physics 2203, fall 2012 modern physics
TRANSCRIPT
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!