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Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
Vu4 light&matter2009
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Vu4 light&matter2009

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  • 1. VCE Physics Unit 4 Topic 2 Interactions of Light & Matter
  • 2. Unit Outline To achieve the outcome the student should demonstrate the knowledge and skills to : • Explain the production of incoherent light from wide spectrum light sources including the Sun, light bulbs and candles (descriptive) in terms of thermal motion of electrons. • Explain the results of Young’s double slit experiment as evidence for the wave like nature of light including: – constructive and destructive interference in terms of path difference – qualitative effect of wavelength on interference patterns • Interpret the pattern produced by light when it passes through a gap or past an obstacle in terms of the diffraction of waves and the significance of the magnitude of the λ/w ratio • interpret the photoelectric effect as evidence of the particle like nature of light including the KE of emitted photoelectrons in terms of the energy of incident photons Ekmax = hf – W, using energy units of both joules and electron-volts, effects of intensity of incident irradiation on the emission of photoelectrons. • interpret electron diffraction patterns as evidence of the wavelike nature of matter expressed as the De Broglie wavelength λ = h/p • compare momentum of photons and of particles of the same wavelength including calculations using p = h/λ • interpret atomic absorption and emission spectra, including those from metal vapour lamps in terms of the quantised energy level model of the atom, including calculations of the energy of photons emitted or absorbed. ∆E = hf • explain a model of quantised energy levels of the atom in which electrons are found in standing wave states • use safe and responsible practises when working with light sources, lasers and related equipment.
  • 3. Chapter 1 Topics covered: • The Nature of Light. • Interference. • Incoherent Light. • Coherent Light • Young’s Experiment. • Path Difference • Single Slit Diffraction • Diffraction around Objects
  • 4. 1.0 The Nature of Light EMR is a self propagating wave consisting of mutually perpendicular, varying ELECTRIC and MAGNETIC FIELDS. EMR travels through a vacuum at 300,000 kms-1 , (3.0 x 108 ms-1 ) Direction of Electromagnetic Wave Movement Changing Magnetic Field Changing Electric Field Light is a form of ENERGY. It is described as ELECTRO - MAGNETIC RADIATION (EMR).
  • 5. 1.1 Superposition When two wave trains interact with one another they also undergo SUPERPOSITION and will either: 1. Add together to produce a larger wave - a process called CONSTRUCTIVE INTERFERENCE or 2. Subtract from one another to produce no wave – a process called DESTRUCTIVE INTERFERENCE Single waves, called pulses. have the ability to pass through one another and, while occupying the same space, add together in a process called SUPERPOSITION. Once the superposition is complete the pulses continue their journey unaffected. CrestTrough A series of pulses together form a wave train with alternating crests and troughs Constructive interference occurs when the two wave trains are in phase Destructive interference occurs when the two wave trains are 1800 out of phase
  • 6. 1.2 Interference Waves, in this case water waves, when passed through two narrow slits, “interfere” or interact with one another to produce areas of large disturbances (Constructive Interference) Constructive Interference Destructive Interference Light behaves in a similar manner. When light (with certain properties) is passed through two narrow slits, an “interference pattern” is produced, showing constructive (light bands) and destructive (dark bands) interference. Bright bands occur where a crest and a crest (or a trough and a trough) arrive at the screen at the same time Dark bands occur when a crest and a trough arrive at the screen at the same time Bright Band Dark Band or areas of no disturbance (Destructive Interference) Crest Trough
  • 7. 1.3 Incoherent Light Light is generated by luminous bodies, eg, The Sun, light globes, burning candles. Typical light sources such as those mentioned above have an inbuilt irregularity in the way they produce light. Light is produced when atoms of the filaments or source become electrically excited and produce an electromagnetic or light wave. Since the excitations occur in an unpredictably random fashion, the light waves are NOT produced in regular repeating manner and so do not maintain a constant “phase relationship” with one other. About once every 10-8 sec, a source will randomly alter its phase. This leads to the sources giving off a broad spectrum of white light composed of all colours in the rainbow. Each of the millions of colours have waves that are random to each other. This is called INCOHERENT LIGHT. Incoherent light, when combined, produces rapidly moving areas of constructive and destructive interference and therefore do not produce a stable, visible interference pattern.
  • 8. Light & Matter Revision Question Type: Q1: The light from a candle can best be described as A. coherent, arising from the vibrations of electrons. B. incoherent, arising only from the transition of electrons in excited energy levels falling to lower energy levels. C. coherent, arising only from the transition of electrons in excited energy levels falling to lower energy levels. D. incoherent, arising from the vibrations of electrons. Incandescent Light
  • 9. Light & Matter Revision Question Type: The spectrum of wavelengths produced by a particular incandescent light globe is shown in Figure 1 below. Q3: Describe the mechanism by which light is produced in an incandescent light globe. Q2: The light produced by an incandescent light globe can best be described as A. coherent. B. incoherent. C. monochromatic. D. in phase. Incandescent Light The thermal/random excitation of electrons in the filament leads to the emission of a broad/continuous spectrum.
  • 10. 1.4 Coherent Light Two waves that are coherent can be combined to produce an unmoving distribution of constructive and destructive interference (a visible interference pattern) depending on the relative phase of the waves at their meeting point. Coherence is a property of waves that measures their ability to interfere with each other. Coherent waves (constant phase difference) Lasers generate light at a single wave length and frequency and all of the waves (and PHOTONS) are in phase. This is called COHERENT LIGHT. Shown are monochromatic (single colour) light waves of the same frequency. They are coherent and in phase, and will combine constructively to produce bright white light Coherent waves (zero phase difference)
  • 11. Light & Matter Revision Question Type: Q4: Which of the following light sources will produce coherent, monochromatic light A. Sunlight B. LED C. Light globe with filter D. LASER Coherent Light
  • 12. Double Slits 1.5 Young’s Experiment Screen Long wavelength, RED light produces a pattern with wider red bands which are spread farther apart. First performed by Thomas Young in the early 1800’s this experiment proved light was a wave. It has been voted the most elegant experiment ever devised. BLUE Light using the same slits RED Light When light of a single frequency (colour) is passed through a pair of closely spaced, narrow slits an “interference pattern” is produced. This pattern has a series of equally spaced coloured and black bands spread across the screen onto which it is projected. The width of the coloured bands and their spacing depends on the wavelength of the light used. Short wavelength, BLUE light produces a pattern with narrow blue bands which are closely spaced. Thomas Young 1773 – 1829 Incident Light c
  • 13. Light & Matter Revision Question Type: Thomas Young’s double slit experiment has been replicated in the experimental arrangement shown in Figure 4. Q5: Explain using the wave theory of light why a series of bright and dark bars are observed on the screen. The slits are now moved further from the screen. Q6: What effect would this have on the pattern observed on the screen? Slits provide 2 coherent sources leading to superposition which produces constructive and destructive interference leading to bright and dark bars on screen Young’s Experiment The pattern will spread further across the screen.
  • 14. Light & Matter Revision Question Type: A physics teacher has apparatus to show Young’s double slit experiment. The apparatus is shown in Figure 4. The pattern of bright and dark bands is observed on the screen. Q7: Which one of the following actions will increase the distance, Δx, between dark bands in this double slit interference pattern? A. decrease the slit width B. decrease the slit separation C. decrease the slit screen distance D. decrease the wavelength of the light Young’s Experiment
  • 15. S1 S2 Slits S1 S2 Light travels the same distance from S1 and S2 to reach the central bright band, BC The path difference S1BC – S2BC = 0. 1.6 Path Difference So a crest and a crest (or trough and trough) arrive at AC at the same time, leading to constructive interference or a bright band The dark band (D1) on either side of AC occurs because a crest and a trough are arriving at the same time leading to destructive interference. Path difference: S1D1 – S2D1 = ½ λ For the next bright band path difference S1B1 – S2B1 = 1 λ For the next bright band path difference would be: S1B2 – S2B2 = 2λ So bright bands occur when path difference = nλ where n = 0,1,2,3,etc The next dark band (D2) has path difference = 1½λ BC D1D1 B1B1 D2D2 So dark bands occur when path difference = (n + ½)λ where n = 0,1,2,3, etc
  • 16. Light & Matter Revision Question Type: In the following diagram, laser light of wavelength 600 nm is shone onto a pair of parallel slits and a pattern of alternating light/dark bands is projected onto a wall. Q8: Explain how the observed pattern on the wall supports a wave model for light. The alternating pattern represents an interference pattern which is a wave phenomenon. The bright bands represent regions of constructive interference where the difference in path length from each source is a multiple of wavelengths. Waves therefore arrive from both sources in phase. The dark regions represent bands of destructive interference with the waves being half a wavelength out of phase due to path differences. Q9: Estimate the difference in length between P1 and P2. The indicated position is on the sixth antinode from the centre. The path difference is 6λ = 6 × 600nm = 3.6 ×10−6 m Interference – Path Difference
  • 17. Light & Matter Revision Question Type: Students have set up an experiment similar to that of English physicist Thomas Young. The students’ experiment uses microwaves of wavelength λ = 2.8 cm instead of light. The beam of microwaves passes through two narrow slits shown as S1 and S2 in Figure 3. The students measure the intensity of the resulting beam at points along the line shown and determine the positions of maximum intensity. These are shown as filled circles and marked P0, P1 . . . Q10: What is the difference in length (S2P2 - S1P2) where P2 is the second maximum away from the central axis. Path difference to 2nd max = 2λ = 5.6 cm Interference – Path Difference
  • 18. 1.7 Single Slit Diffraction Diffraction patterns for blue and red light show that the shorter wavelength blue light produces the more “compact” pattern, When light of a single wavelength shines through a narrow gap or single slit, a “diffraction pattern” is produced. THE EXTENT OF DIFFRACTION DEPENDS UPON THE RATIO λ/w, WHERE λ = WAVELENGTH AND w = SLIT WIDTH. Appreciable diffraction will occur if the ratio λ/w is between about 0.1 and 50. Outside this range, diffraction is not observed. The pattern consists of a rather wide, coloured central maximum with a series of thinner coloured and dark bands spreading out from the centre. Screen Single Slit Width = w Incident Light Wavelength = λ BLUE light using same slit RED light while the longer wavelength red produces a more spread out pattern In other words, the spacing between the lines is wavelength dependent and patterns with the same line spacing were made by light of the same wavelength Note: The wider the slit the narrower central maximum
  • 19. Light & Matter Revision Question Type: In an experiment, monochromatic laser light of wavelength 600 nm shines through a narrow slit, and the intensity of the transmitted light is recorded on the screen some distance away as shown in Figure 2a. The intensity pattern seen on the screen is shown in Figure 2b. Q11: Which one of the intensity patterns (A-D) below best indicates the pattern that would be seen if a wider slit was used? Single Slit Diffraction
  • 20. 1.8 Diffraction around Objects In addition to diffraction occurring when light passes through gaps, it can also occur when light passes around objects Sunlight diffracting around a car tail light lens The shadow of a hand holding a coin illuminated by a He – Ne Laser. Red Light Diffracting around a pinhead You can see a diffraction pattern through your fingers Point your hand toward a light source and view through this gap
  • 21. Chapter 2 Topics covered: • Models for Light Behaviour. • Particle like Properties of Light. • Wave like Properties of Light.
  • 22. 2.0 Models for Light Behaviour • Light is/has energy and thus can do work. We see this in things like plants growing or the chemical reactions on photographic films. • It is quite easy to list or show what light DOES but more difficult to say what light IS ! • Much time and effort has been put into trying to find a THEORY or MODEL which will explain ALL the properties and behaviours exhibited by light. Two separate models have been proposed: 1. The Particle Model: This says that all properties of light can be explained by assuming light is made up of a stream of individual particles. 2. The Wave Model: This says that all properties of light can be explained by assuming light is made up of a stream of waves.
  • 23. 2.1 Particle like Properties of Light • Most properties of light can be explained using the “Wave Model”. • However, some properties can only be explained by assuming light is made up of a “stream of particles”. • These particle like properties are: 1. Light exerts pressure. 2. Light can travel through a vacuum as well as through transparent media. 3. Beams of light are bent by gravity. 5. Light reflects from shiny surfaces (both a wave and particle like property). 6. Different colours correspond to different energies of light (both a wave and particle property). 4. Certain colours of light can cause electrons to be emitted from some metal surfaces. This is called the Photoelectric Effect.
  • 24. 2.2 Wave Like Properties of Light • Light exhibits many properties which can be demonstrated by other forms of waves eg. Water waves or waves on springs or strings. • These wavelike properties of light are: 1. Light travels very fast, but its speed does depend upon the medium through which it moves. 2. Light undergoes Reflection according to the Laws of Reflection. 3. Light undergoes Refraction according to the Laws of Refraction. 4. Light undergoes Diffraction when passing through narrow gaps. 5. Light beams can pass through one another unaffected. 6. Different colours of light correspond to different energies. 7. Light forms interference patterns when passed through a pair of closely spaced, narrow slits.
  • 25. Light & Matter Revision Question Type: Q12. Light sometimes behaves as a particle and sometimes as a wave. Which one or more of the following properties does light sometimes show? A. mass B. momentum C. charge D. energy Light Properties
  • 26. Light & Matter Revision Question Type: Q 13: The table below contains some predictions for the behaviour of light incident on a shiny metal sheet. Complete the table by placing a .Y. (Yes) or .N. (No) in the appropriate boxes if the prediction is supported by the wave and/or particle model of light. Some answers have already been provided. It is possible for predictions to be supported by both models Y N N Y Models for Light
  • 27. Chapter 3 Topics covered: • The Photoelectric Effect. • Electron Energies • Energy Units • Practical Applications
  • 28. 3.0 The Photoelectric Effect Einstein called these individual light particles PHOTONS The most important of the particle like properties of light is the Photoelectric Effect. Simply put, the photoelectric effect occurs when light shines on a metal surface causing electrons to be ejected from the metal. There are, however, certain restrictions: (a) Only certain metals respond. (b) Light must be above a certain frequency - the Cut Off Frequency. Albert Einstein won his Nobel Prize for his explanation of this phenomena not for his work on Relativity Louis De Broglie working in the 1920’s suggested a photon looked like this.
  • 29. 3.1 The Photoelectric Effect (2) Incident Photon G Rheostat + Galvanometer e Ejected Electron The photoelectric effect is best studied using the demonstration circuit shown. An incident photon of the right frequency (that is above the Cut Off Frequency) striking a susceptible metal (such as aluminium, copper, lead, zinc and many others), will free an electron. The electron will cross the gap in the evacuated tube, due to electrostatic attraction. So a stream of incident photons striking the metal will release a stream of electrons, setting up a current in the circuit which will be detected by the galvanometer (a sensitive ammeter able to detect currents in the μA range). Evacuated Glass Tube +
  • 30. 3.2 The Photoelectric Effect (3) The size of the current DOES depend upon the NUMBER of photons striking the plate. The NUMBER of photons depends upon the INTENSITY of the incident light. The current (of ejected photoelectrons) does NOT depend upon the VOLTAGE between the plates. Current (I) Voltage (V) Bright Light Dim Light Low Intensity or Dim Light of a particular colour or frequency will produce a small current. Evacuated Glass Tube + Incident Photon e Ejected Electron High Intensity or Bright Light of the same frequency will produce a large current.
  • 31. 3.3 The Photoelectric Effect (4) Evacuated Glass Tube + G Rheostat + Galvanometer All e- still have enough energy to cross the gap Evacuated Glass Tube + + G Rheostat + Galvanometer Some e- now don’t have enough energy to cross the gap G Rheostat + Galvanometer Evacuated Glass Tube + + + + No more electrons cross the gap thus Current = 0 The slider is again moved until no electrons have enough energy to cross the gap and the current drops to zero. This increasing voltage stops the least energetic electrons crossing the gap, reducing the current in the circuit. Moving the slider on the rheostat increases the voltage between the plates. The power supply has been reversed, reversing the polarity of the plates. To study the energy possessed by the electrons ejected from the metal the circuit is changed. VC Current (I) Voltage (V) At this point the “Cut Off Voltage” (VC) has been reached.
  • 32. -Vc The fact that the current falls to zero only slowly as the reverse voltage increases indicates the photoelectrons are ejected from the metal surface with VARYING AMOUNTS OF ENERGY. Different Intensities of the same colour (frequency) have the same cut off voltages Different frequencies and/or different metals will have differing cut off voltages 3.4 The Photoelectric Effect (5) Current (I) Voltage (V)
  • 33. Light & Matter Revision Question Type: A student carries out an experiment to investigate the photoelectric effect. She shines a monochromatic light onto a metal plate (P) inside a sealed glass chamber as shown in Figure 2. The current in the circuit changes as the voltage is varied as shown in Figure 3. Q14: What is the maximum kinetic energy (in eV) of the electrons ejected from the plate P? 1.7 eV Photoelectric effect Q15: What is the subsequent maximum speed of these electrons ejected from plate P? KE = ½ mv2 v = 7.73 x 105 ms-1 KEMAX = 1.7 eV = (1.7)(1.6 x 10-19 ) J = 2.72 x 10-19 J
  • 34. Light & Matter Revision Question Type: Blue light of wavelength 360 nm was incident on a potassium metal plate. The photo electrons emitted from the potassium plate were collected by a cathode and anode at varying voltages to obtain the curve in Figure 3. Potassium has a work function of 2.3 eV. Q16: Determine the threshold frequency of potassium. W = hfo fo = W/h = 2.3/4.14 x 10-15 = 5.56 x 1014 Hz Photoelectric effect Q17: UV light of 200 nm was now shone onto the potassium plate at the same intensity striking the cathode. Sketch the resulting curve on the graph in Figure 3 above. Q18: Which of the following (A-D) would occur if the frequency was decreased to less than the threshold frequency? A Increased photocurrent B Decreased photocurrent C Lower stopping potential D No signal.
  • 35. 3.5 Electron Energy When an electron is placed in a region where a voltage difference exists (ie. an electric field), When initially placed into the field, the electron will possess Electrical Potential Energy given by the product of the charge on the electron (q) and the size of the voltage difference or accelerating voltage (V) So, E.P.E. = qV Before proceeding with more on the Photoelectric Effect a short detour into the world of electron energy is required. Plate 1, V = 0 Plate 2, V = +V EPE KE = 0 e - EPE KE e - KE EPE e - EPE = 0 KE e - EPE KE e - it will undergo an energy conversion in passing through that voltage difference. At this point the electron is stationary so its KE =0 The electron arrives at Plate 2 where all its initial E.P.E. has been converted to K.E. Thus E.P.E.at start = K.E.at finish Mathematically: qV = ½mv2
  • 36. 3.6 Energy Units • Electrons are tiny and the amount of energy they carry, even when exposed to accelerating voltages of tens of millions of volts, is also tiny. • In order to deal with this situation the normal energy unit of Joules (a very large unit) can be replaced by a special energy unit called the “electron-Volt” (eV). • The electron-Volt is defined as the energy change experienced by an electron in passing through a Voltage of 1 Volt. • Mathematically: E.P.E. = qV = (1.6 x 10-19 )(1) = 1.6 x 10-19 J Thus 1 eV = 1.6 x 10-19 J Plate 1, V = 0 Plate 2, V = +1V e - K Ee - EPE At Plate 1 the E.P.E. of the e- = qV = 1.6 x 10-19 J = 1 eV On arrival at Plate 2 the K.E. of the e- has increased by 1 eV and an electron passing through 10 million volts will have increased its energy by 10 MeV Thus an electron passing through 1 thousand volts will have increased its energy by 1keV
  • 37. 3.7 The Photoelectric Effect (6) • The ENERGY of the incoming photons depends upon the FREQUENCY OF THE LIGHT. So, blue photons are more energetic than red ones. • Mathematically: E = hf where E = Energy (J or eV) h = Planck’s Constant f = Frequency (Hz) • Planck’s constant can have 2 values depending upon the energy units used: If Joules: h = 6.63 x 10-34 J s; if eV: h = 4.14 x 10-15 eVs Some of the photon energy is used free the electron from the metal lattice and the rest is converted into the Kinetic Energy of the ejected electron. This can be summarised mathematically as: KEMAX = hf - W where: KEMAX = Maximum KE of ejected electron (J or eV) hf = Photon Energy (J or eV) W = Work Function (J or eV) The term, W, called the Work Function, is the energy needed to bring the electron from within the metal lattice to the surface before it can be ejected. The Energy carried by the incoming Photon is TOTALLY ABSORBED by the electron with which it collides.
  • 38. 3.8 The Photoelectric Effect (7) 5. All metals will produce graphs with the same slope = h (Planck’s constant) 4. Different metals have different work functions. 3. Different metals have different cut off frequencies. 2. When KEMAX = 0, hf = W and this frequency is called the “cut off frequency” (fO) for that particular metal. 1. Remember KE MAX = hf - W, thus a graph of KE MAX vs frequency is a straight line graph of the form y = mx + c with m = h and c = W A plot of the KEMAX of the ejected electron versus the frequency of the incoming photons is shown below. This graph has a number of important features: K.E.MAX (eV or J ) Frequency (Hz) W fo Slope = h fo Slope = h W Light below this frequency, no matter how intense, will never liberate electrons from this metal.
  • 39. Light & Matter Revision Question Type: Susan and Peter conducted a photo- electric experiment in which they used a light source and various filters to allow light of different frequencies to fall on the metal plate of a photo-electric cell. The maximum kinetic energy of any emitted photo-electrons was determined by measuring the voltage required, VS (stopping voltage), to just stop them reaching the collector electrode. The apparatus is shown in Figure 2. Figure 3 shows the stopping voltage, VS, as a function of the frequency (f) of the light falling on the plate. Photoelectric effect Table 1 shows the work functions for a series of metals.
  • 40. Light & Matter Revision Question Type: Photoelectric effect Q19: Use the information above to identify the metal surface used in Susan and Peter’s experiment. Q20: Use the results in Figure 3 to calculate the value for Planck’s constant that Susan and Peter would have obtained from the data. You must show your working. The work function corresponding to sodium could be found by drawing a line of best fit through the points on the graph to determine the y-intercept. The gradient of the line of best fit gave Planck’s constant as approximately 4.5 x 10-15 eV s.
  • 41. Light & Matter Revision Question Type: Measurements of the kinetic energy of electrons emitted from potassium metal were made at a number of frequencies. The results are shown in Figure 1. Q21: What is the minimum energy of a light photon that can eject an electron from potassium metal? 1.8 eV Q22: Which one of the graphs (A- D) would best describe the result if the experiment was repeated with silver metal instead of potassium metal? The work function for silver metal is higher than the work function for potassium metal. Photoelectric effect
  • 42. Light & Matter Revision Question Type: Some students are investigating the photoelectric effect. They shine light of different wavelengths onto a rubidium plate. They measure the maximum kinetic energy of photoelectrons emitted from the plate. Their data of maximum kinetic energy of ejected photoelectrons as a function of the frequency of incident light is shown in Figure 1. In answering the following questions, you must use the data from the graph. Take the speed of light to be 3.0 × 108 m s-1 Q23: From the data on the graph, what is the minimum energy, W, required to remove photoelectrons from the rubidium plate? 2.0 eV Photoelectric effect The students shine light of wavelength λ = 400 nm onto the rubidium plate. Q24: From the graph, with what maximum kinetic energy would the photoelectrons be emitted? 1.0 eV
  • 43. 3.9 The Photoelectric Effect (8) PHOTOELECTRIC EFFECTS OBSERVED WAVE MODEL PREDICTIONS 3. The photoelectrons are emitted IMMEDIATELY the metal is exposed to light above the cut off frequency. 3. Because of the nature of the energy delivery process, electrons should take SOME TIME to build enough energy to be ejected. A TIME DELAY should exist. 2. The KEmax of the ejected photoelectrons is INDEPENDENT of the INTENSITY of the incident light, but DEPENDENT on the FREQUENCY of that light. 2. The energy of the light beam is completely described by its INTENSITY, so the KEMAX of the ejected photoelectrons should be INDEPENDENT of Frequency. These anomalies between the observed and predicted results meant the death of the wave theory as a complete explanation for light behaviour. A new theory was required. 1. The energy of the light beam arrives uniformly and continuously at the metal surface. If the intensity increases, the energy available to the electrons increases, so KEMAX of the ejected electrons should also increase. 1. Below a certain frequency, different for different metals, NO photoelectrons are observed, no matter how INTENSE the light.
  • 44. Light & Matter Revision Question Type: Photoelectric effect The students use a light source that emits a large range of frequencies. They use filters which allow only certain frequencies from the source to shine onto the plate. Most of the students filters produce frequencies below the cut-off frequency. Alice says that if they increase the intensity of light, these frequencies below the cut-off frequency will be able to produce emitted photoelectrons. They experiment and find Alice is incorrect. Q25: Comment whether this experimental evidence supports the wave like or the particle-like theory of light. The experimental evidence supported the particle theory. The wave theory predicts that photoelectrons would be emitted at any frequency if the intensity was sufficient. the particle theory, the energy of the photons is related to the frequency, not the intensity, so there would be no electrons emitted below the threshold frequency.
  • 45. 3.10 The Photoelectric Effect (9) Individual Photon Light beam Light Beams can be regarded as a series of photons radiating out from the source. As mentioned previously it was in 1905 that Einstein proposed that light travels as a series of discrete units or particles or “quanta” which he called “photons” and each carried a discrete amount of energy dependent upon the light’s frequency. (ie E = hf). Photons are neither waves nor particles, having properties similar to particles when travelling through a vacuum and when in a gravitational field, while also having properties similar to waves when refracting and interfering. Photons can be visualised as a series of individual particles each of which displays some wave like properties.
  • 46. 3.11 Practical Applications The photoelectric effect has practical applications in many areas: Solar Cells Night Vision Goggles Spacecraft The photoelectric effect will cause spacecraft exposed to sunlight to develop a positive charge. This can get up to the tens of volts. This can be a major problem, as other parts of the spacecraft in shadow develop a negative charge (up to several kilovolts) from nearby plasma, and the imbalance can discharge through delicate electrical components.
  • 47. Chapter 4 Topics covered: • Investigating the Electron • Electrons and Matter as Waves. • Electron Interference • Electron Diffraction • Electrons and X Rays • Photon Momentum. • Photons as Waves.
  • 48. 4.0 Investigating the Electron • The electron is the smallest of the 3 fundamental particles which make up all atoms. • It carries the basic unit of electric charge (1.6 x 10-19 C) • It has a mass of 9.1 x 10-31 kg • When in an atom, the electron circulates around the nucleus BUT only in certain allowed positions or orbits. • This restriction on position means that electrons are behaving more like waves than the particles we know them to be. • It was not until the development of “Quantum Mechanics” that the problem of restricted position was satisfactorily explained. • Electrons are stripped from atoms and then used as individual particles in many applications. • One important application is in Cathode Ray Tubes - the basis for Cathode Ray Oscilloscopes (CRO’s), T.V. and non LCD video display units. • These devices rely on a stream of “energetic” electrons, which are “boiled off” a hot wire filament. • Their energy is then increased by being accelerated through a large Voltage. • Their trajectory can be manipulated by using magnetic or electric fields.
  • 49. 4.1 Electrons and Matter as Waves • DeBroglie suggested that matter (with mass, m, moving with velocity, v), will have a wavelength (called the DeBroglie wavelength) associated with its motion. • This wavelength is calculated from the momentum equation: λ = h/p = h/mv where λ = DeBroglie wavelength (m) h = Planck’s Constant p = Particle Momentum (kgms-1 ) m = Mass (kg) v = Velocity (ms-1 ) A racing car, mass 800 kg, travelling at 200 kmh-1 (55.6 ms-1 ) has a DeBroglie wavelength of: λ = h/p = h/mv = 6.63 x 10-34 /(800)(55.6) = 1.5 x 10-38 m An indescribably small number. Too small to measure or even notice. An electron of mass 9.1 x 10-31 kg travelling at 1.4 x 104 ms-1 has a DeBroglie wavelength of: λ = h/p = h/mv = 6.63 x 10-34 /(9.1 x 10-31 )(1.4 x 104 ) = 5.2 x 10-8 m A measurable quantity, so electrons are able to display wave like properties like interference and diffraction. v In 1924 French Physicist Louis DeBroglie suggested that “the universe is symmetrical” so, if radiation (light) displayed both particle and wave like properties simultaneously, matter should also display similar behaviours.
  • 50. Light & Matter Revision Question Type: A sketch of a cathode ray tube (CRT) is shown in Figure 5. In this device, electrons of mass 9.10 × 10-31 kg are accelerated to a velocity of 2.0 × 107 m s-1 . A fine wire mesh in which the gap between the wires is w = 0.50 mm has been placed in the path of the electrons, and the pattern produced is observed on the fluorescent screen. Q26: Calculate the de Broglie wavelength of the electrons. You must show your working. λ= h/mv the de Broglie wavelength was found to be 3.64 x 10-11 m. De Broglie Wavelength Q27: Explain, with reasons, whether or not the students would observe an electron diffraction pattern on the fluorescent screen due to the presence of the mesh. For diffraction, the gap width must be the same order of magnitude as the wavelength. Since the wavelength was much smaller than the gap, no diffraction pattern would be observed.
  • 51. Light & Matter Revision Question Type: Neutrons are subatomic particles and, like electrons, can exhibit both particle-like and wave-like behaviour. A nuclear reactor can be used to produce a beam of neutrons, which can then be used in experiments. The neutron has a mass of 1.67 × 10-27 kg. The neutrons have a de Broglie wavelength of 2.0 × 10-10 m. Q28: Calculate the speed of the neutrons. Q29: The neutron beam is projected onto a metal crystal with interatomic spacing of 3.0 × 10-10 m. Would you expect to observe a diffraction pattern? Explain your answer. λ= h/mv gives 2.0 × 103 ms–1 (or 1985 ms–1 ) There would be a diffraction pattern because the wavelength is of the same order of magnitude as the interatomic spacing. De Broglie Wavelength
  • 52. 3.11 Practical Applications The photoelectric effect has practical applications in many areas: Solar Cells Night Vision Goggles Spacecraft The photoelectric effect will cause spacecraft exposed to sunlight to develop a positive charge. This can get up to the tens of volts. This can be a major problem, as other parts of the spacecraft in shadow develop a negative charge (up to several kilovolts) from nearby plasma, and the imbalance can discharge through delicate electrical components.
  • 53. 4.2 Electron Interference The now familiar Young’s double slit experiment is performed with the light source replaced by an “electron gun” which fires a single electron at a time. 8 electrons 270 electrons 2000 electrons 6000. electrons Initially, with only small numbers, electrons arrive at the screen in a random fashion. . ... . Even as the numbers reach the thousands still nothing unusual appears on the screen. However as the numbers climb past 5000 the familiar light and dark bands begin to appear The total exposure time from picture (a) to the picture (d) was 20 min. This result shows particles (in this case electrons) can display wave like properties.
  • 54. 4.3 Electron Diffraction Just as waves diffract when hitting a solid object so electrons can diffract from a well-ordered arrangement of atoms on the surface of a sample. Electrons which have been accelerated through a potential of 30 to 500 volts (i.e., have K.E.’s of 30 to 500 eV), have a de Broglie wavelength between 2.2 x 10-10 m and 0.5 x 10-10 m. This fits nicely into the range of distances between atoms in solids and can therefore strongly diffract from them. The arrangement of the spots is interpreted to provide information about the ordered arrangement of atoms on the surface and the distances between the spots gives information on the distance between the atoms. Platinum Sample e- energy = 65 eV
  • 55. Light & Matter Revision Question Type: A beam of electrons, pass through two very thin slits (approximately 10-10 m) and are detected on a electron detector screen as shown in Figure 5. Q30: Explain whether the particle model or the wave model best explains the expected observations from this experiment. Electron Diffraction The wave model best explains this. The wave model is better: Interference pattern (a wave phenomenon) will be observed. The Particle model would not predict the interference pattern, rather two zones where the electrons would strike the screen.
  • 56. 4.4 Electrons and X Rays High energy electron beam Al foil Diffraction Pattern A high energy electron beam is fired at an Aluminium target X Ray Beam When the electron beam is replaced with an X Ray beam a somewhat similar diffraction is produced The distance between the bright lines in both patterns is the same, meaning the wavelengths of both beams was the same.The diffraction pattern shown is produced. Hence if λX Ray is known, then the De Broglie wavelength of the electrons is also known.
  • 57. Light & Matter Revision Question Type: A beam of X-rays, wavelength λ = 250 pm (250 × 10-12 m), is directed onto a thin aluminium foil as shown in Figure 4a. The X-rays scatter from the foil onto the photographic film. Q31 : Calculate the energy, in keV, of these X-rays. E=hc/λ = (4.14 x 10-15)(3.0 x 108 )/ 250 x 10-12 ) = 4.97 x 103 ev = 4.97 keV or 5.0 keV X ray Energy
  • 58. After the X-rays pass through the foil, a diffraction pattern is formed as shown in Figure 4b. In a later experiment, the X-rays are replaced with a beam of energetic electrons. Again, a diffraction pattern is observed which is very similar to the X-ray diffraction pattern. This is shown in Figure 4c. Light & Matter Revision Question Type: X ray Diffraction Electrons have a de Broglie wavelength which was the same (or very similar) to the wavelength of the X-rays. Q32: Explain why the electrons produce a diffraction pattern similar to that of the X-rays. Q33: Assuming the two diffraction patterns are identical, estimate the momentum of the electrons. Include the unit. p = h / λ = (6.63 x 10-34)/(250 x 10-12 ) = 2.7 x 10-24 kgms-1 The diffraction pattern were the same, so the wavelength of the electrons must equal that of the X- rays.
  • 59. Light & Matter Revision Question Type: In 1927, G.I. Thomson fired a beam of electrons through a very thin metal foil and the emerging electrons formed circular patterns, similar to the pattern obtained when the exercise was repeated with a beam of X-rays. The following diagram shows a comparison of the patterns obtained. Q34: How does this evidence support the concept of matter waves? Q35: If the X-rays have frequency of 1.5 x 1019 Hz, what is the wavelength of the electrons? Q36: What is the momentum of an X- ray photon? The circular pattern is a diffraction pattern. With X-rays, the bright lines represent regions of constructive interference alternating with regions of destructive interference. The corresponding electron pattern shows similar diffraction properties which suggests electrons have wave properties, as diffraction is a wave phenomenon. X ray & electron Diffraction λ = c/f = 3.0 x 108 /1.5 x 1019 = 2.0 x 10-11 m p = h/λ = 6.63 x 10-34 /2.0 x 10-11 = 3.3 x 10-23 kgms-1
  • 60. 4.5 Evidence for Photons as Waves Young’s double slit experiment is well known as evidence for the wave like nature of light. This experiment creates difficulties for the particle (photon) model. How can individual photons, which must pass through one or other of the slits, interact with themselves to produce an interference pattern ? When the experiment is carried out at extremely low light intensities, it is found that, for a while, no interference pattern is noticed. However, as time passes the number of photons arriving at the screen builds and an interference pattern does emerge with more photons going to the areas of bright bands and less to areas of dark bands. It appears trying to predict the fate of individual photons is not possible, but as their numbers build the wave model allows the prediction of their average behaviour with great accuracy. Thus photons (en masse) can display wave like properties although they are generally regarded as discrete, individual particles.
  • 61. 4.6 Photon Momentum • When a light beam strikes a surface, the individual photons will transfer MOMENTUM as they are absorbed or reflected and so will exert a pressure on that surface. • The size of the MOMENTUM can be calculated from p = E/c …………(1) where, p = Photon Momentum (kgms-1 ) E = Photon Energy (J or eV) c = Speed of Light (ms-1 ) • Using Photon Energy, E = hf, Equation 1 can be rewritten as: p = hf/c …………(2) • Using the Wave Equation, c = fλ • Equation 2 can be rewritten as: p = hf/f λ p = h/λ
  • 62. Light & Matter Revision Question Type: A monochromatic violet light of wavelength 390 nm is emitted by a light source with a power of 200 Watt. Q37: How many photons leave this light source in a 10 second interval? p = h/λ = 1.7 x 10-27 kgms-1 Q38: What is the momentum of each of these photons? P = W/t = E/t  E = Pt = (200)(10) = 2000 J Each photon has Energy = hc/λ = = (6.63 x 10-34 )(3.0 x 108 )/(390 x 10-9 ) = 5.1 x 10-19 J No of photons in 2000 J = 2000/(5.1 x 10-19 ) = 3.92 x 1021 Photon energy
  • 63. Chapter 5 • Topics covered: • Quantised Energy Levels for Atoms. • Emission Spectra • Absorption Spectra • Solar Spectrum. • Quantised Energy Levels • Wave Particle Duality.
  • 64. 5.0 Quantised Energy Levels for Atoms • The Bohr model for the atom, restricting electrons to certain regions or orbits with fixed energy levels, worked well in describing atomic structure and behaviour. • But it was not until DeBroglie suggested electrons had a wavelength, was there a satisfactory explanation of why fixed orbits and energies existed. • DeBroglie suggested that for an electron to survive in an orbit, it must form a standing wave which represents a stable state for the electron. • Only certain radius orbits allow the standing wave to be set up, so electrons can only exist certain regions. Allowed Orbit Disallowed orbit The 1st five electron shells with their standing waves
  • 65. 5.1 Emission Line Spectra Sodium Emission Spectra When atoms of substances are exposed to certain kinds of energy sources, the electrons surrounding the nucleus can gain energy and move up to higher energy levels, called “excited states”. The energy absorbed corresponds to jumps between the fixed energy levels of the electron shells within the atom The excited electrons return to their ground state by emitting a photon of a certain frequency (colour) The emitted photons from all electrons transitions form “a picture” called an “Emission Spectrum”. These Spectra are unique to each element and are used to identify unknowns in a field called spectroscopy. Helium Emission Spectra
  • 66. Light & Matter Revision Question Type: i Continuous (or broad) ii Discrete (or individual lines) A class looks at spectra from two sources. i. an incandescent light globe ii. a mercury vapour lamp They observe that the spectra are of different types. Q39: State the type of spectrum seen from each source. i. Incandescent light globe ii. Mercury vapour lamp Emission Spectra Q40: State the electron mechanism, in each of the sources below, that produces each spectrum. i. Incandescent light globe ii. Mercury vapour lamp i Thermal motion of free electrons ii Energy transition of bound electrons
  • 67. 5.2 Absorption Line Spectra If you look at the spectrum from a common incandescent light bulb, you'll see a relatively smooth spectrum with no part being brighter or darker than any other part. This type of a spectrum is called a continuous spectrum. Electrons in an atom can jump between discreet energy levels by absorbing a photon. So certain atoms under certain conditions will absorb very specific wavelengths of light dependent on the configuration of their electrons. This will remove energy and leave blank spaces (dark lines) in the spectrum. An emission spectrum is simply the reverse of an absorption spectrum. The pattern of absorption lines in a spectrum will be unique to a chemical element so we can use absorption lines to detect the presence of specific elements in astronomical objects.
  • 68. Light & Matter Revision Question Type: Figure 5a shows part of the emission spectrum of hydrogen in more detail. With a spectroscope, Val examines the spectrum of light from the sun. The spectrum is continuous, with colours ranging from red to violet. However there were black lines in the spectrum, as shown in Figure 5b. Q41: Explain why these dark lines are present in the spectrum from the sun. The dark lines exist because photons are absorbed which correspond to the energy levels in hydrogen. This indicates the presence of hydrogen. Absorption Spectra
  • 69. 5.3 The Solar Spectrum In 1802 English chemist William Wollaston was the first to note the appearance of a number of dark lines in the solar spectrum. It was later discovered that each chemical element was associated with a set of spectral lines, and that the dark lines in the solar spectrum were caused by absorption by those elements in the upper layers of the sun. Some of the observed features are also caused by absorption in oxygen molecules in Earth’s atmosphere. Solar Spectrum When the “White Light” arriving at the Earth’s surface is passed through a prism, it is broken up into its constituent colours. This produces the “solar spectrum” In 1814, Joseph Von Fraunhofer independently rediscovered the lines and began a systematic study and careful measurement of the wavelength of these features. In all, he mapped over 570 lines.
  • 70. 5.4 Quantised Energy Levels Mercury Energy Levels Ground State n = 10 eV n = 24.9 eV n = 36.7 eV 8.8 eV n = 4 n = 510.4 eV Ionisation The Ground State (n = 1) represents the lowest electron energy level States n = 2 to n = 5 represent allowed energy levels for electrons in Mercury atoms. Electrons absorbing more than 10.4 eV of energy from a photon collision will be stripped completely from the Mercury atom leaving it ionised Electrons in excited state n = 3 can return to the ground state by: 1. Emitting a single 6. 7 eV photon OR 2. By emitting a 1.8 eV photon followed by a 4.9 eV photon Electrons surrounding the nucleus can “jump” up and down between allowed energy levels by absorbing or emitting a photon. The ABSORPTION and EMISSION SPECTRA for atoms show the energy values associated with the “jumps”. The energy is associated with these “jumps” is given by E = hf where E = Energy of the emitted photon for a downward jump OR the energy absorbed from an incident photon for an upward jump. (eV) h = Planck’s Constant f = Frequency of Photon (Hz)
  • 71. Light & Matter Revision Question Type: The spectrum of photons emitted by excited atoms is being investigated. Shown in Figure 6 is the atomic energy level diagram of the particular atom being studied. Although most of the atoms are in the ground state, some atoms are known to be in n = 2 and n = 3 excited states. Q42: What is the lowest energy photon that could be emitted from the excited atoms? The smallest energy gap was that from n = 3 to n = 2, an energy difference of 1.8 eV. Energy Levels Q43: Calculate the wavelength of the photon emitted when the atom changes from the n = 2 state to the ground state (n = 1). Data: h = 4.14 × 10-15 eV s , c = 3.0 × 108 m s-1 The energy of the photon was 3.4 eV. By substituting this into the equation E =hc/λ, the wavelength was 3.65 x 10-7 m.
  • 72. Light & Matter Revision Question Type: Figure 2 shows the energy levels of a sodium atom. A sodium atom is initially in an n = 4 excited state. Q44: Calculate the highest frequency of light that this sodium atom could emit. ΔE = 3.61 = h f Hence f = 8.7 x 1014 Hz Energy Levels Figure 2 shows that electrons in a sodium atom can only occupy specific energy levels. Q45: Describe how the wave nature of electrons can explain this. Electrons have an equivalent or de Broglie wavelength. Only orbits of an integral number of wavelengths are allowed as these will form a standing wave
  • 73. Light & Matter Revision Question Type: Part of the visible region of the spectrum of light emitted from excited hydrogen gas has three lines as shown in Figure 3. The energy level diagram for the hydrogen atom is shown in Figure 4. The binding energy is 13.6 eV. Q46: What is the energy of the photons with a wavelength 434.1 nm in Figure 3? E = hc/λ = 2.86 eV Q47: A different photon has an energy of 3.0 eV. On Figure 4 indicate with an arrow the electron transition that leads to emission of a photon of light with this energy. Energy Level Diagrams
  • 74. 5.5 The Wave - Particle Duality • As you can see from what has been presented, attempting to characterise light as only wave like or only particle like in nature has failed. • In fact, as of today, light (and matter) are regarded as some part particle like and some part wave like in nature. • This conflict is summarised in the use of the term “wave-particle duality” to describe light’s known behaviour. • This is an unsatisfactory situation, as it defies our need for simple all encompassing explanations for the behaviours we observe around us and in the universe. • It would appear, at this stage, we have not yet unravelled all the details of the behaviour of light and matter. BUT THIS IS WHAT PHYSICS IS ALL ABOUT – PUSHING THE ENVELOPE – EXPLORING THE UNKNOWN – TRYING TO FIT RATIONAL AND LOGICAL EXPLANATIONS TO THAT WHICH HAS BEEN OBSERVED.
  • 75. Ollie Leitl 2005

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