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### Phy 310 chapter 3

1. 1. CHAPTER 3: PHOTOELECTRIC EFFECT Dr Ahmad Taufek Abdul Rahman School of Physics & Material Studies Faculty of Applied Sciences Universiti Teknologi MARA Malaysia Campus of Negeri Sembilan 72000 Kuala Pilah, NS 1 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
2. 2. is a phenomenon where under certain circumstances a particle exhibits wave properties and under other conditions a wave exhibits properties of a particle. Wave properties of particle 2 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
3. 3. Learning Outcome: 26.1 de Broglie wavelength (1 hour) At the end of this chapter, students should be able to:  State and use formulae for wave-particle duality of de Broglie, h  p 3 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
4. 4. 3.1 de Broglie wavelength  From the Planck’s quantum theory, the energy of a photon is given by E hc (10.1)   From the Einstein’s special theory of relativity, the energy of a photon is given by 2 E  mcand mc  p (10.2) E  pc  By equating eqs. (10.1) and (10.2), hence hc particle aspect  pc  h p  where 4 PHY310 – Photoelectric Effect (10.3) wave aspect p : momentum DR.ATAR @ UiTM.NS
5. 5.  From the eq. (10.3), thus light has momentum and exhibits particle property. This also show light is dualistic in nature, behaving is some situations like wave and in others like particle (photon) and this phenomenon is called wave particle duality of light.  Table 10.1 shows the experiment evidences to show wave particle duality of light. Wave Particle Young’s double slit experiment Photoelectric effect Diffraction experiment Compton effect Table 1  Based on the wave particle duality of light, Louis de Broglie suggested that matter such as electron and proton might also have a dual nature. 5 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
6. 6.  He proposed that for any particle of momentum p should have a wavelength  given by h h (10.4)   p mv where  : de Broglie wavelength h : Planck's constant m : mass of a particle v : velocityof a particle Eq. (10.4) is known as de Broglie relation (principle).  This wave properties of matter is called de Broglie waves or matter waves.  The de Broglie relation was confirmed in 1927 when Davisson and Germer succeeded in diffracting electron which shows that electrons have wave properties. 6 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
7. 7. Example 1 : In a photoelectric effect experiment, a light source of wavelength 550 nm is incident on a sodium surface. Determine the momentum and the energy of a photon used. (Given the speed of light in the vacuum, c =3.00108 m s1 and Planck’s constant, h =6.631034 J s) 7
8. 8. Example 1 : In a photoelectric effect experiment, a light source of wavelength 550 nm is incident on a sodium surface. Determine the momentum and the energy of a photon used. (Given the speed of light in the vacuum, c =3.00108 m s1 and Planck’s constant, h =6.631034 J s) Solution :   550 10 9 m By using the de Broglie relation, thus h  p 6.63  10 34 550  10 9  p p  1.21 10 27 kg m s 1 and the energy of the photon is given by E hc  6.63 10 3.00 10  E 34 550  10 9 E  3.62  10 19 J 8 8
9. 9. Example 2 : Calculate the de Broglie wavelength for a. a jogger of mass 77 kg runs with at speed of 4.1 m s1. b. an electron of mass 9.111031 kg moving at 3.25105 m s1. (Given the Planck’s constant, h =6.631034 J s) 9
10. 10. Example 2 : Calculate the de Broglie wavelength for a. a jogger of mass 77 kg runs with at speed of 4.1 m s1. b. an electron of mass 9.111031 kg moving at 3.25105 m s1. (Given the Planck’s constant, h =6.631034 J s) Solution : a. Given m  77 kg; v  4.1 m s 1 The de Broglie wavelength for the jogger is 6.63  10 34  77 4.1 36   2.110 m b. Given m  9.11 10 31 kg; v  3.25  10 5 m s 1 h  mv The de Broglie wavelength for the electron is 6.63  10 34  9.11  10 31 3.25  10 5      2.24 10 9 m 10
11. 11. Example 3 : An electron and a proton have the same speed. a. Which has the longer de Broglie wavelength? Explain. b. Calculate the ratio of e/ p. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, mp=1.671027 kg and e=1.601019 C) 11
12. 12. Example 3 : An electron and a proton have the same speed. a. Which has the longer de Broglie wavelength? Explain. b. Calculate the ratio of e/ p. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, mp=1.671027 kg and e=1.601019 C) Solution : a. From de Broglie relation, ve  vp  v h  mv the de Broglie wavelength is inversely proportional to the mass of the particle. Since the electron lighter than the mass of the proton therefore the electron has the longer de Broglie wavelength. 12
13. 13. Solution : ve  vp  v Therefore the ratio of their de Broglie wavelengths is  h   m v e  e    p  h     mp v    mp  me 1.67  10 27  9.11  10 31 e  1833 p 13
14. 14. Learning Outcome: 26.2 Electron diffraction (1 hour) At the end of this chapter, students should be able to:  Describe Davisson-Germer experiment by using a schematic diagram to show electron diffraction.  Explain the wave behaviour of electron in an electron microscope and its advantages compared to optical microscope. 14 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
15. 15. 3.2 Electron diffraction Davisson-Germer experiment  Figure 10.1 shows a tube for demonstrating electron diffraction by Davisson and Germer. graphite film anode screen diffraction pattern e +4000 V cathode electron diffraction Figure 10.1: electron diffraction tube  A beam of accelerated electrons strikes on a layer of graphite which is extremely thin and a diffraction pattern consisting of rings is seen on the tube face. 15 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
16. 16.  This experiment proves that the de Broglie relation was right and the wavelength of the electron is given by h  mv where (10.5) m : mass of an electron v : velocity of an electron  If the velocity of electrons is increased, the rings are seen to become narrower showing that the wavelength of electrons decreases with increasing velocity as predicted by de broglie (eq. 10.5).  The velocity of electrons are controlled by the applied voltage V across anode and cathode i.e. U K 16 PHY310 – Photoelectric Effect 1 2 eV  mv 2 2eV v m (10.6) DR.ATAR @ UiTM.NS
17. 17.  By substituting the eq. (10.6) into eq. (10.5), thus  Note:    17 h  2eV m  m  h  2meV     (10.7) Electrons are not the only particles which behave as waves. The diffraction effects are less noticeable with more massive particles because their momenta are generally much higher and so the wavelength is correspondingly shorter. Diffraction of the particles are observed when the wavelength is of the same order as the spacing between plane of the atom. PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
18. 18. Example 4 : a. An electron is accelerated from rest through a potential difference of 2000 V. Determine its de Broglie wavelength. b. An electron and a photon has the same wavelength of 0.21 nm. Calculate the momentum and energy (in eV) of the electron and the photon. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and e=1.601019 C) 18
19. 19. Example 4 : a. An electron is accelerated from rest through a potential difference of 2000 V. Determine its de Broglie wavelength. b. An electron and a photon has the same wavelength of 0.21 nm. Calculate the momentum and energy (in eV) of the electron and the photon. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and e=1.601019 C) Solution : a. Given V  2000 V The de Broglie wavelength for the electron is  h 2meV   6.63  10 34   2 9.11  10 31 1.60  10 19 2000   2.75 10 11 m 19
20. 20. Solution : b. Given e  p  0.21  10 9 m For an electron, h Its momentum is p  and its energy is 6.63  10 34 p 0.21  10 9 e p  3.16 10 24 kg m s 1 p 1 2 K  me vand v  me 2 2 p  2me  24 2 3.16  10  2 9.11  10 31 5.48  10 18  1.60  10 19 20 K  34.3 eV    
21. 21. Solution : b. Given e  p  0.21  10 9 m For a photon, Its momentum is p  3.16  10 24 and its energy is hc kg m s 1 E p 6.63  10 34 3.00  108  0.21  10 9 9.47  10 16  1.60  10 19 E  5919 eV   21 
22. 22. Example 5 : Compare the de Broglie wavelength of an electron and a proton if they have the same kinetic energy. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, mp=1.671027 kg and e=1.601019 C) 22
23. 23. Example 5 : Compare the de Broglie wavelength of an electron and a proton if they have the same kinetic energy. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, mp=1.671027 kg and e=1.601019 C) Solution : Ke  Kp  K By using the de Broglie wavelength formulae, thus   h 2meV h and eV  K 2mK 23
24. 24. Solution : Ke  Kp  K Therefore the ratio of their de Broglie wavelengths is  h    e  2me K    p  h     2m K  p   mp  me 1.67  10 27  9.11  10 31 e  42.8 p 24
25. 25. Electron microscope  A practical device that relies on the wave properties of electrons     25 is electron microscope. It is similar to optical compound microscope in many aspects. The advantage of the electron microscope over the optical microscope is the resolving power of the electron microscope is much higher than that of an optical microscope. This is because the electrons can be accelerated to a very high kinetic energy giving them a very short wavelength λ typically 100 times shorter than those of visible light. Therefore the diffraction effect of electrons as a wave is much less than that of light. As a result, electron microscopes are able to distinguish details about 100 times smaller. PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
26. 26.  In operation, a beam of electrons falls on a thin slice of sample.  The sample (specimen) to be examined must be very thin (a few     26 micrometres) to minimize the effects such as absorption or scattering of the electrons. The electron beam is controlled by electrostatic or magnetic lenses to focus the beam to an image. The image is formed on a fluorescent screen. There are two types of electron microscopes:  Transmission – produces a two-dimensional image.  Scanning – produces images with a three-dimensional quality. Figures 10.2 and 10.3 are diagram of the transmission electron microscope and the scanning electron microscope. PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
27. 27. 27 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
28. 28. Exercise 26.1 : Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and e=1.601019 C 1. a. An electron and a photon have the same wavelengths and the total energy of the electron is 1.0 MeV. Calculate the energy of the photon. b. A particle moves with a speed that is three times that of an electron. If the ratio of the de Broglie wavelength of this particle and the electron is 1.813104, calculate the mass of the particle. ANS. : 1.621013 J; 1.671027 kg 2. a. An electron that is accelerated from rest through a potential difference V0 has a de Broglie wavelength 0. If the electron’s wavelength is doubled, determine the potential difference requires in terms of V0. b. Why can an electron microscope resolve smaller objects than a light microscope? (Physics, 3rd edition, James S. Walker, Q12 & Q11, p.1029) 28
29. 29. Learning Outcome: 3.1 The photoelectric effect (3 hours) At the end of this chapter, students should be able to:  Explain the phenomenon of photoelectric effect.  Define threshold frequency, work function and stopping potential.  Describe and sketch diagram of the photoelectric effect experimental set-up.  Explain by using graph and equations the observations of photoelectric effect experiment in terms of the dependence of :  kinetic energy of photoelectron on the frequency of light; 29 1 2 mv max  eVs  hf  hf0 2 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
30. 30. Learning Outcome: 3.1 The photoelectric effect (3 hours) At the end of this chapter, students should be able to:  photoelectric current on intensity of incident light;  work function and threshold frequency on the types of metal surface. W0  hf 0  30 Explain the failure of wave theory to justify the photoelectric effect. PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
31. 31. 3.1 The photoelectric effect  is defined as the emission of electron from the surface of a metal when the EM radiation (light) of higher frequency strikes its surface.  Figure 1 shows the emission of the electron from the surface of the metal after shining by the light. - EM radiation - - - - - - photoelectron - - - - Metal 31 Free electrons Figure 1  Photoelectron is defined as an electron emitted from the surface of the metal when the EM radiation (light) strikes its surface. PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
32. 32. 3.1.1 Photoelectric experiment  The photoelectric effect can be studied through the experiment made by Franck Hertz in 1887.  Figure 2 shows a schematic diagram of an experimental arrangement for studying the photoelectric effect. EM radiation (light) cathode anode - photoelectron - - vacuum glass G V power supply 32 rheostat Figure 2 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
33. 33.  The set-up apparatus as follows:  Two conducting electrodes, the anode (positive electric potential) and the cathode (negative electric potential) are encased in an evacuated tube (vacuum).  The monochromatic light of known frequency and intensity is incident on the cathode. Explanation of the experiment  When a monochromatic light of suitable frequency (or wavelength) shines on the cathode, photoelectrons are emitted.  These photoelectrons are attracted to the anode and give rise to the photoelectric current or photocurrent I which is measured by the galvanometer.  When the positive voltage (potential difference) across the cathode and anode is increased, more photoelectrons reach the anode , thus the photoelectric current increases. 33 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
34. 34.  As positive voltage becomes sufficiently large, the photoelectric current reaches a maximum constant value Im, called saturation current.  Saturation current is defined as the maximum constant value of photocurrent when all the photoelectrons have reached the anode.  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 anode and the photoelectric current flows is I0.  Finally, when the voltage is made negative by reversing the power supply terminal as shown in Figure 2, the photoelectric current decreases even further to very low values since most photoelectrons are repelled by anode which is now negative electric potential. 34 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
35. 35. EM radiation (light) cathode anode - photoelectron - - vacuum glass G V power supply rheostat Figure 3: reversing power supply terminal  As the potential of the anode becomes more negative, less photoelectrons reach the anode thus the photoelectric current drops until its value equals zero which the electric potential at this moment is called stopping potential (voltage) Vs.  Stopping potential is defined as the minimum value of negative voltage when there are no photoelectrons reaching the anode. 35 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
36. 36. U due to this retarding voltage Vs now equals the maximum kinetic energy Kmax of the photoelectron.  The potential energy U  K max 1 2 (1) eVs  mv max 2 where m : mass of theelectron  The variation of photoelectric current I as a function of the voltage V can be shown through the graph in Figure 9.4c. Photoelectric current, I Im I0  Vs Figure 4 36 PHY310 – Photoelectric Effect After 0 Voltage ,V Before reversing the terminal DR.ATAR @ UiTM.NS
37. 37. 3.1.2 Einstein’s theory of photoelectric effect  A photon is a ‘packet’ of electromagnetic radiation with particle-like characteristic and carries the energy E given by E  hf and this energy is not spread out through the medium. Work function W0 of a metal  Is defined as the minimum energy of EM radiation required to emit an electron from the surface of the metal.  It depends on the metal used.  Its formulae is and Emin  hf 0 W0  Emin W0  hf 0 (2) where f0 is called threshold frequency and is defined as the minimum frequency of EM radiation required to emit an electron from the surface of the metal. 37 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
38. 38.  Since c=f then the eq. (9.6) can be written as hc W0  (3) 0 where 0 is called threshold wavelength and is defined as the maximum wavelength of EM radiation required to emit an electron from the surface of the metal.  Table 1 shows the work functions of several elements. Element Aluminum 2.3 Copper 4.7 Gold 5.1 Silver PHY310 – Photoelectric Effect 4.3 Sodium 38 Work function (eV) 4.3 Table 1 38 DR.ATAR @ UiTM.NS
39. 39. Einstein’s photoelectric equation  In the photoelectric effect, Einstein summarizes that some of the energy E imparted by a photon is actually used to release an electron from the surface of a metal (i.e. to overcome the binding force) and that the rest appears as the maximum kinetic energy of the emitted electron (photoelectron). It is given by E  K max  W0 where 1 2 hf  mv max  W0 2 E  hf and K max 1 2  mv max 2 (4) where eq. (4) is known as Einstein’s photoelectric equation.  Since Kmax=eVs then the eq. (4) can be written as hf  eVs  W0 where (5) Vs : stoppingvoltage e : magnitude for charge of electron
40. 40. Note:  1st case: hf  WOR 0 f  f0 hf 2nd case: hf -  WOR 0  3rd case: Electron is emitted with maximum kinetic energy. f  f0 hf Figure 5b K max W0 Metal Figure 5a  - vmax Metal hf  WOR 0 - K max  0 v0 W0 Electron is emitted but maximum kinetic energy is zero. f  f0 hf No electron is emitted. Figure 5c 40 PHY310 – Photoelectric Effect Metal - W0 DR.ATAR @ UiTM.NS
41. 41. Example 3 : Cadmium has a work function of 4.22 eV. Calculate a. its threshold frequency, b. the maximum speed of the photoelectrons when the cadmium is shined by UV radiation of wavelength 275 nm, c. the stopping potential. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and e=1.601019 C) 41
42. 42. Solution : a. By using the equation of the work function, thus   W0  4.22 1.60 10 19  6.75 10 19 J W0  hf 0   6.75 10 19  6.63 10 34 f 0 15 f 0  1.02 10 Hz 42
43. 43.   Solution : W  4.22 1.60  10 19  6.75  10 19 J 0 b. Given   275  10 9 m By applying the Einstein’s photoelectric equation, thus E  K max  W0 hc 1 2  mv max  W0  2 6.63  10 34 3.00  108 1 2  9.11 10 31 vmax  6.75  10 19 2 275  10 9 vmax  3.26 105 m s 1      c. The stopping potential is given by K max 1 2  mv max 2 1 2 eVs  mv max 2 1 19 1.60  10 Vs  9.11 10 31 3.26 10 5 2 43 Vs  0.303 V      2
44. 44. Example 4 : A beam of white light containing frequencies between 4.00 1014 Hz and 7.90 1014 Hz is incident on a sodium surface, which has a work function of 2.28 eV. a. Calculate the threshold frequency of the sodium surface. b. What is the range of frequencies in this beam of light for which electrons are ejected from the sodium surface? c. Determine the highest maximum kinetic energy of the photoelectrons that are ejected from this surface. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and e=1.601019 C) 44
45. 45.  Solution : W  2.28 1.60  10 19 0 a. The threshold frequency is   3.65 10 19 J W0  hf 0 3.65 10 19  6.63 10 34 f 0 f 0  5.51 1014 Hz   b. The range of the frequencies that eject electrons is 5.51 1014 Hz and 7.90 1014 Hz c. For the highest Kmax, take f  7.90  1014 Hz By applying the Einstein’s photoelectric equation, thus E  K max  W0 1 2 hf  mv max  W0 2 6.63 10 7.90 10   K 34 14 max  3.65 10 19 K max  1.59 10 19 J 45
46. 46. Exercise 3.1 : Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and e=1.601019 C 1. 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, electrons are emitted with a maximum kinetic energy of 1.10 eV. Calculate the work function of this metal in joules. ANS. : 553 nm; 1.841019 J 2. In a photoelectric effect experiment it is observed that no current flows when the wavelength of EM radiation is greater than 570 nm. Calculate a. the work function of this material in electron-volts. b. the stopping voltage required if light of wavelength 400 nm is used. (Physics for scientists & engineers, 3rd edition, Giancoli, Q15, p.974) ANS. : 2.18 eV; 0.92 V 46
47. 47. Exercise 3.1 : 3. In an experiment on the photoelectric effect, the following data were collected. Wavelength of EM radiation,  (nm) Stopping potential, Vs (V) 350 1.70 450 0.900 a. Calculate the maximum velocity of the photoelectrons when the wavelength of the incident radiation is 350 nm. b. Determine the value of the Planck constant from the above data. ANS. : 7.73105 m s1; 6.721034 J s 47
48. 48. 3.2 Graph of photoelectric experiment Variation of photoelectric current I with voltage V  for the radiation of different intensities but its frequency is fixed. I 2I m Im  Vs Intensity 2x Intensity 1x 0 V Figure 6 Reason: From the experiment, the photoelectric current is directly proportional to the intensity of the radiation as shown in Figure 6. 48 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
49. 49. I 2I m Im 0 1 2 Light intensity Figure 7  for the radiation of different frequencies but its intensity is fixed. I Im f2 > f1 f2 f1  Vs2 Vs1 0 49 PHY310 – Photoelectric Effect V Figure 8 49 DR.ATAR @ UiTM.NS
50. 50. Reason: From the Einstein’s photoelectric equation, Stopping voltage ,Vs hf  eV  W s 0 W0 h Vs    f  e e y m x c Vs2 Vs1 0 W0  e f 0 f1 f 2 frequency, f If Vs=0, hf  e(0)  W0 W0  hf f 0 Figure 9 50 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
51. 51.  For the different metals of cathode but the intensity and frequency of the radiation are fixed. I Im W02 > W01 W01 Figure 10 W02 V  Vs1  Vs20 Reason: From the Einstein’s photoelectric equation, Vs hf e hf  eVs  W0 Vs1 Vs2 0  1  hf  Vs    W0     e  e  y m x  c W01 W02 hf  E W0 Energy of a photon in EM radiation Figure 11 51 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
52. 52. Variation of stopping voltage Vs with frequency f of the radiation  for different metals of cathode but the intensity is fixed. Vs W01 W02 W03 W03 >W02 > W01 Figure 12 0 f 01 f 02 f 03 f W0  f 0 Threshold (cut-off) Reason: Since W0=hf0 then frequency W0 If Vs=0, h hf  e(0)  W0 Vs    f  hf  eVs  W0 e e W0  hf f 0 y m x c 52 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
53. 53. 3.3 Failure of wave theory of light  Table 2 shows the classical predictions (wave theory), photoelectric experimental observation and modern theory explanation about photoelectric experiment. Classical predictions Experimental observation Modern theory Emission 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 53 53 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
54. 54. Classical predictions Experimental observation Modern theory The higher the intensity, the greater the energy imparted to the metal surface for emission of photoelectrons. When the intensity is low, the energy of the radiation is too small for emission of electrons. Very low intensity but high frequency radiation could emit photoelectrons. The maximum kinetic energy of photoelectrons is independent of light intensity. The intensity of light is the number of photons radiated per unit time on a unit surface area. Based on the Einstein’s photoelectric equation: 54 PHY310 – Photoelectric Effect K max  hf  W0 The maximum kinetic energy of photoelectron depends only on the light frequency and the work function. If the light intensity is doubled, the number of electrons emitted also doubled but the maximum kinetic energy remains unchanged. DR.ATAR @ UiTM.NS
55. 55. Classical predictions Experimental observation Modern theory Light 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. 55 PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
56. 56. Classical predictions Experimental observation Modern theory Energy of light depends only on amplitude ( or intensity) and not on frequency. Energy of light depends on frequency. According to Planck’s quantum theory which is E=hf Energy of light depends on its frequency. Table 2 Note:    56 Experimental observations deviate from classical predictions based on wave theory of light. 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. PHY310 – Photoelectric Effect DR.ATAR @ UiTM.NS
57. 57. Example 5 : a. Why does the existence of a threshold frequency in the photoelectric effect favor a particle theory for light over a wave theory? b. In the photoelectric effect, explains why the stopping potential depends on the frequency of light but not on the intensity. 57
58. 58. Example 5 : Solution : a. Wave theory predicts that the photoelectric effect should occur at any frequency, provided the light intensity is high enough. However, as seen in the photoelectric experiments, the light must have a sufficiently high frequency (greater than the threshold frequency) for the effect to occur. b. The stopping voltage measures the kinetic energy of the most energetic photoelectrons. Each of them has gotten its energy from a single photon. According to Planck’s quantum theory , the photon energy depends on the frequency of the light. The intensity controls only the number of photons reaching a unit area in a unit time. 58
59. 59. Example 6 : 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 9.10. f 1014 Hz 4.83 0 K max (eV) Figure 13 Based on the graph, for the light of frequency 7.141014 Hz, calculate a. the threshold wavelength, b. the maximum speed of the photoelectron. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and e=1.601019 C) 59
60. 60. Solution : f  7.14  1014 Hz a. By rearranging Einstein’s photoelectric equation, f 1014 Hz hf  K max  W0 4.83 0 K max (eV) W0 1 f    K max  h h 1 f    K max  f 0 h y m x  c From the graph, f 0  4.83  1014 Hz Therefore the threshold wavelength is given by c 0  f0 3.00  108  4.83  1014 0  6.21 10 7 m 60
61. 61. Solution : f  7.14  1014 Hz b. By using the Einstein’s photoelectric equation, thus  1 2 hf  mv max  W0 2 1 2 hf  mv max  hf0 2 1 2 mv max  h f  f 0  2   1 2 9.11 10 31 vmax  6.63 10 34 7.14 1014  4.83 1014 2 vmax  5.80 105 m s 1 61 
62. 62. Exercise 25.2 : Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and e=1.601019 C 1. A photocell with cathode and anode made of the same metal connected in a circuit as shown in the Figure 14. 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 Figure 15. I (nA) 365 nm 5 G V 1 Figure 14 0 Figure 15 V ( V) 62
63. 63. Exercise 25.2 : 1. a. Calculate the maximum kinetic energy of photoelectron. b. Deduce the work function of the cathode. c. If the experiment is repeated with monochromatic light of wavelength 313 nm, determine the new intercept with the V-axis for the new graph. ANS. : 1.601019 J, 3.851019 J; 1.57 V 2. When EM radiation falls on a metal surface, electrons may be emitted. This is photoelectric effect. a. Write Einstein’s photoelectric equation, explaining the meaning of each term. b. Explain why for a particular metal, electrons are emitted only when the frequency of the incident radiation is greater than a certain value? c. Explain why the maximum speed of the emitted electrons is independent of the intensity of the incident radiation? (Advanced Level Physics, 7th edition, Nelkon&Parker, Q6, p.835) 63