#### Transcript photoelectric effect

UNIT 24 : QUANTIZATION OF LIGHT 3 hours 24.1 Planck’s Quantum Theory 24.2 The Photoelectric Effect 1 SUBTOPIC : 24 .1 Planck’s Quantum Theory ½ hour LEARNING OUTCOMES : At the end of this lesson, the students should be able to : a) Distinguish between Planck’s quantum theory and classical theory of energy. b) Use Einstein’s formulae for hc E hf a photon energy, . 2 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 black body from the interior walls until it is 3 completely absorbed. • The spectrum of electromagnetic radiation emitted by the black body (experimental result) is shown in figure 1. Experimental result Rayleigh Jeans theory Wien’s theory Classical physics 4 Figure 1 : Black Body Spectrum • 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 does not depend on its frequency or wavelength. Energy of the e.m. radiation is continuously. 5 • 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 emitted by the black body is a discrete (separate) packets of energy known as quanta. This means the energy of e.m. radiation is quantised. The energy size of the radiation depends on its frequency. 6 Comparison between Planck’ quantum theory and classical theory of energy. Planck’s Quantum Theory Energy of the e.m radiation is quantised. (discrete) Classical theory Energy of the e.m radiation is continously. Photon Energy of e.m radiation depends on its frequency or wavelength E hf Energy of e.m radiation does not depend on its frequency or wavelength (depends on Intensity) I A2 Eclassical k BT k B Boltzman' s constant T temperature 7 • According to this assumptions, the quantum E of the energy for radiation of frequency f is given by E hf where c f h : Planck constant 6.63 1034 J s hc E Planck’s quantum theory 8 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” 9 • In equation form, photon energy (energy of photon) is c E hf h • 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 : 1 eV 1.60 1019 J • 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 10 have particle property. Example 24.1 Calculate the energy of a photon of blue light, 450 nm . (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) 11 Example 24.2 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) 12 SUBTOPIC : 24 .2 The Photoelectric Effect 2 ½ hours LEARNING OUTCOMES : At the end of this lesson, the students should be able to : a) Explain the phenomenon of photoelectric effect. b) Define and determine threshold frequency, work function and stopping potential. c) Describe and sketch diagram of the photoelectric effect experimental set-up. d) Explain the failure of wave theory to justify the photoelectric effect. 13 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 14 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). em radiation photoelectron (light) - - - - -- - - - - Metal surface 15 Free electrons 9.2 The photoelectric effect • The photoelectric effect can be measured using a device like that pictured in figure below. e.m. radiation (incoming light) Cathode (emitter or target metal) Anode(collector) photoelectron - - vacuum glass A V A power supply rheostat The photoelectric effect’s experiment 16 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. 17 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 as the maximum constant value of photocurrent in which when all the photoelectrons have reached the anode. 18 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 . Photoelect ric current , I Im Graph of photoelectric current I0 against voltage for photoeclectric effect’s experiment Vs 0 B (After) Voltage,V A (Before reversing the terminal) 19 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) photoelectron - - vacuum glass A V power supply B rheostat Reversing power supply terminal (to determine the stopping potential) 20 • 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 1 2 eVs mv 2 21 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. 22 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 , E W K .E o max 1 2 hf W0 mvmax 2 hf W0 eVs but 1 2 eVs mv 2 23 Einstein’s theory of Photoelectric Effect E Wo K .Emax 1 2 hf W0 mvmax 2 E hf h c Einstein’s photoelectric equation = photon energy f = frequency of em radiation /incoming light K .Emax 1 2 mv max= maximum kinetic energy of 2 ejected electron. vmax = maximum speed of the photoelectron 24 Einstein’s theory of Photoelectric Effect E Wo K .Emax 1 2 hf W0 mvmax 2 Wo hfo hc o Wo = the work function of a metal. = the minimum energy required (needed) to eject an electron from the surface of target metal. fo = threshold frequency. = minimum frequency of e.m. radiation c required to eject an electron from the fo o surface of the metal. o = threshold wavelength. = maximum wavelength of e.m. radiation required to eject an electron from the surface of the target metal. 25 Einstein’s theory of Photoelectric Effect vmax hf - Metal Electron is emitted hf - v=0 W0 - Metal 1 2 hf W0 mvmax 2 Electron is ejected. - W0 hf Wo hf > Wo hf Metal No electron is ejected. - hf < Wo W0 26 Example 24 .3 The work function for a silver surface is Wo = 4.74 eV. Calculate the a) minimum frequency that light must have to eject electrons from the surface. b) 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) a)Wo hfo f o 1.14x1015 Hz b) o 263nm 27 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) K.Emax = E – Wo 28 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 a. W0 hf0 29 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. hc W0 1 mv2 2 c. eVs 1 mv2 2 30 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) 31 Solution 24.6 a) W0 8.55 1019 J b) 0 2.33107 m 32 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) 33 Solution 24.7 Ans. : 553 nm, 1.84 x 10-19 J 34 Graphs in Photoelectric Effect Generally, Einstein’s photoelectric equation; E Wo K .Emax K .Emax E Wo K .Emax hf Wo K.Emax m x c y f ↑ K.Emax ↑ f 0 Wo fo 35 K .Emax hf Wo Graphs in Photoelectric Effect eVs hf Wo Wo h Vs f e e y mx c Stopping voltage ,Vs 0 W0 e fo f ↑ Vs ↑ frequency , f 36 Graphs in Photoelectric Effect Variation of stopping voltage Vs with frequency f of the radiation for different metals but the intensity is fixed. Stopping voltage ,Vs W01 W02 K .Emax hf Wo eVs hf Wo Wo h Vs f e e y mx c 0 f 01 frequency , f f 02 W0 f0 W02 > W01 f02 > f01 37 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. Photoelect ric current , I Intensity 2x Intensity 1x Vs 0 Voltage,V 38 Notes: Classical physics Light intensity , I energy time area Quantum physics number of photons Light intensity , I time area Light intensity number of photons Light intensity ↑ , number of photons ↑ , number of electrons ↑ , current ↑ . (If light intensity ↑, photoelectric current ↑). 39 Graphs in Photoelectric Effect Variation of photoelectric current I with voltage V for the radiation of different frequencies but its intensity and metal are fixed. Photoelect ric current , I Im f2 > f1 f2 Vs2 > Vs1 f1 Voltage,V Vs 2 Vs10 K .Emax hf Wo eVs hf Wo Wo h Vs f e e f ↑ Vs ↑ 40 Graphs in Photoelectric Effect Variation of photoelectric current I with voltage V for the different metals but the intensity and frequency of the radiation are fixed. Photoelect ric current , I Im W01 W02 Vs1 Vs 20 K .Emax hf Wo eVs hf Wo Wo h Vs f e e Wo ,Vs Voltage,V W02 > W01 Vs1 > Vs2 41 Example 24.8 K.Emax (x 10-19 J) f(x 1014 )Hz 0 4 .8 Use the graph above to find the value of i) work function and ii) the threshold wavelength. 42 K .Emax E Wo Solution 24.8 K .Emax hf Wo K.Emax (x 10-19 J) y m x c f(x 1014 )Hz 0 4 .8 From the graph w hen K .Emax 0, Wo hfo f o 4.8 x1014 Hz 3.18 x 10 -19 J Threshold w avelength, o c fo 6.25 x 10-7 m 43 OBSERVATIONS of the photoelectric effects experiment 1. Electrons are emitted immediately 2. Stopping potential does not depend on the intensity of light. 3. Threshold frequency of light is different for different target metal. 4. Number of electrons emitted of the photoelectron current depend on the intensity of light. 44 EXPLAIN the failure of classical theory to justify the photoelectric effect. 1. MAXIMUM KINETIC ENERGY OF PHOTOELECTRON Clasiccal prediction Experimental Result Modern Theory The 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. Based on Einstein’s photoelectric equation: Very low intensity but high frequency radiation could emit photoelectrons. The maximum kinetic energy of photoelectrons is independent of light intensity. K max hf W0 The maximum kinetic energy of photoelectron depends only on the light frequency . The maximum kinetic energy of photoelectrons DOES NOT depend on light intensity. 45 2. EMISSION OF PHOTOELECTRON ( energy ) Clasiccal prediction Emission of photoelectro ns occur for all frequencies of light. Energy of light is independent of frequency. Experimental Result Emission of photoelectrons occur only when frequency of the light exceeds the certain frequency which value is characteristic of the material being illuminated. Modern Theory 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. 46 3. EMISSION OF PHOTOELECTRON ( time ) Clasiccal prediction Experimental Result Light energy is spread Photoelectrons over the wavefront, the are emitted amount of energy from the incident on any one surface of the electron is small. An metal almost electron must gather instantaneously sufficient energy before after the emission, hence there is surface is time interval between illuminated, absorption of light even at very energy and emission. low light Time interval increases intensities. if the light intensity is low. Modern Theory 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 47 emission. 4. ENERGY OF LIGHT Clasiccal prediction Experiment Modern Theory al Result 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. 48 • Experimental 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. 49 SUMMARY : Comparison between classical physics and quantum physics about photoelectric effect experiment Feature Classical physics Quantum physics Threshold frequency An 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 energy of photoelectrons Depends on the light intensity. Depends only on the light frequency . Emission of photoelectrons There should be some delays to emit electrons from a metal surface. Electrons are emitted spontaneously. Energy of light Depends on the light intensity. Depends only on the light 50 frequency . Exercise (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) 1. Find the energy of the photons in a beam whose wavelength is 500 nm. ( 3.98 x 10 -19 J) 2. Determine the vacuum wavelength corresponding to a -ray energy of 1019 eV. (1.24 x10-25 m) 3. A sodium surface is illuminated with light of wavelength 300 nm. The work function for sodium metal is 2.46 eV. Calculate a) the kinetic energy of the ejected photoelectrons b) the cutoff wavelength for sodium c) maximum speed of the photoelectrons. -1) (1.68 eV, 505 nm, 7.68 x 105 ms51 4. 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) 5. 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 0.063 V. When light of wavelength 440 nm is used, the h stopping potential becomes 0.865 V. Determine the ratio . e ( 6.35 x 10-15 J s C-1) 52 7. In an experiment on the photoelectric effect, the following data were collected. Wavelength of e.m. radiation, (nm) 350 450 Stopping potential, Vs (V) 1.70 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. (7.73 x 105 m s-1, 6.72 x 10-34 J s) 53 8. In a photoelectric effect experiment it is observed that no current flows unless the wavelength is less than 570 nm. Calculate a. 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) 9. 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. f 1014 Hz 2.0 Kmax (eV ) Based on the graph, for the light frequency of 6.00 x 1014 Hz, calculate a. the threshold frequency. b. the maximum kinetic energy of the photoelectron. c. the maximum velocity of the photoelectron. f 0 4.83 1014 Hz , K max 7.78 1020 J , v 4.13 10 5 m s -1 55 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. I (nA) 365 nm 5 G V 1 0 V (V ) 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 of wavelength 313 nm, determine the new intercept with the V-axis for the new graph. 56 -19 -19 (1.60 x 10 J, 3.85 x 10 J, -1.57 V)