Presentation on theme: "UNIT 24 : QUANTIZATION OF LIGHT"— Presentation transcript:
1UNIT 24 : QUANTIZATION OF LIGHT 3 hours24.1 Planck’s Quantum Theory24.2 The Photoelectric Effect
224 .1 Planck’s Quantum Theory ½ hour SUBTOPIC :Planck’s Quantum Theory½ hourLEARNING OUTCOMES :At the end of this lesson, the students shouldbe able to :Distinguish between Planck’s quantumtheory and classical theory of energy.b) Use Einstein’s formulae fora photon energy,
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 bodyradiation.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.black body
4The spectrum of electromagnetic radiation emitted by the black body (experimental result) is shown in figure 1.Experimental resultRayleigh -Jeans theoryWien’s theoryClassical physicsFigure 1 : Black Body Spectrum
5Rayleigh-Jeans and Wien’s theories (classical physics) 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 not depend on its frequency or wavelength.Energy of the e.m. radiation is continuously.
6In 1900, Max Planck proposed his theory that is fit with the experimental curve in figure 1 at all wavelengths known as Planck’s quantum theory.The assumptions made by Planck in his theory are :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.
7Planck’s Quantum Theory Comparison between Planck’ quantum theory and classical theory of energy.Planck’s Quantum TheoryClassical theoryEnergy of the e.m radiation is quantised. (discrete)Energy of the e.m radiation is continously.Energy of e.m radiation depends on its frequency or wavelengthEnergy of e.m radiation does not depend on its frequency or wavelength (depends on Intensity)Photon
8According to this assumptions, the quantum E of the energy for radiation of frequency f is given by wherePlanck’s quantum theory
9Photons In 1905, Albert Einstein proposed that light comes in bundle 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”
10In equation form, photon energy (energy of photon) isUnit 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 :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.
11Example 24.1 Calculate the energy of a photon of blue light, . (Given c = 3.00 x 108 m s-1, h = 6.63 x J s1 eV=1.60 x J, me = 9.11 x kg, e = 1.60 x C)
12Example 24.2A photon have an energy of 3.2 eV. Calculate the frequency, vacuum wavelength and energy in 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 J s1 eV=1.60 x J, me = 9.11 x kg, e = 1.60 x C)
1324 .2 The Photoelectric Effect 2 ½ hours SUBTOPIC :The Photoelectric Effect2 ½ hoursLEARNING OUTCOMES :At the end of this lesson, the students shouldbe able to :Explain the phenomenon of photoelectric effect.Define and determine threshold frequency, work function and stopping potential.Describe and sketch diagram of the photoelectric effect experimental set-up.Explain the failure of wave theory to justify the photoelectric effect.
1424 .2 The Photoelectric Effect 2 ½ hours SUBTOPIC :The Photoelectric Effect2 ½ hoursLEARNING OUTCOMES :At the end of this lesson, the students should be able to :e) Explain by using graph and equations the observations of photoelectric effect experiment in terms 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 – Wo
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)photoelectronMetal surfaceFree electrons
16The photoelectric effect can be measured using a device like that pictured in figure below.Anode(collector)Cathode (emitter or target metal)photoelectronglass-rheostatpower supplye.m. radiation (incoming light)vacuumAVAThe photoelectric effect’s experiment
17A negative electrode (cathode or target metal or 9.2 The photoelectric effectA 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.
18When the positive voltage (potential difference) 9.1 The photoelectric effectWhen 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.Saturation current is defined as the maximum constant value of photocurrent in which when all the photoelectrons have reached the anode.
19If the positive voltage is gradually decreased, the 9.2 The photoelectric effectIf the positive voltage is gradually decreased, thephotoelectric current I also decreases slowly.Even at zero voltage there are still somephotoelectrons with sufficient energy reach thecollector and the photoelectric current flows is Io .Graph of photoelectric current against voltage for photoeclectric effect’s experimentB (After)A (Before reversing the terminal)
20When the voltage is made negative by reversing 9.2 The photoelectric effectWhen 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.Cathode (emitter or target metal)vacuumAVe.m. radiation (incoming light)Anode(collector)-photoelectronglassReversing power supply terminal(to determine the stopping potential)power supplyrheostatB
21If this reverse voltage is small enough, the fastest electrons 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
22Einstein’s theory of Photoelectric Effect According 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.
23If the frequency f of the incoming light is so low Einstein’s theory of Photoelectric EffectIf 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 ,but
24f = frequency of em radiation /incoming light Einstein’s theory of Photoelectric EffectEinstein’s photoelectric equation= photon energyf = frequency of em radiation /incoming light= maximum kinetic energy ofejected electron.vmax = maximum speed of the photoelectron
25Wo = the work function of a metal. Einstein’s theory of Photoelectric EffectWo = the work function of a metal.= the minimum energy required (needed) toeject an electron from the surface oftarget metal.fo= threshold frequency.= minimum frequency of e.m. radiationrequired to eject an electron from thesurface of the metal.= threshold wavelength.= maximum wavelength of e.m. radiationrequired to eject an electron from thesurface of the target metal.
26hf > Wo hf < Wo vmax hf hf - v=0 - W0 W0 Metal hf W0 - Metal Einstein’s theory of Photoelectric Effect-hfv=0MetalW0-hfvmaxMetalW0Electron is ejected.Electron is emittedhf > WohfW0Metal-hf < WoNo electron is ejected.
27Example 24 .3The work function for a silver surface is Wo = 4.74 eV. Calculate theminimum frequency that light must have to eject electrons from the surface.maximum wavelength that light must have to eject electrons from the surface.(Given c = 3.00 x 108 m s-1, h = 6.63 x J s1 eV=1.60 x J, me = 9.11 x kg, e = 1.60 x C)
28Example 24.4What is the maximum kinetic energy of electrons ejected from calcium by 420 nm violet light, given the work function for calcium metal is 2.71 eV?(Given c = 3.00 x 108 m s-1, h = 6.63 x J s1 eV=1.60 x J, me = 9.11 x kg, e = 1.60 x C)K.Emax = E – Wo
29Example 24.5 Solution 24.5 Sodium 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.(Given c = 3.00 x 108 m s-1, h = 6.63 x J s1 eV=1.60 x J, me = 9.11 x kg, e = 1.60 x C)Solution 24.5a.
30Solution 24.5 b. c. (Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s 1 eV=1.60 x J, me = 9.11 x kg, e = 1.60 x C)b.c.
31Example 24.6In an experiment of photoelectric effect, no current flows through the circuit when the voltage across the anode and cathode is V. Calculatea. the work function, andb. the threshold wavelength of the metal (cathode) if it is illuminated by ultraviolet radiation of frequency 1.70 x 1015 Hz.(Given : c = 3.00 x 108 m s-1,h = 6.63 x J s , 1 eV=1.60 x J,me = 9.11 x kg, e = 1.60 x C)
33Example 24.7The energy of a photon from an electromagnetic wave is 2.25 eVa. Calculate its wavelength.b. If this electromagnetic wave shines on a metal, photoelectrons are emitted with a maximum kinetic energy of 1.10 eV. Calculate the work function of this metal in joules.(Given c = 3.00 x 108 m s-1, h = 6.63 x J s ,1 eV=1.60 x J, mass of electron m = 9.11 x kg, e = 1.60 x C)
44of the photoelectric effects experiment OBSERVATIONSof the photoelectric effects experimentElectrons are emitted immediatelyStopping potential does not depend on the intensity of light.Threshold frequency of light is different for different target metal.Number of electrons emitted of the photoelectron current depend on the intensity of light.
45EXPLAIN the failure of classical theory to justify the photoelectric effect.1. MAXIMUM KINETIC ENERGY OF PHOTOELECTRONClasiccal predictionExperimental ResultModern TheoryThe higher the intensity, the greater the energy imparted to the metal surface for emission of photoelectrons.The higher the intensity of light the greater the kinetic energy maximum of photoelectrons.Very low intensity but high frequency radiation could emit photoelectrons. The maximum kinetic energy of photoelectrons is independent of light intensity.Based on Einstein’s photoelectric equation:The maximum kinetic energy of photoelectron depends only on the light frequency .The maximum kinetic energy of photoelectrons DOES NOT depend on light intensity.
46Clasiccal prediction Experimental Result Modern Theory 2. EMISSION OF PHOTOELECTRON ( energy )Clasiccal predictionExperimental ResultModern TheoryEmission of photoelectrons occur for all frequencies of light. Energy of light is independent of frequency.Emission of photoelectrons occur only when frequency of the light exceeds the certain frequency which value is characteristic of the material being illuminated.When the light frequency is greater than threshold frequency, a higher rate of photons striking the metal surface results in a higher rate of photoelectrons emitted. If it is less than threshold frequency no photoelectrons are emitted. Hence the emission of photoelectrons depend on the light frequency.
47Clasiccal prediction Experimental Result Modern Theory 3. EMISSION OF PHOTOELECTRON ( time )Clasiccal predictionExperimental ResultModern TheoryLight energy is spread over the wavefront, the amount of energy incident on any one electron is small. An electron must gather sufficient energy before emission, hence there is time interval between absorption of light energy and emission. Time interval increases if the light intensity is low.Photoelectrons are emitted from the surface of the metal almost instantaneously after the surface is illuminated, even at very low light intensities.The transfer of photon’s energy to an electron is instantaneous as its energy is absorbed in its entirely, much like a particle to particle collision. The emission of photoelectron is immediate and no time interval between absorption of light energy and emission.
48Energy of light depends only on amplitude Clasiccal predictionExperimental ResultModern TheoryEnergy of light depends only on amplitude( or intensity) and not on frequency.Energy of light depends on frequencyAccording to Planck’s quantum theory which isE=hfEnergy of light depends on its frequency.
49Experimental observations deviate from classical predictions based on Maxwell’s e.m. theory. Hence the classical physics cannot explain the phenomenon of photoelectric effect.The modern theory based on Einstein’s photon theory of light can explain the phenomenon of photoelectric effect.It is because Einstein postulated that light is quantized and light is emitted, transmitted and reabsorbed as photons.
50Electrons are emitted SUMMARY : Comparison between classical physics and quantum physics about photoelectric effect experimentFeatureClassical physicsQuantum physicsThreshold frequencyAn incident light of any frequency can eject electrons (does not has threshold frequency), as long as the beam has sufficient intensity.To eject an electron, the incident light must have a frequency greater than a certain minimum value, (threshold frequency) , no matter how intense the light.Maximum kinetic energyof photoelectronsDepends on the light intensity.Depends only on the light frequency .Emission of photoelectronsThere should be some delays to emit electrons from a metal surface.Electrons are emittedspontaneously.Energy of light
51Exercise (Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s 1 eV=1.60 x J, me = 9.11 x kg, e = 1.60 x C)1. Find the energy of the photons in a beam whosewavelength is 500 nm. ( 3.98 x 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)
524. Radiation of wavelength 600 nm is incidents upon the surface of a metal. Photoelectrons are emitted from the surface with maximum speed 4.0 x 105 ms-1. Determine the threshold wavelength of the radiation. (7.7 x 10-7 m)Determine the maximum kinetic energy, in eV, of photoelectrons emitted from a surface which has a work function of 4.65 eV when electromagnetic radiation of wavelength 200 nm is incident on the surface. (1.57 eV)6. When light of wavelength 540 nm is incident on the cathode of photocell, the stopping potential obtained is V. When light of wavelength 440 nm is used, the stopping potential becomes V. Determine the ratio( 6.35 x J s C-1)
53In an experiment on the photoelectric effect, the following data 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)Stopping potential, Vs (V)3501.704500.900(7.73 x 105 m s-1, 6.72 x J s)
548. In a photoelectric effect experiment it is observed that no current flows unless the wavelength is less than 570 nm. Calculatea. the work function of this material in electronvolts.b. the stopping voltage required if light of wavelength 400 nm is used.(2.18 eV, 0.92 V)
559. In a photoelectric experiments, a graph of the light frequency f is plotted against the maximum kinetic energy Kmax of the photoelectron as shown in figure below.Based on the graph, for the light frequency of 6.00 x 1014 Hz, calculatea. the threshold frequency.b. the maximum kinetic energy of the photoelectron.c. the maximum velocity of the photoelectron.
56a. Calculate the maximum kinetic energy of the photoelectron. 10. A photocell with cathode and anode made of the same metal connected in a circuit as shown in the figure below. Monochromatic light of wavelength 365 nm shines on the cathode and the photocurrent I is measured for various values of voltage V across the cathode and anode. The result is shown 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 nmVG(1.60 x J, 3.85 x J, V)