3Dview,

whatevertheportionstakenintheco-ordinateplanes-Octant

Consideracube-perpendicularcutand

bisectit-Octant

AnoctantisoneoftheeightdivisionsofaEuc

lidean3Dcoordinate

system

45


46/65

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


47/65

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

/

/


48/65

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

/

/


49/65

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


50/65

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


51/65

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)


52/65

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)


53/65

= =

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.


54/65

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.


55/65

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


56/65

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


57/65

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


58/65

58


59/65

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.


60/65

Particlein1dimensionalbox

Sincethe

wallsareofinfinitepotential,theparticledoesnot

penetrateoutfromthe

box.

Wemay

expressthefactbysayingth

atoutsidethe

box,

the

potentialenergyisfiniteasshowninfig

ure.

Inotherwordswe

canwrite

60

V(x)=0

when

0

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