UNIT 24 : QUANTIZATION OF LIGHT (3 hours)

1. UNIT 24 : QUANTIZATION OF LIGHT (3 hours) 24.1 Planck’s Quantum Theory 24.2 The Photoelectric Effect

2. SUBTOPIC : 24 .1 Planck’s Quantum Theory ½ hour LEARNING OUTCOMES : At the end of this lesson, the students should be able to : • Distinguish between Planck’s quantum • theory and classical theory of energy. • b) Use Einstein’s formulae for • a photon energy, .

3. Einstein memorial in Washington, D.C Fig 39-CO, p.1244

4. Albert Einstein German-American Physicist (1879–1955) p.1251

5. Max Planck German Physicist (1858–1947) p.1288

6. 24.1 Planck’s Quantum Theory • The foundation of the Planck’s quantum theory is a • theory of black body radiation. • Black body is defined as an ideal system or object • that absorbs and emits all the em radiations that is • incident on it. • The electromagnetic radiation emitted • by the black body is called black body • radiation. • In an ideal black body, incident light is • completely absorbed. • Light that enters the cavity through the • small hole is reflected multiple times • from the interior walls until it is • completely absorbed. black body

7. Experimental result Rayleigh -Jeans theory Classical physics Wien’s theory • The spectrum of electromagnetic radiation emitted by the black body (experimental result) is shown in figure 1. Figure 1 : Black Body Spectrum

8. Rayleigh-Jeans and Wien’s theories (classical • physics) failed to explain the shape of the black • body spectrum or the spectrum of light emitted by • hot objects. • Classical physics predicts a black body radiation • curve that rises without limit as the f increases. • The classical ideas are : • Energy of the e.m. radiation is related to the amplitude and doesnot depend on its frequency or wavelength. • An object at any temperature emits radiation (energy) • Energy of the e.m. radiation is continuously.

9. In 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 emittedor absorb by the black body not in a continuous stream of waves but in discrete little bundles (separate) packets of energy or quanta, known as photon. • This means the energy of e.m. radiation is quantised. Not all values of energy are possible

10. The energy size of the radiation depends • on its frequency. • The energy is proportional to the frequency (E f ) of the radiation. The greater the frequency , the greater the quantum of energy (f ↑, E↑). • Each discrete energy value represents a different quantum state for the molecule. • When molecule is in n = 1 quantum state, its energy is hf. When molecule is in n = 2 quantum state, its energy is 2hf and so on.

11. Photon Comparison between Planck’ quantum theory and classical theory of energy.

12. According to this assumptions, the quantum E of the energy for radiation of frequency f is given by where Planck’s quantum theory

13. Photons • In 1905, Albert Einstein proposed that light comes in • bundle of energy (light is transmitted as tiny • particles), called photons. • Photon is defined as a particle with zero mass • consisting of a quantum of electromagnetic • radiation where its energy is concentrated. Quantum means “fixed amount”

14. In equation form, photon energy (energy of photon) • is • Unit of photon energy is J or eV. • The electronvolt (eV) is a unit of energy that can • be defined as the kinetic energy gained by an • electron in being accelerated by a potential • difference (voltage) of 1 volt. • Unit conversion : • Photons travel at the speed of light in a vacuum. • Photons are required to explain the photoelectric • effect and other phenomena that require light to • have particle property.

15. Higher frequency photons have more energy. A photon of blue light is more energetic than a photon of red light. A photon travels at the speed of light (c) in vacuum

17. Example 24.1 Calculate the energy of a photon of blue light, . (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)

18. Example 24.2 (a) What is the energy of a photon of red light of wavelength 650 nm ? (b) What is the wavelength of a photon of energy 2.40 eV ? Solution

19. (b) Converting energy in eV to Joule (J)

20. Exercise A 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 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)

21. SUBTOPIC : 24 .2 The Photoelectric Effect 2 ½ hours LEARNING OUTCOMES : At the end of this lesson, the students should be 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.

22. SUBTOPIC : 24 .2 The Photoelectric Effect 2 ½ hours LEARNING 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 frequency of light; ½ mvmax2 = eVs = hf – hfo ii ) photoelectric current on intensity of incident light; iii) work function and threshold frequency on the types of metal surface; Wo =hfo f) Use Einstein’s photoelectric effect equation, Kmax= eVs = hf – Wo

23. - - - - - - - - - - - photoelectron em radiation (light) - Metal surface Free electrons 24 .2 The photoelectric effect • The photoelectric effect is the emission of electrons • from the metal surface when electromagnetic • radiation of enough frequency falls/strikes/ • incidents /shines on it. • A photoelectron is an electron ejected due to • photoelectric effect (an electron emitted from • the surface of the metal when light strikes its surface).

24. e.m. radiation (incoming light) Cathode (emitter or target metal) glass vacuum Anode(collector) photoelectron - - - V A power supply rheostat 9.2 The photoelectric effect • The photoelectric effect can be measured using a • device like that pictured in figure below. A The photoelectric effect’s experiment

25. 9.2 The photoelectric effect • A negative electrode (cathode or target metal or • emitter) and a positive electrode (anode or • collector) are placed inside an evacuated glass • tube. • The monochromatic light (UV- incoming light) of • known frequency is incident on the target metal. • The incoming light ejects photoelectrons from a • target metal. • The photoelectrons are then attracted to the • collector. • The result is a photoelectric current flows in • the circuit that can be measured with an ammeter.

26. 9.1 The photoelectric effect • When the positive voltage (potential difference) • is increased, more photoelectrons reach the • collector , hence the photoelectric current also • increases. • As positive voltage becomes sufficiently large, the • photoelectric current reaches a maximum • constant value Im, called saturation current. Saturation current is defined asthe maximum constant value of photocurrent in which when all the photoelectrons have reached the anode.

27. 9.2 The photoelectric effect • If the positive voltage is gradually decreased, the • photoelectric current I also decreases slowly. • Even at zero voltage there are still some • photoelectrons with sufficient energy reach the • collector and the photoelectric current flows is Io . Graph of photoelectric current against voltage for photoeclectric effect’s experiment B (After) A (Before reversing the terminal)

28. glass vacuum photoelectron - - - V A 9.2 The photoelectric effect • When the voltage is made negative by reversing • the power supply terminal as shown in figure • below, the photoelectric current decreases since • most photoelectrons are repelled by the collector • which is now negative electric potential. Cathode (emitter or target metal) e.m. radiation (incoming light) Anode(collector) Reversing power supply terminal (to determine the stopping potential) power supply rheostat B

29. If this reverse voltage is small enough, the fastest • electrons will still reach the collector and there will • be the photoelectric current in the circuit. • If the reverse voltage is increased, a point is • reached where the photoelectric current reaches • zero – no photoelectrons have sufficient kinetic • energy to reach the collector. • This reverse voltage is called the stopping • potential , 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

30. Einstein’s theory of Photoelectric Effect • According to Einstein’s theory, an electron is ejected/emitted from the target metal by a collision with a single photon. • In this process, all the photon energy is transferred to the electron on the surface of metal target. • Since electrons are held in the metal by attractive forces, some minimum energy,Wo(work function, which is on the order of a few electron volts for most metal) is required just enough to get an electron out through the surface.

31. Einstein’s theory of Photoelectric Effect • If the frequency f of the incoming light is so low that is hf < Wo , then the photon will not have enough energy to eject any electron at all. • If hf > Wo , then electron will be ejected and energy will be conserved (the excess energy appears as kinetic energy of the ejected electron). • This is summed up by Einstein’s photoelectric equation , but

32. Einstein’s theory of Photoelectric Effect Einstein’s photoelectric equation = photon energy f = frequency of em radiation /incoming light = maximum kinetic energy of ejected electron. vmax = maximum speed of the photoelectron

33. vmax hf hf v=0 - - - - - W0 W0 Metal Metal Einstein’s theory of Photoelectric Effect hf > Wo Electron is emitted Electron is ejected. hf W0 hf < Wo Metal No electron is ejected.

34. Terms in photoelectric effect Work Function, is the minimum energy required (needed) to eject an electron from the surface of target metal where = Planck’s Constant = Threshold frequency

35. W0 depends on the type metal used

36. Terms in photoelectric effect Threshold Frequency, Is the minimum frequency of e.m required to eject an electrons from surface of the target metal If the frequency of the incident radiation is less than the threshold frequency (f < f0 ) then electrons would not be removed from the metal surface. Depends on the type of metal used threshold wavelength. maximum wavelength of e.m. radiation required to eject an electron from the surface of the target metal.

37. Terms in photoelectric effect Maximum Kinetic Energy, Kmax where; 1eV = 1.6 x 10-19 J Kmax of the photoelectrons depends on the frequency of the incident radiation Kmax of the photoelectrons increases when the frequency of the incident radiation increases

38. Example 24 .3 • The work function for a silver surface is Wo = 4.74 eV. Calculate the • minimum 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 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)

39. Example 24.4 What 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 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)

40. Example 24.5 Sodium has a work function of 2.30 eV. Calculate a. its threshold frequency, b. the maximum speed of the photoelectrons produced when the sodium is illuminated by light of wavelength 500 nm, c. the stopping potential with light of this wavelength. (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) Solution 24.5

41. Solution 24.5 (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) b. c.

42. Example 24.6 In an experiment of photoelectric effect, no current flows through the circuit when the voltage across the anode and cathode is -1.70 V. Calculate a. the work function, and b. 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 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)

43. Solution 24.6

44. Example 24.7 The energy of a photon from an electromagnetic wave is 2.25 eV a. 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 10-34 J s , 1 eV=1.60 x 10-19 J, mass of electron m = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)

45. Solution 24.7 Ans. : 553 nm, 1.84 x 10-19 J

46. Graphs in Photoelectric Effect • Generally, Einstein’s photoelectric equation; K.Emax f↑ K.Emax↑ f 0

47. Graphs in Photoelectric Effect f↑ Vs↑

48. W01 W02 Graphs in Photoelectric Effect Variation of stopping voltage Vswith frequencyfof the radiation for different metals but the intensity isfixed. W02>W01 f02>f01 • Kmax of photoelectron varies linearly with light frequency. • Different threshold frequency for different metal

49. Intensity 2x Intensity 1x Graphs in Photoelectric Effect Variation of photoelectric current I with voltage V for the radiation of different intensities but its frequency and metal are fixed. Vs When intensity is increased the maximum current attained is higher showing that more e- are emitted. Vs remains the same shows that the Kmax of photoelectron independent of intensity of light

50. Notes: Classical physics Light intensity , Quantum physics Light intensity , Light intensity ↑ , number of photons ↑ , number of electrons ↑ , current ↑ . (If light intensity ↑, photoelectric current ↑).