Physics2203,Fall2012ModernPhysics

.

 Friday,Sept.7th,2012:‐‐Chapter4:QuanAzaAonoflight‐‐‐StartChapter5:QuanAzaAonofAtomicEnergyLevels

 Announcements:‐‐FirstcomputerexercisenextWednesday.‐‐Homework#3onCh.4and5willbepostedtoday.‐‐IwillbeinWashington,DC,MondayandWednesday. ProfessorZhangwillteachthisclass

JamesClerkMaxwell(1831‐1879)ScoYshPhysicist,MathemaAcian.

IsaacNewton(1643‐1727)EnglishPhysicist,MathemaAcian

Physicistsmighthaverewri[enthebiblicalstoryofcreaAonasfollows!

InthebeginningHecreatedtheheavenandtheearth‐‐F ≡G mm 'r2

= maAndHesaid“lettherebelight”

E

idA=Qε0

∫ E

ids= −

dΦB

dt∫

BidA = 0 Bids = µ0I +ε0µ0dΦE

dt ∫ ∫

ThefirsthintthattherewassomethingwrongwiththispictureofthecreaAoncamefromthetheintensityofradiaAoncomingforaheatedcavity(blackbody)asafuncAonofthewavelength.

1000K

Ultravioletcatastrophe

MaxPlanck(1858‐1947)German,Nobelprizein1918

QuanAzedLight

MaxPlanck

1918NobelPrize

TheobjecAveistoexplaintheamountoflightintensitythatisradiatedfromanobjectasafuncAonoftemperature.Weconsiderwhatisdescribedasablackbody.Thewallsareblackwithaverysmallhole,whichistheblackbody

BlackBodyRadiaBon:Createacavitywithonlyaverysmallopening.ThispicturesshowslightenteringthecavitybutsinceemissionandabsorpAonareconnectedbyKirchhoff’sTheoremaperfectabsorberisanidealradiator.

JosefStefan(1835‐1893)observedthatthetotalpoweretotalperunitareawasproporAonaltoT4

etotal = ef0

∞

∫ df =σT 4

In1893WilhelmWienproposedageneralformoftheblackbodyradiaAonformula,whichgavethecorrectdependenceofthemaximiuminthedistribuAon–Wien’sdisplacementlaw

λmaxT = 2.898x10−3miK

u( f ,T ) = Af 3e−β f /T

u is energy density/f

TheSun’sradiusisRs=7.0x108mandtheaverageEarth‐SundistanceisR=1.5x1011m.ThepowerperunitareafromtheSunis1400W/m2.WhatisthetemperatureoftheSun?

etotal (Rs ) =σT4

σ = 5.67x10−8W im−2iK −4

We need to connect R s (sun) to R on earth: Conservation of energy etotal (Rs )i4πRs

2 = etotal (R)i4πR2

etotal (Rs ) =etotal (R)iR2

Rs2

T =etotal (R)iR2

σRs2

1/4

= 5800 K

ObservaBon:Thepeakintheefficiencyofthehumaneyeoccursatawavelengthof500nm.WhattemperatureareyoumostsensiAveto?

T =2.898x10−3miK

λmax=2.898x10−3miK500x10−9m

= 5800KWHY!

VerycarefulmeasurementsininfraredandfarinfraredregionsofthespectrumshowedthatWien’slawfailedinthisregion.

u(λ,T ) = Aλ−3e−β /λT

LookattheWienequaAonforlongwavelengths.

u(λ,T ) = Aλ−3e−β /λT ≈ Aλ 31− β

λT

Aλ 3

ExperimentbyRubensshowedaTdependence!!

OnaSundayeveninginOctoberof1900Planckdiscoveredthefamousblackbodyformula,whichusheredinthequantumtheory.

‐‐‐earlierinthedayhehadvisitedRubensandfoundoutabouttheTdependenceofu(f,T).

u( f ,T ) = 8π f3

c31

ehf /kBT −1

Planck’sconstanthh=6.26x10‐24J/Kh=4.135x10‐15eV/KBoltzman’sconstantkB=1.380x10‐34JsHigh frequency limit.

u( f ,T ) = 8π f 3

c3

1ehf /kBT −1

≈

8π f 3

c3 e−hf /kBT

Wien 's formula

Low frequency limit.

u( f ,T ) = 8π f 3

c3

1ehf /kBT −1

≈

8π f 3

c3

11+ hf / kBT + -1

=

8π f 2

c3

kBTh

Linear

PlanckbelievedthatblackbodyradiaAonwasproducedbyvibraAngsubmicroscopicelectriccharges‐‐‐calledResonators.

InclassicalMaxwelliantheoryaResonatorscouldoscillateatanyfrequencyandcouldloseanyfracAonofitsenergy.

PlanckhadtoassumetheResonatorscouldonlyhavediscretevaluesoftheEnergyandthatthechangeinenergywasquanBzed.

Eresonator = nhf n=1, 2, 3----

ΔE = hf

Angular frequency ω=f/2πEresonator = nω n=1, 2, 3----

ΔE = ω

1000K

Ultravioletcatastrophe

ClassicalRayleigh‐Jeansblackbodyu(λ,T )dλ = 8π

λ 4kBTdλ

u( f ,T )df = 8π f2

c3kBTdf

Planck’squantumblackbody

u(λ,T )dλ = 8πhcdλλ5 ehc /λkBT −1( )

u( f ,T )dλ = 8πhf 3dfc3 ehf /kBT −1( )

In1905EinsteinpublishedaseminalpaperontheparAclenatureoflight—Photoelectriceffect.

MillikansetouttoconfirmthesepredicAons.R.A.Millikan:Phys.Rev.3355(1916).

TheobservaAonsconsistofthemeasuringmagnitudeofthephotoemi[edcurrentasafuncAonof:• Frequencyofthelightf.• TheintensityofthelightI.• Thebiasvoltagebetweenthecathodeandanode.• thenatureofthemetal.

ClassicalInterpretaBon:TheenergyinalightwaveisproporAonaltotheIntensityI,so•  Theemi[edelectronshouldhavemoreenergywithlargerIntensity.

•  IftheIntensityIislargeenoughtherewillalwaysbeelectronsemi[ed,independentoff.

•  ThereshouldbeAmedependenceinthesignal.

Kmax =12mevmax

2 = eVs

ObservaAon#2:KmaxisalinearfuncAonoffrequency.ObservaAon#3:Thereisathresholdfrequencyf0.

ObservaAon#1:Kmaxdoesnotdependuponintensity.

ObservaAon#4:ThereisnoAmelagbetweenthestartofilluminaAonandthestartofthephotocurrent.

h[p://phet.colorado.edu/en/simulaAon/photoelectric

WhydotheI‐Vcurvesinthefigurerisegraduallybetween–Vsand0?Whyisn’tthereanabruptraiseinthephotocurrentat–Vs?

WhatdoestheslopeofthiscurvetellyouaboutthedistribuAonofelectronsinsidethemetal?

h[p://phet.colorado.edu/en/simulaAon/photoelectric

Einsteinproposedthatlightcomesinquanta,calledaPhoton.ThesePhotonsbehavelikeparAclesandcanonlybeproducedandabsorbedascompeteunits.

Energy of a Photon: E=hf=ω: h is Planck's constant.

Momentum of a photon: p= Ec=hfc=hλ

AphotonisabsorbedbyexciAnganelectroninthesolid.AswewilllearnthemaximumenergyelectronsareattheFermienergy.TheexcitedelectronnowhasakineAcenergyinsidegivenbyE=hf+EF,butithastoescapeforthesolid.ItlosseskineAcenergyequaltotheworkfuncAonφ.

mvmax2

2= hf −φ

eV0 (cut off)=0hf0 = φ

f0 =φh

ω0 =φ

h[p://phet.colorado.edu/en/simulaAon/photoelectric

PlanckviewedthistheorywithskepAcismandMillikansetouttoprovehimwrong.

MillikansetouttomeasurethecutoffasafuncAonoff.TheslopeshouldbePlanck’sconstantandtheintercepttheworkfuncAon.In1916hepublisheddataprovingEinsteinright

Themeasuredslopewas4.1x10‐15eV.s,ingoodagreementwiththevalueofh.Theinterceptisabout2eVwhichisOKforanalkalimetal.

Supposethatlightoftotalintensity1.0µW/cm2fallsonacleanironsample1.0cm2inarea.Assumethatthesamplereflects96%ofthelightonly3%oftheincidentlightliesinthevioletregionofthespectrumabovethethresholdfrequency.a)Whatintensityisactuallyavailableforthephotoelectriceffect?

I = (0.03)(0.040)I0 = (0.03)(0.040)(1.0µW / cm2 ) =1.2nW / cm2

b)AssumingthatallthephotonsinthevioletregionhaveaneffecAvewavelengthof250nm,howmanyelectronswillbeemi[edpersecond?

N(electrons / sec) = 1.2x10−9W

hf=λ 1.2x10−9( )

hc

N =250x10−9m( ) 1.2x10−9 J / s( )6.6x10−34 J is( ) 3x108m / s( )2

=1.5x109

Supposethatlightoftotalintensity1.0µW/cm2fallsonacleanironsample1.0cm2inarea.Assumethatthesamplereflects96%ofthelightonly3%oftheincidentlightliesinthevioletregionofthespectrumabovethethresholdfrequency.c)Calculatethecurrentinthephototubeinamperes.

d)IftheworkfuncAonis4.5eV,whatisf0?

N =1.5x109

I = N(1.6x10−19C) = 2.4x10−10A

f0 =φh=

4.5eV4.14x10−15eV is

=1.1x1015Hz

e)FindthestoppingvoltageVsifthelightis250nm.

eVs = hf −φ =hcλ−φ =

4.14x10−15eV is( ) 3.0x108m / s( )250x10−9m

− 4.5eV = 0.46V

WilhelmRöntgen(1845‐1923

1901NobelPrize“inrecogniAonofthe

extraordinaryserveshehasrenderedbythediscoveryof

theremarkablerayssubsequentlynamedawer

him.”

CharacterisAcRadiaAonBremsstrahlung“breakingradiaAon”

λmin

Bremsstrahlung

Voltage on tube V0

Maximum energy of photon is eV0

eV0 = Emax = hfmax =hcλmin

= ω

Duane‐HuntRule

CharacterisAcRadiaAon

λmin =1.24x10−6V •m

V0

300 150 100 50 0

15 10 5 0

O 2

p-R

u 4

d

Sr

4p

Sr

4s

Ru

4p

Ru

4s

Sr

3d

5/2

Sr

3d

3/2

Sr

3p

3/2

Sr

3p

1/2

Ru

3d

5/2

Ru

3d

3/2

Sr3Ru

2O

7

Inte

nsi

ty (

a.u

.)

Binding Energy (eV)

!E~0.15eV

Binding Energy (eV)

In

ten

sity

(a.

u.)

Femi Edge

ArthurCompton(1892‐1962)

NobelPrize1927“forhisdiscoveryoftheeffectnameda8erhim,”Comptoneffect

ThisisaverybeauAfulsca[eringexperiment.Anincidentx‐rayorgammarayissca[eredthroughanangleθbyacollisionwithanelectrona[achedtoanucleus.Theelectronissca[eredthroughanangleφ. ΑcalculaAonofconservaAonofmomentumandenergyisneededtodeterminethechangeinthewavelengthofthesca[eredphoton.

The momentum of the photon is

p p= Ec=hλ

The momentum of the electron is

pe =E2 − mc2( )2

c

Conservation of E: E+mec2 = E '+ Ee

Conservation of momentum: p=p'cosθ+pecosφp'sinθ=pe sinφ

Eliminate φpe

2 = p '2+ p2 − 2pp 'cosθ

Ee = hf − hf '+mec2

Alotofalgebra!

λ '− λ0 =hmec

1− cosθ( )

ArthurCompton(1892‐1962)

NobelPrize1927“forhisdiscoveryoftheeffectnamesa8erhim,”Comptoneffect

λ2 − λ1 =hmc

1− cosθ( ) = λC 1− cosθ( )

λC =hmc

=hcmc2 =

1.24x103eV •nm5.11x105eV

= 0.00243 nm

hc =1.24x103eV inm

X‐raysofwavelengthλ=0.200nmareaimedatablockofcarbon.Thesca[eredx‐raysareobservedatanangleof45othetheincidentbeam.Calculatetheincreasedwavelengthofthesca[eredx‐raysatthisangle.

λ '− λ0 =hmec

1− cosθ( ) = hcmec

2 1− cosθ( ) = λC 1−12

Δλ = 0.00243 nm( ) 0.293( ) = 0.00071nmλ ' = Δλ + λ0 = 0.20071nm

(a)Whyarex‐rayphotonsusedintheComptonexperiment,ratherthanvisible‐lightphotons?

Δλ = 0.0243 Å( ) 1( ) = 0.0243 Å

For all incident wavelengths

ToanswerthisletusfindΔλ forthreedifferentincidentphotonwavelengths.(1)veryhighenergyγrayswithλ=0.0106Å,(2)x‐rayswithλ=0.712Å,and(3)greenlightfromHglamp,withλ=5461Å.Note10Å=1nm

Now calculate Δλλ0

γ − ray Δλλ0

=0.0243Å0.0106Å

= 2.29

x − ray Δλλ0

=0.0243Å0.712Å

= 0.0341

green Δλλ0

=0.0243Å5461ÅÅ

= 4.45x10−6

(b)TheelectronsinChavea4eVbindingenergy.WhycanthisbeignoredintheComptonequaAonforx‐rays?Lookatλ=0.712Å.

Energy E=hf= hcλ

E = 12400eV iÅ0.712Å

=17,400 eVResolving power → λ

Δλ

ExperimentalisthadmeasuredverycarefullytheemissionandabsorpAonspectraofH.TheyfoundanempiricalformulatodescribetheirobservaAons.

In 1885 Balmer described the visible spectrum: Balmer Series1λ= R 1

22 −1n2

: n= 3, 4, 5, .....

R=0.0110 nm

In general1λ= R

1n '2

−1n2

Lymanisn’=1

Paschenisn’=3 E =hcλ

E n,n '( ) = hcR 1n '2

−1n2

E = hf =hcλ

ΔE n,n '( ) = hcR 1n '2

−1n2

Rydberg EnergyER = hcR = 13.6eV

Ionization potential n'=1 to n=∞

I = 13.6 11− 0

= 13.6ev

Define

En = −ER

n2

h[p://web.phys.ksu.edu/vqm/sowware/online/vqm/html/h2spec.html

Classicalpicture

Visible

Hspectra,noAcethatmostofthelinesarenotinthevisible!