Quantum Physicsphysics AAA Powerpointpowerpointpowerpoint Presentationpresentationpresentation Bybyby Paulpaulpaul E.E.E

ChapterChapter 38B38B -- QuantumQuantum PhysicsPhysics AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, ProfessorProfessorProfessor ofofof PhysicsPhysicsPhysics SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity

•• DiscussDiscuss thethe meaningmeaning ofof quantumquantum physicsphysics andand PlanckPlanck’’ss constantconstant forfor thethe descriptiondescription ofof mattermatter inin termsterms ofof waveswaves oror particles.particles.

•• DemonstrateDemonstrate youryour understandingunderstanding ofof thethe photoelectricphotoelectric effecteffect,, thethe stoppingstopping potentialpotential,, andand thethe deBrogliedeBroglie wavelengthwavelength..

•• ExplainExplain andand solvesolve problemsproblems similarsimilar toto thosethose presentedpresented inin thisthis unit.unit. PlankPlank’’ss ConstantConstant InIn hishis studiesstudies ofof blackblack--bodybody radiation,radiation, MaxwellMaxwell PlanckPlanck discovereddiscovered thatthat electromagneticelectromagnetic energyenergy isis emittedemitted oror absorbedabsorbed inin discretediscrete quantities.quantities.

Planck’s E = hf (h = 6.626 x 10-34 J s) Equation:

Apparently,Apparently,Apparently, lightlightlight consistsconsistsconsists ofofof Photon tinytinytiny bundlesbundlesbundles ofofof energyenergyenergy calledcalledcalled photonsphotonsphotons,,, eacheacheach havinghavinghaving aaa wellwellwell--- defineddefineddefined quantumquantumquantum ofofof energy.energy.energy. E = hf EnergyEnergy inin ElectronElectron--voltsvolts PhotonPhoton energiesenergies areare soso smallsmall thatthat thethe energyenergy isis betterbetter expressedexpressed inin termsterms ofof thethe electronelectron--voltvolt..

OneOneOne electron-voltelectron-volt (eV)(eV) isisis thethethe energyenergyenergy ofofof ananan electronelectronelectron whenwhenwhen acceleratedacceleratedaccelerated throughthroughthrough aaa potentialpotentialpotential differencedifferencedifference ofofof oneoneone volt.volt.volt.

1 eV = 1.60 x 10-19 J 1 keV = 1.6 x 10-16 J 1 MeV = 1.6 x 10-13 J ExampleExample 1:1: WhatWhat isis thethe energyenergy ofof aa photonphoton ofof yellowyellow--greengreen lightlight (( == 555555 nmnm)?)? FirstFirst wewe findfind ff fromfrom wavewave equation:equation: c = f chc fEhf;   (6.626 x 1034 J s)(3 x 108 m/s) E  555 x 10-9 m

EE== 3.583.58 xx 1010-19-19 JJ OrOr EE== 2.242.24 eVeV

SinceSince 11 eVeV == 1.601.60 xx 1010-19 JJ UsefulUseful EnergyEnergy ConversionConversion SinceSince lightlight isis oftenoften describeddescribed byby itsits wavelengthwavelength inin nanometersnanometers (nm)(nm) andand itsits energyenergy EE isis givengiven inin eVeV,, aa conversionconversion formulaformula isis useful.useful. (1(1 nmnm == 11 xx 1010-9 m)m) hc E(in Joules) ; 1 eV 1.60 x 10-19 J  hc(1 x 109 nm/m) E(in eV)  (1.6 x 10-19 J/eV) IfIf  isis inin nmnm,, thethe energyenergy inin eVeV isis foundfound from:from: 1240 Verify the answer E  Verify the answer  inin ExampleExample 11 ...... TheThe PhotoPhoto--ElectricElectric EffectEffect

Incident light WhenWhen lightlight shinesshines onon Cathode Anode thethe cathodecathode CC ofof aa C A photocell,photocell, electronselectrons areare ejectedejected fromfrom AA andand Ammeter -- ++ attractedattracted byby thethe positivepositive A potentialpotential duedue toto battery.battery.

ThereThereThere isisis aaa certaincertaincertain thresholdthresholdthreshold energy,energy,energy, calledcalledcalled thethethe workworkwork functionfunctionfunction W,WW,, thatthatthat mustmustmust bebebe overcomeovercomeovercome beforebeforebefore anyanyany electronselectronselectrons cancancan bebebe emitted.emitted.emitted. PhotoPhoto--ElectricElectric EquationEquation Incident light hc 1 2 Cathode Anode EWmv  2 C A

Threshold hc Ammeter W  -- ++ wavelength  A  0

TheTheThe conservationconservationconservation ofofof energyenergyenergy demandsdemandsdemands thatthatthat thethethe energyenergyenergy ofofof thethethe incomingincomingincoming lightlightlight hc/hc/hc/bebebe equalequalequal tototo thethethe workworkwork functionfunctionfunction WWW ofofof thethethe surfacesurfacesurface plusplusplus thethethe kinetickinetickinetic energyenergyenergy ½½mvmv22 ofofof thethethe emittedemittedemitted electrons.electrons.electrons. ExampleExample 2:2: TheThe thresholdthreshold wavelengthwavelength ofof lightlight forfor aa givengiven surfacesurface isis 600600 nmnm.. WhatWhat isis thethe kinetickinetic energyenergy ofof emittedemitted electronselectrons ifif lightlight ofof wavelengthwavelength 450450 nmnm shinesshines onon thethe metal?metal? hc  = 600 nm WK  hc hc K 0 A

hc hc 1240 1240 K   ;; KK == 2.762.76 eVeV –– 2.072.07 eVeV 0 450 nm 600 nm

KK == 0.6900.690 eVeV OrOr KK == 1.101.10 xx 1010-19-19 JJ StoppingStopping PotentialPotential AA potentiometerpotentiometer isis usedused Incident light toto varyvary toto thethe voltagevoltage VV Cathode Anode betweenbetween thethe electrodes.electrodes.

TheThe stoppingstopping potentialpotential V A isis thatthat voltagevoltage VVo thatthat justjust stopsstops thethe emissionemission ++ -- ofof electrons,electrons, andand thusthus Potentiometer equalsequals theirtheir originaloriginal K.E.K.E. KKmax == eVeVo PhotoelectricPhotoelectric equation:equation: hW Ehf WeV Vf0  0 ee SlopeSlope ofof aa StraightStraight LineLine (Review)(Review) TheThe generalgeneral equationequation forfor The slope of a line: aa straightstraight lineline is:is: y y = mx + b y = mx + b Slope y x TheThe xx--interceptintercept xxo occursoccurs whenwhen lineline crossescrosses xx axisaxis xo x oror whenwhen yy == 00..

The slope of the line is y The slope of the line is Slope  m thethe riserise overover thethe run:run: x FindingFinding PlanckPlanck’’ss Constant,Constant, hh UsingUsing thethe apparatusapparatus onon thethe previousprevious slide,slide, wewe determinedetermine thethe stoppingstopping potentialpotential forfor aa numbernumber ofof incidentincident lightlight frequencies,frequencies, thenthen plotplot aa graph.graph.

hW Finding h constant Vf0  ee Stopping V potential h Slope  Slope e y x fo NoteNote thatthat thethe xx--interceptintercept ffo isis thethe thresholdthreshold frequency.frequency. Frequency ExampleExample 3:3: InIn anan experimentexperiment toto determinedetermine PlanckPlanck’’ss constant,constant, aa plotplot ofof stoppingstopping potentialpotential versusversus frequencyfrequency isis made.made. TheThe slopeslope ofof thethe curvecurve isis 4.134.13 xx 1010-15 V/HzV/Hz.. WhatWhat isis PlanckPlanck’’ss constant?constant? Stopping V potential hW Vf0  ee Slope y f x h o Slope 4.13 x 10-15 V/Hz Frequency e hh == ee(slope)(slope) == (1.6(1.6 xx 1010-19C)(4.13C)(4.13 xx 1010-15 V/Hz)V/Hz)

ExperimentalExperimental Planck’sPlanck’s hh == 6.616.61 xx 1010-34-34 J/Hz J/Hz ExampleExample 4:4: TheThe thresholdthreshold frequencyfrequency forfor aa givengiven surfacesurface isis 1.091.09 xx 101015 HzHz.. WhatWhat isis thethe stoppingstopping potentialpotential forfor incidentincident lightlight whosewhose photonphoton energyenergy isis 8.488.48 xx 1010-19 JJ?? Incident light PhotoelectricPhotoelectric Equation:Equation: Cathode Anode

Ehf WeV0 V A

eV00 E W; W  hf ++ -- WW == (6.63(6.63 xx 1010-34 Js)(1.09Js)(1.09 xx 101015 Hz)Hz) =7.20=7.20 xx 1010-19 JJ -19 -19 -19 eV0 8.48 x 10 J 7.20 x 10 J 1.28 x 10 J -19 1.28 x 10 J Stopping V  V = 0.800 V 0 1.6 x 10-19 J potential: o TotalTotal RelativisticRelativistic EnergyEnergy RecallRecall thatthat thethe formulaformula forfor thethe relativisticrelativistic totaltotal energyenergy waswas givengiven by:by:

222 Total Energy, E Emcpc()0

For a particle with zero For a particle with zero E = m c2 momentummomentum pp == 00:: o

AA lightlight photonphoton hashas mmo == 0,0, butbut itit doesdoes E = pc havehave momentummomentum pp:: WavesWaves andand ParticlesParticles WeWe knowknow thatthat lightlight behavesbehaves asas bothboth aa wavewave andand aa particle.particle. TheThe restrest massmass ofof aa photonphoton isis zero,zero, andand itsits wavelengthwavelength cancan bebe foundfound fromfrom momentum.momentum. hc h Epc Wavelength    of a photon: p

AllAll objectsobjects,, notnot justjust EMEM waves,waves, havehave wavelengthswavelengths whichwhich cancan bebe foundfound fromfrom theirtheir momentummomentum h de Broglie   Wavelength: mv FindingFinding MomentumMomentum fromfrom K.E.K.E. InIn workingworking withwith particlesparticles ofof momentummomentum pp == mvmv,, itit isis oftenoften necessarynecessary toto findfind thethe momentummomentum fromfrom thethe givengiven kinetickinetic energyenergy K.K. RecallRecall thethe formulas:formulas: KK == ½mv2 ; p = mv

MultiplyMultiply firstfirst mKmK == ½m2v2 = ½p2 EquationEquation byby mm::

Momentum from K: p  2mK ExampleExample 5:5: WhatWhat isis thethe dede BroglieBroglie wavelengthwavelength -31 ofof aa 9090--eVeV electron?electron? ((mme == 9.19.1 xx 1010 kg.)kg.) -19 1.6 x 10 J -17 - K 90 eV 1.44 x 10 J 1 eV ee- 9090 eVeV Next,Next, wewe findfind momentummomentum fromfrom thethe kinetickinetic energy:energy: p  2mK p  2(9.1 x 10-31 kg)(1.44 x 10-17 J) hh   pp == 5.125.12 xx 1010-24 kgkg m/sm/s p mv h 6.23 x 10-34 J   == 0.1220.122 nmnm p 5.12 x 10-24 kg m/s

SummarySummary

Planck’s E = hf (h = 6.626 x 10-34 J s) Equation:

The Electron-volt: 1 eV = 1.60 x 10-19 J 1 keV = 1.6 x 10-16 J 1 MeV = 1.6 x 10-13 J SummarySummary (Cont.)(Cont.)

Incident light hc 1 2 Cathode Anode EWmv  2 C A

Threshold hc Ammeter W  -- ++ wavelength  A  0

IfIf  isis inin nmnm,, thethe energyenergy inin eVeV isis foundfound from:from: WavelengthWavelength inin nm;nm; 1240 E  EnergyEnergy inin eVeV  SummarySummary (Cont.)(Cont.) Stopping PlanckPlanck’’ss Experiment:Experiment: V potential

Incident light Slope y Cathode Anode x fo Frequency

V A hW Vf0  ++ -- ee Potentiometer h KKmax == eVeVo Slope  e SummarySummary (Cont.)(Cont.)

QuantumQuantumQuantum physicsphysicsphysics worksworksworks forforfor waveswaveswaves ororor particles:particles:particles:

For a particle with zero For a particle with zero E = m c2 momentummomentum pp == 0:0: o

AA lightlight photonphoton hashas mmo == 00,, butbut itit doesdoes E = pc havehave momentummomentum pp::

h h Wavelength   de Broglie   of a photon: p Wavelength: mv CONCLUSION:CONCLUSION: ChapterChapter 38B38B QuantumQuantum PhysicsPhysics