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C24 quantization-of-light


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  • Light and its nature have caused a lot of ink to flow during these last decades. Its dual behavior is partly explained by (1)Double-slit experiment of Thomas Young - who represents the photon’s motion as a wave - and also by (2)the Photoelectric effect in which the photon is considered as a particle. A Revolution: SALEH THEORY solves this ambiguity and this difficulty presenting a three-dimensional trajectory for the photon's motion and a new formula to calculate its energy. More information on
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C24 quantization-of-light

  1. 1. 1Chapter 29 Waves and Particles:29.1 Planck’s Quantum Theory29.2 The Photoelectric Effect
  2. 2. 229 .1 Planck’s Quantum TheorySUBTOPIC :LEARNING OUTCOMES :a) Distinguish between Planck’s quantumtheory and classical theory of energy.b) Use Einstein’s formulae fora photon energy, .At the end of this lesson, the students shouldbe able to :hcE hf
  3. 3. 324.1 Planck’s Quantum Theory• The foundation of the Planck’s quantum theory is atheory of black body radiation.• Black body is defined as an ideal system or objectthat absorbs and emits all the em radiations that isincident on it.• The electromagnetic radiation emittedby the black body is called black body• In an ideal black body, incident light iscompletely absorbed.• Light that enters the cavity through thesmall hole is reflected multiple timesfrom the interior walls until it iscompletely absorbed.
  4. 4. 4ExperimentalresultRayleigh -Jeans theoryWien’s theoryClassicalphysics• The spectrum of electromagnetic radiation emittedby the black body (experimental result) is shown infigure 1.Figure 1 : Black Body Spectrum
  5. 5. 5• Rayleigh-Jeans and Wien’s theories (classicalphysics) failed to explain the shape of the blackbody spectrum or the spectrum of light emitted byhot objects.• Classical physics predicts a black body radiationcurve that rises without limit as the f increases.• The classical ideas are : Energy of the e.m. radiation does notdepend on its frequency or wavelength. Energy of the e.m. radiation iscontinuously.
  6. 6. 6• In 1900, Max Planck proposed his theory that isfit with the experimental curve in figure 1 at allwavelengths known as Planck’s quantumtheory.• The assumptions made by Planck in his theoryare : The e.m. radiation emitted by the black bodyis a discrete (separate) packets of energyknown as quanta. This means the energy ofe.m. radiation is quantised. The energy size of the radiation dependson its frequency.
  7. 7. 7Comparison between Planck’ quantum theory and classical theory of energy.Planck’s QuantumTheoryClassical theoryEnergy of the e.m radiation isquantised. (discrete)Energy of the e.m radiation iscontinously.Energy of e.m radiationdepends on its frequency orwavelengthEnergy of e.m radiation doesnot depend on its frequencyor wavelength (depends onIntensity)Photon2AIhfETkE BclassicaletemperaturconstantsBoltzmanTkB
  8. 8. 8• According to this assumptions, the quantum E ofthe energy for radiation of frequency f is givenbyhfEwhere constantPlanck:h J s10636 34.Planck’s quantumtheoryfchcE
  9. 9. 9Photons• In 1905, Albert Einstein proposed that light comes inbundle of energy (light is transmitted as tinyparticles), called photons.• Photon is defined as a particle with zero massconsisting of a quantum of electromagneticradiation where its energy is concentrated.Quantum means “fixed amount”
  10. 10. 10• Photons travel at the speed of light in a vacuum.• Photons are required to explain the photoelectriceffect and other phenomena that require light tohave particle property.• In equation form, photon energy (energy of photon)is chhfE• Unit of photon energy is J or eV.• The electronvolt (eV) is a unit of energy that canbe defined as the kinetic energy gained by anelectron in being accelerated by a potentialdifference (voltage) of 1 volt.• Unit conversion : J10601eV1 19.
  11. 11. 11Example 24.1Calculate the energy of a photon of blue light,.nm450(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
  12. 12. 12Example 24.2A photon have an energy of 3.2 eV. Calculatethe frequency, vacuum wavelength and energyin joule of the photon.(7.72 x 1014 Hz ,389 nm, 5.12 x10-19 J)(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
  13. 13. 1329 .2 The Photoelectric EffectSUBTOPIC :LEARNING OUTCOMES :At the end of this lesson, the students shouldbe able to :a) Explain the phenomenon of photoelectric effect.b) Define and determine threshold frequency, workfunction and stopping potential.c) Describe and sketch diagram of the photoelectriceffect experimental set-up.d) Explain the failure of wave theory to justify thephotoelectric effect.
  14. 14. 14LEARNING OUTCOMES :At the end of this lesson, the students should be able to :e) Explain by using graph and equations theobservations of photoelectric effect experiment interms of the dependence of :i ) kinetic energy of photoelectron on the frequencyof light;½ mvmax2 = eVs = hf – hfoii ) photoelectric current on intensity of incident light;iii) work function and threshold frequency on thetypes of metal surface; Wo =hfof) Use Einstein’s photoelectric effect equation,Kmax = eVs = hf – WoThe Photoelectric Effect
  15. 15. 1524 .2 The photoelectric effect• The photoelectric effect is the emission of electronsfrom the metal surface when electromagneticradiation of enough frequency falls/strikes/incidents /shines on it.• A photoelectron is an electron ejected due tophotoelectric effect (an electron emitted fromthe surface of the metal when light strikes its surface).-em radiation(light)photoelectron- - - - - - - - - -Metal surfaceFree electrons-
  16. 16. 16• The photoelectric effect can be measured using adevice like that pictured in figure below.9.2 The photoelectric effectAnode(collector)Cathode (emitteror target metal)photoelectronglass---rheostatpower supplye.m. radiation (incoming light)vacuum AVThe photoelectric effect’s experimentA
  17. 17. 179.2 The photoelectric effect• A negative electrode (cathode or target metal oremitter) and a positive electrode (anode orcollector) are placed inside an evacuated glasstube.• The monochromatic light (UV- incoming light) ofknown frequency is incident on the target metal.• The incoming light ejects photoelectrons from atarget metal.• The photoelectrons are then attracted to thecollector.• The result is a photoelectric current flows inthe circuit that can be measured with an ammeter.
  18. 18. 18• When the positive voltage (potential difference)is increased, more photoelectrons reach thecollector , hence the photoelectric current alsoincreases.• As positive voltage becomes sufficiently large, thephotoelectric current reaches a maximumconstant value Im, called saturation current.9.1 The photoelectric effectSaturation current is defined as the maximumconstant value of photocurrent in which whenall the photoelectrons have reached theanode.
  19. 19. 199.2 The photoelectric effect• When the voltage is made negative by reversingthe power supply terminal as shown in figurebelow, the photoelectric current decreases sincemost photoelectrons are repelled by the collectorwhich is now negative electric potential.Anode(collector)Cathode (emitteror target metal)photoelectronglass---rheostatpower supplye.m. radiation (incoming light)vacuum AVBReversing powersupply terminal(to determine thestopping potential)
  20. 20. 20• If this reverse voltage is small enough, the fastestelectrons will still reach the collector and there willbe the photoelectric current in the circuit.• If the reverse voltage is increased, a point isreached where the photoelectric current reacheszero – no photoelectrons have sufficient kineticenergy to reach the collector.• This reverse voltage is called the stoppingpotential , Vs.Vs is defined as the minimum reverse potential (voltage)needed for electrons from reaching the collector.• By using conservation of energy :(loss of KE of photoelectron = gain in PE) ;K.Emax = eVs 2s mv21eV
  21. 21. 21According to Einstein’s theory, an electron isejected/emitted from the target metal by acollision with a single photon.In this process, all the photon energy istransferred to the electron on the surface of metaltarget.Since electrons are held in the metal by attractiveforces, some minimum energy,Wo (work function,which is on the order of a few electron volts formost metal) is required just enough to get anelectron out through the surface.Einstein’s theory of Photoelectric Effect
  22. 22. 22If the frequency f of the incoming light is so lowthat is hf < Wo , then the photon will not haveenough energy to eject any electron at all.If hf > Wo , then electron will be ejected andenergy will be conserved (the excess energyappears as kinetic energy of the ejected electron).This is summed up by Einstein’s photoelectricequation ,Einstein’s theory of Photoelectric Effect2max021mvWhfmax.EKWE o2s mv21eVseVWhf 0but
  23. 23. 23= photon energy2maxmax21. mvEKEinstein’s theory of Photoelectric Effect= maximum kinetic energy ofejected electron.f = frequency of em radiation /incoming lightvmax = maximum speed of the photoelectronEinstein’sphotoelectricequationchhfE2max021mvWhfmax.EKWE o
  24. 24. 24fo = threshold frequency.= minimum frequency of e.m. radiationrequired to eject an electron from thesurface of the metal.oWo = the work function of a metal.= the minimum energy required (needed) toeject an electron from the surface oftarget metal.Einstein’s theory of Photoelectric Effect= threshold wavelength.= maximum wavelength of e.m. radiationrequired to eject an electron from thesurface of the target metal.oocfmax.EKWE o2max021mvWhfooohchfW
  25. 25. 25-hfvmax-MetalW0-hfv=0-MetalW0hf-MetalW0oWhf2max021mvWhfhf < Wohf > WoElectron isemittedElectron isejected.No electron is ejected.Einstein’s theory of Photoelectric Effect
  26. 26. 26Example 24 .3The work function for a silver surface is Wo = 4.74 eV.Calculate thea) minimum frequency that light must have to ejectelectrons from the surface.b) maximum wavelength that light must have to ejectelectrons from the surface.nmb)Hz1.14x10a)15263oooofhfW(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
  27. 27. 27Example 24.4What is the maximum kinetic energy of electronsejected from calcium by 420 nm violet light, giventhe work function for calcium metal is 2.71 eV?K.Emax = E – Wo(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
  28. 28. 28Example 24.5Sodium has a work function of 2.30 eV. Calculatea. its threshold frequency,b. the maximum speed of the photoelectronsproduced when the sodium is illuminated bylight of wavelength 500 nm,c. the stopping potential with light of thiswavelength.00 hfWa.(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)Solution 24.5
  29. 29. 29Solution 24.5b. 2021mvWhc221mveVsc.(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
  30. 30. 30In an experiment of photoelectric effect, no currentflows through the circuit when the voltage acrossthe anode and cathode is -1.70 V. Calculatea. the work function, andb. the threshold wavelength of the metal (cathode)if it is illuminated by ultraviolet radiation offrequency 1.70 x 1015 Hz.(Given : c = 3.00 x 108 m s-1,h = 6.63 x 10-34 J s , 1 eV=1.60 x 10-19 J,me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)Example 24.6
  31. 31. 31Solution 24.6JWa 190 1055.8)mb 70 1033.2)
  32. 32. 32Example 24.7The energy of a photon from an electromagneticwave is 2.25 eVa. Calculate its wavelength.b. If this electromagnetic wave shines on ametal, photoelectrons are emitted with amaximum kinetic energy of 1.10 eV. Calculate thework function of this metal in joules.(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s ,1 eV=1.60 x 10-19 J, mass of electron m = 9.11 x10-31 kg, e = 1.60 x 10-19 C)
  33. 33. 33Solution 24.7Ans. : 553 nm, 1.84 x 10-19 J
  34. 34. 34Graphs in Photoelectric EffectGenerally, Einstein’s photoelectric equation;max.EKWE ocmxyWhfEKWEEKoomaxmax..f ↑ K.Emax ↑K.EmaxfofoW0
  35. 35. 35Graphs in Photoelectric Effectf,frequencysV,voltageStoppingeW00cmxyeWfehVWhfeVWhfEKososomax.f ↑ Vs ↑of
  36. 36. 36Graphs in Photoelectric EffectVariation of stopping voltage Vs with frequency f ofthe radiation for different metals but the intensityis fixed.f,frequencysV,voltageStopping001fW0102fW02cmxyeWfehVWhfeVWhfEKososomax.00 fWW02 > W01f02 > f01
  37. 37. 37Graphs in Photoelectric EffectIntensity 2xIntensity 1xI,currentricPhotoelectV,Voltage0Variation of photoelectric current I with voltage V forthe radiation of different intensities but itsfrequency and metal are fixed.Vs
  38. 38. 38Notes:Classical physicsLight intensity ,areatimeenergyIQuantum physicsLight intensity ,areatimephotonsofnumberIphotonsofnumberintensityLightLight intensity ↑ ,number of photons ↑ ,number of electrons ↑ ,current ↑ .(If light intensity ↑, photoelectric current ↑).
  39. 39. 39Graphs in Photoelectric EffectVariation of photoelectric current I with voltage Vfor the radiation of different frequencies but itsintensity and metal are fixed.f2mI1sVI,currentricPhotoelectV,Voltage0f1f2 > f12sVeWfehVWhfeVWhfEKososomax.f ↑ Vs ↑Vs2 > Vs1
  40. 40. 40Graphs in Photoelectric EffectVariation of photoelectric current I with voltage Vfor the different metals but the intensity andfrequency of the radiation are fixed.W011sVmII,currentricPhotoelectV,Voltage02sVW02W02 > VW ,Vs1 > Vs2
  41. 41. 41Example 24.8K.Emax (x 10-19 J)f(x 1014 )Hz8.40Use the graph above to find the value ofi) work function andii) the threshold wavelength.
  42. 42. 42Solution 24.8K.Emax (x 10-19 J)f(x 1014 )Hz8.40cmxyWhfEKWEEKoomaxmax..m10x6.25h,wavelengtThreshold10x3.18x104.8whengraphtheFrom7-19-14ooooofcJHzfhfWEK ,0. max
  43. 43. 43OBSERVATIONSof the photoelectric effects experiment1.Electrons are emitted immediately2.Stopping potential does not depend on theintensity of light.3.Threshold frequency of light is different fordifferent target metal.4.Number of electrons emitted of thephotoelectron current depend on the intensity oflight.
  44. 44. 44EXPLAIN the failure of classical theory to justify thephotoelectric effect.Clasiccal prediction ExperimentalResultModern TheoryThe higher theintensity, thegreater the energyimparted to themetal surface foremission ofphotoelectrons.•The higher theintensity of light thegreater the kineticenergy maximumof photoelectrons.Very low intensitybut highfrequencyradiation couldemitphotoelectrons.The maximumkinetic energy ofphotoelectrons isindependent oflight intensity.Based on Einstein’sphotoelectric equation:The maximum kineticenergy of photoelectrondepends only on thelight frequency .The maximum kineticenergy ofphotoelectrons DOESNOT depend on lightintensity.0WhfKmax1. MAXIMUM KINETIC ENERGY OF PHOTOELECTRON
  45. 45. 45ClasiccalpredictionExperimentalResultModern TheoryEmission ofphotoelectrons occur forallfrequenciesof light.Energy oflight isindependentoffrequency.Emission ofphotoelectronsoccur onlywhenfrequency ofthe lightexceeds thecertainfrequencywhich value ischaracteristicof the materialbeingilluminated.When the light frequency isgreater than thresholdfrequency, a higher rate ofphotons striking the metalsurface results in a higherrate of photoelectronsemitted. If it is less thanthreshold frequency nophotoelectrons areemitted. Hence theemission of photoelectronsdepend on the lightfrequency.2. EMISSION OF PHOTOELECTRON ( energy )
  46. 46. 46Clasiccal prediction ExperimentalResultModern TheoryLight energy is spreadover the wavefront, theamount of energyincident on any oneelectron is small. Anelectron must gathersufficient energy beforeemission, hence there istime interval betweenabsorption of lightenergy and emission.Time interval increasesif the light intensity islow.Photoelectronsare emittedfrom thesurface of themetal almostinstantaneouslyafter thesurface isilluminated,even at verylow lightintensities.The transfer ofphoton’s energy toan electron isinstantaneous as itsenergy is absorbedin its entirely, muchlike a particle toparticle collision.The emission ofphotoelectron isimmediate and notime intervalbetween absorptionof light energy andemission.3. EMISSION OF PHOTOELECTRON ( time )
  47. 47. 47ClasiccalpredictionExperimental ResultModern TheoryEnergy of lightdepends onlyon amplitude( or intensity)and not onfrequency.Energy oflightdependsonfrequencyAccording to Planck’squantum theory which isE=hfEnergy of light dependson its frequency.4. ENERGY OF LIGHT
  48. 48. 48• Experimental observations deviate fromclassical predictions based on Maxwell’se.m. theory. Hence the classical physicscannot explain the phenomenon ofphotoelectric effect.• The modern theory based on Einstein’sphoton theory of light can explain thephenomenon of photoelectric effect.• It is because Einstein postulated that light isquantized and light is emitted, transmittedand reabsorbed as photons.
  49. 49. 49Feature Classical physics Quantum physicsThresholdfrequencyAn incident light of anyfrequency can ejectelectrons (does nothas thresholdfrequency), as long asthe beam has sufficientintensity.To eject an electron, theincident light must have afrequency greater than acertain minimum value,(threshold frequency) , nomatter how intense thelight.Maximum kineticenergyof photoelectronsDepends on the lightintensity.Depends only on the lightfrequency .Emission ofphotoelectronsThere should be somedelays to emitelectrons from a metalsurface.Electrons are emittedspontaneously.Energy of light Depends on the lightintensity.Depends only on the lightfrequency .SUMMARY : Comparison between classical physicsand quantum physics about photoelectric effect experiment
  50. 50. 50Exercise1. Find the energy of the photons in a beam whosewavelength is 500 nm. ( 3.98 x 10 -19 J)2. Determine the vacuum wavelength corresponding to a -rayenergy of 1019 eV. (1.24 x10-25 m)3. A sodium surface is illuminated with light of wavelength300 nm. The work function for sodium metal is 2.46 eV.Calculatea) the kinetic energy of the ejected photoelectronsb) the cutoff wavelength for sodiumc) maximum speed of the photoelectrons.(1.68 eV, 505 nm, 7.68 x 105 ms-1)(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
  51. 51. 514. Radiation of wavelength 600 nm is incidents upon thesurface of a metal. Photoelectrons are emitted from thesurface with maximum speed 4.0 x 105 ms-1. Determinethe threshold wavelength of the radiation. (7.7 x 10-7 m)5. Determine the maximum kinetic energy, in eV, ofphotoelectrons emitted from a surface which has a workfunction of 4.65 eV when electromagnetic radiation ofwavelength 200 nm is incident on the surface. (1.57 eV)6. When light of wavelength 540 nm is incident on thecathode of photocell, the stopping potential obtained is0.063 V. When light of wavelength 440 nm is used, thestopping potential becomes 0.865 V. Determine the ratio( 6.35 x 10-15 J s C-1).he
  52. 52. 527. In an experiment on the photoelectric effect, the followingdata were collected.a. Calculate the maximum velocity of the photoelectronswhen the wavelength of the incident radiation is 350 nm.b. Determine the value of the Planck constant from theabove data.Wavelength of e.m.radiation, (nm)Stoppingpotential, Vs (V)350 1.70450 0.900(7.73 x 105 m s-1, 6.72 x 10-34 J s)
  53. 53. 8. In a photoelectric effect experiment it is observed thatno current flows unless the wavelength is less than 570nm. Calculatea. the work function of this material in electronvolts.b. the stopping voltage required if light of wavelength400 nm is used.(2.18 eV, 0.92 V)
  54. 54. 549. In a photoelectric experiments, a graph of the light frequency fis plotted against the maximum kinetic energy Kmax of thephotoelectron as shown in figure below.Based on the graph, for the light frequency of 6.00 x 1014Hz, calculatea. the threshold frequency.b. the maximum kinetic energy of the photoelectron.c. the maximum velocity of the photoelectron.Hz10f 1402.)(max eVK,1083.4 140 Hzf ,1078.7 20max JK 1-5sm10134v .
  55. 55. 5510. A photocell with cathode and anode made of the same metalconnected in a circuit as shown in the figure below.Monochromatic light of wavelength 365 nm shines on thecathode and the photocurrent I is measured for various valuesof voltage V across the cathode and anode. The result isshown in the graph.a. Calculate the maximum kinetic energy of the photoelectron.b. Deduce the work function of the cathode.c. If the experiment is repeated with monochromatic light ofwavelength 313 nm, determine the new intercept with theV-axis for the new graph.365 nmVG51)(nAI)(VV0(1.60 x 10-19 J, 3.85 x 10-19 J, -1.57 V)