Vu4 light&matter2009


Published on

1 Comment
No Downloads
Total Views
On Slideshare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Vu4 light&matter2009

  1. 1. VCE Physics Unit 4 Topic 2Interactions of Light & Matter
  2. 2. Unit OutlineTo 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 Ek max = 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. 3. Chapter 1Topics covered:• The Nature of Light.• Interference.• Incoherent Light.• Coherent Light• Young’s Experiment.• Path Difference• Single Slit Diffraction• Diffraction around Objects
  4. 4. 1.0 The Nature of Light EMR is a self propagating waveLight is a form of ENERGY. consisting of mutuallyIt is described as ELECTRO - perpendicular, varyingMAGNETIC RADIATION (EMR). ELECTRIC and MAGNETIC FIELDS. EMR travels through a vacuum at 300,000 kms-1, (3.0 x 108 ms-1) Changing Magnetic Field Direction of Electromagnetic Wave Movement Changing Electric Field
  5. 5. 1.1 Superposition Single waves, called Once the pulses. have the ability superposition is to pass through one complete the pulses another and, while continue their occupying the same journey unaffected. space, add together in a process called A series of pulses SUPERPOSITION. together form a Trough Crest wave train with alternating crestsWhen two wave trains interact and troughswith one another they also Constructive interference occurs whenundergo SUPERPOSITION and the two wave trains are in phasewill either:1. Add together to produce alarger wave - a process calledCONSTRUCTIVE INTERFERENCE or2. Subtract from one another to Destructive interference occurs whenproduce no wave – a process called the two wave trains are 1800 out ofDESTRUCTIVE INTERFERENCE phase
  6. 6. 1.2 Interference Light behaves in a similar manner. When light (with certain properties) Destructive is passed through two narrow slits, Interference an “interference pattern” is Constructive produced, showing constructive Interference (light bands) and destructive (dark bands) interference. Bright bands occur where a crest and aWaves, in this case water crest (or a trough and awaves, when passed trough) arrive at thethrough two narrow slits, screen at the same time“interfere” or interact withone another to produceareas of large disturbances Dark bands Crest Bright(Constructive Interference) Band occur when a Troughor areas of no disturbance crest and a(Destructive Interference) trough arrive at Dark the screen at Band the same time
  7. 7. 1.3 Incoherent LightLight is generated by luminous bodies, eg,The Sun, light globes, burning candles.Light is produced when atoms of thefilaments or source become electricallyexcited and produce an electromagneticor light wave. Since the excitations occur in an unpredictably random fashion, theTypical light sources such as those light waves are NOT produced inmentioned above have an inbuilt regular repeating manner and so doirregularity in the way they produce light. not maintain a constant “phase relationship” with one other. About once every 10-8 sec, a source will randomly alter its phase. Incoherent light, when combined,This leads to the sources giving off a produces rapidly moving areas ofbroad spectrum of white light composed constructive and destructiveof all colours in the rainbow. interference and therefore do notEach of the millions of colours have produce a stable, visible interferencewaves that are random to each other. pattern.This is called INCOHERENT LIGHT.
  8. 8. Light & Matter RevisionQuestion Type: Incandescent Light 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.
  9. 9. Light & Matter RevisionQuestion Type: Incandescent Light The spectrum of wavelengths produced by a particular incandescent light globe is shown in Figure 1 below. Q2: The light produced by an incandescent light globe can best be described as A. coherent. B. incoherent. C. monochromatic. D. in phase. Q3: Describe the mechanism by which light is produced in an incandescent light globe. The thermal/random excitation of electrons in the filament leads to the emission of a broad/continuous spectrum.
  10. 10. 1.4 Coherent Light Coherence is a property of waves that measures their ability to interfere with Coherent waves (zero phase each other. difference)Shown are monochromatic (singlecolour) light waves of the samefrequency. Coherent waves (constantThey are coherent and in phase, and will phase difference)combine constructively to produce bright Two waves that are coherent canwhite light 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.Lasers generate light at a singlewave length and frequency and allof the waves (and PHOTONS) are inphase.This is called COHERENT LIGHT.
  11. 11. Light & Matter RevisionQuestion Type: Coherent Light Q4: Which of the following light sources will produce coherent, monochromatic light A. Sunlight B. LED C. Light globe with filter D. LASER
  12. 12. 1.5 Young’s Experiment First performed by This pattern has a series of equally Thomas Young Thomas Young in the spaced coloured and black bands 1773 – 1829 early 1800’s this spread across the screen onto experiment proved which it is projected. light was a wave. The width of the coloured bands It has been voted the and their spacing depends on most elegant Incident Light the wavelength of the light used. experiment ever Short wavelength, BLUE light devised. produces a pattern with narrow blueWhen light of a single frequency bands which are closely spaced.(colour) is passed through a pair Long wavelength, RED lightof closely spaced, narrow slits produces a pattern with wideran “interference pattern” is red bands which are spreadproduced. farther apart. BLUE Light using the same slits Screen Double Slits RED Light
  13. 13. Light & Matter RevisionQuestion Type: Young’s Experiment Thomas Young’s double slit experiment The slits are now moved further has been replicated in the experimental from the screen. arrangement shown in Figure 4. Q6: What effect would this have on the pattern observed on the screen? The pattern will spread further across the screen. Q5: Explain using the wave theory of light why a series of bright and dark bars are 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
  14. 14. Light & Matter RevisionQuestion Type: Young’s Experiment 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
  15. 15. 1.6 Path Difference Light travels the same distance from S1 and S2 to reach the central bright band, BC The path difference The dark band (D1) on S1 S2 S1BC – S2BC = 0. either side of AC occurs So a crest and a crest because a crest and a (or trough and trough) trough are arriving at the arrive at AC at the same same time leading to time, leading to destructive interference. constructive Path difference: interference or a bright S1D1 – S2D1 = ½ λ band The next dark band (D2) Slits For the next bright has path difference = 1½λ S1 S2 band path difference S1B1 – S2B1 = 1 λ For the next bright So dark bands occur band path difference when would be: path difference = (n + ½)λ S1B2 – S2B2 = 2λ where n = So bright bands occur 0,1,2,3, etc when path difference = nλ where n = 0,1,2,3,etcD2 B1 D1 B C D1 B1 D2
  16. 16. Light & Matter RevisionQuestion Type: Interference – Path DifferenceIn the following diagram, laser light ofwavelength 600 nm is shone onto a pair of Q9: Estimate the difference inparallel slits and a pattern of alternating length between P1 and P2.light/dark bands is projected onto a wall. The indicated position is on the sixth antinode from the centre. The path difference is 6λ = 6 × 600nm = 3.6 ×10−6 mQ8: Explain how the observed patternon the wall supports a wave model forlight.The alternating pattern represents an interference pattern which is a wavephenomenon. The bright bands represent regions of constructive interferencewhere the difference in path length from each source is a multiple ofwavelengths. Waves therefore arrive from both sources in phase. The darkregions represent bands of destructive interference with the waves being halfa wavelength out of phase due to path differences.
  17. 17. Light & Matter RevisionQuestion Type: Interference – Path Difference 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 cmStudents have set up an experiment similarto that of English physicist Thomas Young.The students’ experiment uses microwavesof wavelength λ = 2.8 cm instead of light.The beam of microwaves passes throughtwo narrow slits shown as S1 and S2 inFigure 3.The students measure the intensity of theresulting beam at points along the lineshown and determine the positions ofmaximum intensity. These are shown asfilled circles and marked P0, P1 . . .
  18. 18. 1.7 Single Slit Diffraction Appreciable diffraction will occur ifWhen light of a single THE EXTENT OF the ratio λ/w is between about 0.1wavelength shines DIFFRACTION and 50. Outside this range,through a narrow gap or DEPENDS UPON diffraction is not observed.single slit, a “diffraction THE RATIO λ /w,pattern” is produced. WHERE λ = Diffraction patterns for blue and WAVELENGTH red light show that the shorterThe pattern consists of AND w = SLIT wavelength blue light produces thea rather wide, coloured WIDTH. more “compact” pattern, while Screencentral maximum with the longer wavelength reda series of thinner produces a more spread outcoloured and dark patternbands spreading out In other words, the spacingfrom the centre. between the lines is Width = w wavelength dependent and Single Slit patterns with the same lineIncident Light spacing were made by lightWavelength = λ of the same wavelength BLUE light using same slit Note: The wider the slit the narrower central maximum RED light
  19. 19. Light & Matter RevisionQuestion Type: Single Slit DiffractionIn an experiment,monochromatic laser light ofwavelength 600 nm shinesthrough a narrow slit, and theintensityof the transmitted light isrecorded on the screen somedistance away as shown inFigure 2a. The intensitypattern seen on the screen isshown in Figure 2b.Q11: Which one of theintensity patterns (A-D) belowbest indicates the pattern thatwould be seen if a wider slitwas used?
  20. 20. 1.8 Diffraction around ObjectsIn addition to diffraction Sunlight diffracting aroundoccurring when light passes a car tail light lensthrough gaps, it can alsooccur when light passesaround objects The shadow of a Red Light hand holding a coin Diffracting around a illuminated by a He – You can see a pinhead Ne Laser. diffraction Point your hand pattern toward a light through your source and view fingers through this gap
  21. 21. Chapter 2Topics covered:• Models for Light Behaviour.• Particle like Properties of Light.• Wave like Properties of Light.
  22. 22. 2.0 Models for Light Behaviour• Light is/has energy and thus Two separate models have can do work. We see this in been proposed: things like plants growing or 1. The Particle Model: the chemical reactions on This says that all properties of photographic films. light can be explained by• It is quite easy to list or show assuming light is made up of what light DOES but more a stream of individual difficult to say what light IS ! particles.• Much time and effort has been put into trying to find a 2. The Wave Model: THEORY or MODEL which This says that all properties of will explain ALL the light can be explained by properties and behaviours assuming light is made up of exhibited by light. a stream of waves.
  23. 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”. 4. Certain colours of light can cause• These particle like properties electrons to be emitted from some are: metal surfaces. This is called the1. Light exerts pressure. Photoelectric Effect.2. Light can travel through a 5. Light reflects from shiny vacuum as well as through surfaces (both a wave and particle like transparent media. property). 6. Different colours correspond to3. Beams of light are bent by different energies of light (both a wave gravity. and particle property).
  24. 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: 4. Light undergoes Diffraction when passing through narrow gaps.1. Light travels very fast, but its speed does depend upon the medium 5. Light beams can pass through one through which it moves. another unaffected.2. Light undergoes Reflection according 6. Different colours of light correspond to the Laws of Reflection. to different energies.3. Light undergoes Refraction according 7. Light forms interference patterns to the Laws of Refraction. when passed through a pair of closely spaced, narrow slits.
  25. 25. Light & Matter RevisionQuestion Type: Light Properties 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
  26. 26. Light & Matter RevisionQuestion Type: Models for LightQ 13: The table below contains some predictions for thebehaviour of light incident on a shiny metal sheet. Complete thetable by placing a .Y. (Yes) or .N. (No) in the appropriate boxes ifthe prediction is supported by the wave and/or particle model oflight. Some answers have already been provided. It is possible forpredictions to be supported by both models Y Y N N
  27. 27. Chapter 3Topics covered:• The Photoelectric Effect.• Electron Energies• Energy Units• Practical Applications
  28. 28. 3.0 The Photoelectric Effect Albert Einstein won his NobelThe most important of the Prize for his explanation of thisparticle like properties of phenomena not for his work onlight is the Photoelectric RelativityEffect.Simply put, the photoelectriceffect occurs when lightshines on a metal surfacecausing electrons to beejected 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. Einstein called these individual light particles PHOTONSLouis De Broglie working in the 1920’ssuggested a photon looked like this.
  29. 29. 3.1 The Photoelectric Effect (2)The photoelectric effect is best studied Incident Photon Evacuatedusing the demonstration circuit shown. Glass Tube eAn incident photon of the right frequency +(that is above the Cut Off Frequency) Ejected Electronstriking a susceptible metal (such asaluminium, copper, lead, zinc and many Galvanometer Gothers), will free an electron. RheostatThe electron will cross the gap in theevacuated tube, due to electrostaticattraction. + 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).
  30. 30. 3.2 The Photoelectric Effect (3)The current (of ejected photoelectrons)does NOT depend upon the VOLTAGEbetween the plates. Incident Photon The size of the current DOES depend Evacuated upon the NUMBER of photons striking Glass Tube e the plate. +The NUMBER of photons depends Ejected Electronupon the INTENSITY of the incidentlight.Low Intensity or Dim Light of a Current (I)particular colour or frequency willproduce a small current. Bright LightHigh Intensity or Bright Light of thesame frequency will produce a large Dim Lightcurrent. Voltage (V)
  31. 31. 3.3 The Photoelectric Effect (4)To study the energy possessed by theelectrons ejected from the metal the Evacuated Evacuated Evacuated Glass Tube + Glass Tube Glass Tube +circuit is changed. + + + + The power supply has been reversed, + reversing the polarity of the plates. Some e-e-electrons crossenough No morenow don’t have the gap All still have enough thus energy to crossgap gap Current = 0 energy to cross the the Galvanometer G Galvanometer Moving the slider on the rheostat increases the voltage between the plates.This increasing voltage stops the least Rheostat Rheostat Rheostatenergetic electrons crossing the gap,reducing the current in the circuit. + +The slider is again moved until no Current (I)electrons have enough energy to crossthe gap and the current drops to zero.At this point the “Cut Off Voltage”(VC) has been reached. Voltage (V) VC
  32. 32. 3.4 The Photoelectric Effect (5) 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 Current (I)colour (frequency) have thesame cut off voltages Different frequencies and/or -Vc Voltage (V) different metals will have differing cut off voltages
  33. 33. Light & Matter RevisionQuestion Type: Photoelectric effectA student carries out an experiment to Q14: What is the maximum kineticinvestigate the photoelectric effect. She energy (in eV) of the electronsshines a monochromatic light onto a ejected from the plate P?metal plate (P) inside a sealed glass 1.7 eVchamber as shown in Figure 2. Q15: What is the subsequent maximum speed of these electrons ejected from plate P? KEMAX = 1.7 eV = (1.7)(1.6 x 10-19) J = 2.72 x 10-19 J KE = ½ mv2 v = 7.73 x 105 ms-1 The current in the circuit changes as the voltage is varied as shown in Figure 3.
  34. 34. Light & Matter RevisionQuestion Type: Photoelectric effectBlue light of wavelength 360 nm was Q16: Determine the thresholdincident on a potassium metal plate. The frequency of electrons emitted from the W = hfopotassium plate were collected by a fo = W/hcathode and anode at varying voltages to = 2.3/4.14 x 10-15obtain the curve in Figure 3. Potassium = 5.56 x 1014 Hzhas a work function of 2.3 eV. 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 photocurrentQ17: UV light of 200 nm was now shone onto the C Lower stoppingpotassium plate at the same intensity striking the potentialcathode. Sketch the resulting curve on the graph in D No signal.Figure 3 above.
  35. 35. 3.5 Electron Energy When initially placed into the field, theBefore proceeding with more on electron will possess Electrical Potentialthe Photoelectric Effect a short Energy given by the product of thedetour into the world of electron charge on the electron (q) and the size ofenergy is required. the voltage difference or acceleratingWhen an electron is placed in a voltage (V) So, E.P.E. = qVregion where a voltage difference At this point the electron is stationaryexists (ie. an electric field), it will so its KE =0undergo an energy conversionin passing through that voltagedifference. The electron arrives at Plate 2 KE KE = 0 KE KE KE where all its initial E.P.E. has e- e- e- e- e- been converted to K.E. Thus EPE EPE EPE EPE start = finish EPE =0 Mathematically: qV = ½mv2 Plate 1, Plate 2, V=0 V = +V
  36. 36. 3.6 Energy Units• Electrons are tiny and the amount of energy they carry, even when exposed to accelerating voltages of tens of K e- E millions of volts, is also tiny. e-• In order to deal with this situation the normal energy unit of Joules (a very large unit) can be replaced by a EPE special energy unit called the “electron-Volt” (eV).• Plate 1, V = 0 Plate 2, V = +1V The electron-Volt is defined as the energy change experienced by an electron in passing through a Voltage On Plate 1 at Plate 2 thethe e-of At arrival the E.P.E. of K.E. of 1 Volt. = qVe- hasx 10-19 J = 1 eV eV the = 1.6 increased by 1• Mathematically: E.P.E. = qV Thus an electron passing through = (1.6 x 10 )(1) -19 1 thousand volts will have = 1.6 x 10-19 J increased its energy by 1keV and Thus 1 eV = 1.6 x 10-19 J an electron passing through 10 million volts will have increased its energy by 10 MeV
  37. 37. 3.7 The Photoelectric Effect (6)• The ENERGY of the incoming photons depends upon the The Energy carried by the incoming FREQUENCY OF THE LIGHT. So, Photon is TOTALLY ABSORBED by blue photons are more energetic the electron with which it collides. than red ones.• Mathematically: Some of the photon energy is used free E = hf the electron from the metal lattice and the where E = Energy (J or eV) rest is converted into the Kinetic Energy h = Planck’s Constant of the ejected electron. This can be f = Frequency (Hz) summarised mathematically as:• Planck’s constant can have 2 values depending upon the energy units KEMAX = hf - W used: If Joules: h = 6.63 x 10-34 J s; where: if eV: h = 4.14 x 10-15 eVs 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.
  38. 38. 3.8 The Photoelectric Effect (7)A plot of the KEMAX 1. Remember KE MAX = hf - W, 3. Different metals haveof the ejected thus a graph of KE MAX vs different cut offelectron versus the frequency is a straight line frequencies.frequency of the graph of the form y = mx + c 4. Different metals haveincoming photons is with m = h and c = W different work functions.shown below. This 2. When KEMAX = 0, hf = Wgraph has a number and this frequency is called 5. All metals will produceof important the “cut off frequency” (fO) graphs with the samefeatures: slope = h (Planck’s for that particular metal. constant) Light below this frequency, no matter how intense, will never liberate electrons from this K.E.MAX metal. (eV or J ) Slope = h Slope = h fo fo Frequency (Hz) W W
  39. 39. Light & Matter RevisionQuestion Type: Photoelectric effectSusan and Peter conducted a photo-electric experiment in which they used alight source and various filtersto allow light of different frequencies to fallon the metal plate of a photo-electric cell.The maximum kinetic energy of any emittedphoto-electrons was determined by Figure 3 shows the stoppingmeasuring the voltage required, VS voltage, VS, as a function of the(stopping voltage), to just stop themreaching the collector electrode. The frequency (f) of the light falling onapparatus is shown in Figure 2. the plate. Table 1 shows the work functions for a series of metals.
  40. 40. Light & Matter RevisionQuestion Type: Photoelectric effectQ19: Use the information above toidentify the metal surface used inSusan and Peter’s experiment.The work function corresponding tosodium could be found by drawing aline of best fit through the points on thegraph to determine the y-intercept.Q20: Use the results in Figure 3 tocalculate the value for Planck’s constantthat Susan and Peter would haveobtained from the data.You must show your working.The gradient of the line of best fit gave Planck’sconstant as approximately 4.5 x 10-15 eV s.
  41. 41. Light & Matter RevisionQuestion Type: Photoelectric effectMeasurements of the kinetic energy ofelectrons emitted from potassiummetal were made at a number offrequencies. The results are shown inFigure 1.Q21: What is the minimum energyof a light photon that can eject anelectron from potassium metal? 1.8 eVThe work function for silvermetal is higher than the workfunction for potassium metal.Q22: Which one of the graphs (A-D) would best describe the resultif the experiment was repeatedwith silver metal instead ofpotassium metal?
  42. 42. Light & Matter RevisionQuestion Type: Photoelectric effectSome students areinvestigating thephotoelectric effect. Theyshine light of differentwavelengths onto arubidium plate. Theymeasure the maximumkinetic energy ofphotoelectrons emittedfrom the plate. Their dataof maximum kinetic energy Q23: From the data on the graph, what is theof ejected photoelectrons minimum energy, W, required to removeas a function of the photoelectrons from the rubidium plate?frequency of incident light 2.0 eVis shown in Figure 1. The students shine light of wavelength λ = 400 nmIn answering the following onto the rubidium plate.questions, you must use Q24: From the graph, with what maximum kineticthe data from the graph. energy would the photoelectrons be emitted?Take the speed of light to 1.0 eVbe 3.0 × 108 m s-1
  43. 43. 3.9 The Photoelectric Effect (8) PHOTOELECTRIC EFFECTS OBSERVED WAVE MODEL PREDICTIONS1. Below a certain frequency, 1. The energy of the light beamdifferent for different metals, NO arrives uniformly and continuously atphotoelectrons are observed, no the metal surface. If the intensitymatter how INTENSE the light. increases, the energy available to the electrons increases, so KEMAX of the ejected electrons should also increase.2. The KEmax of the ejected 2. The energy of the light beam isphotoelectrons is INDEPENDENT completely described by its INTENSITY,of the INTENSITY of the incident so the KEMAX of the ejectedlight, but DEPENDENT on the photoelectrons should be INDEPENDENTFREQUENCY of that light. of Frequency. 3. Because of the nature of the energy3. The photoelectrons are emitted delivery process, electrons should takeIMMEDIATELY the metal is exposed to SOME TIME to build enough energy tolight above the cut off frequency. be ejected. A TIME DELAY should exist.These anomalies between the observed and predicted results meant the death of the wavetheory as a complete explanation for light behaviour. A new theory was required.
  44. 44. Light & Matter RevisionQuestion Type: Photoelectric effect The students use a light source that Q25: Comment whether this emits a large range of frequencies. experimental evidence supports They use filters which allow only the wave like or the particle-like certain frequencies from the source theory of light. to shine onto the plate. Most of the students filters produce frequencies The experimental evidence supported below the cut-off frequency. the particle theory. Alice says that if they increase the The wave theory predicts that intensity of light, these frequencies photoelectrons would be emitted at below the cut-off any frequency if the intensity was frequency will be able to produce sufficient. emitted photoelectrons. They experiment and find Alice is the particle theory, the energy of the incorrect. photons is related to the frequency, not the intensity, so there would be no electrons emitted below the threshold frequency.
  45. 45. 3.10 The Photoelectric Effect (9)As mentioned previously it Photons are neither Photons can bewas in 1905 that Einstein waves nor particles, visualised as a series ofproposed that light travels having properties individual particles eachas a series of discrete units similar to particles when of which displays someor particles or “quanta” travelling through a wave like properties.which he called “photons” vacuum and when in aand each carried a discrete gravitational field, whileamount of energy also having propertiesdependent upon the light’s similar to waves when Individual Photonfrequency. (ie E = hf). refracting and interfering. Light beam Light Beams can be regarded as a series of photons radiating out from the source.
  46. 46. 3.11 Practical ApplicationsThe photoelectric effect haspractical applications in manyareas: Solar Cells 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 Night Vision Goggles 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. 47. Chapter 4Topics covered:• Investigating the Electron• Electrons and Matter as Waves.• Electron Interference• Electron Diffraction• Electrons and X Rays• Photon Momentum.• Photons as Waves.
  48. 48. 4.0 Investigating the Electron • 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• The electron is the smallest of the 3 (CRO’s), T.V. and non LCD video fundamental particles which make display units. up all atoms. • These devices rely on a stream of• It carries the basic unit of electric “energetic” electrons, which are charge (1.6 x 10-19 C) “boiled off” a hot wire filament.• It has a mass of 9.1 x 10-31 kg • Their energy is then increased by• When in an atom, the electron being accelerated through a large circulates around the nucleus BUT Voltage. only in certain allowed positions or • Their trajectory can be manipulated orbits. by using magnetic or electric fields.• 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.
  49. 49. 4.1 Electrons and Matter as Waves In 1924 French Physicist Louis DeBroglie suggested that “the universe is symmetrical” so, if radiation (light) A racing car, mass 800 kg, travelling at displayed both particle and wave 200 kmh-1 (55.6 ms-1) has a DeBroglie like properties simultaneously, wavelength of: matter should also display similar λ = h/p = h/mv = 6.63 x 10-34/(800)(55.6) behaviours. = 1.5 x 10-38 m• DeBroglie suggested that matter (with An indescribably small number. Too mass, m, moving with velocity, v), will small to measure or even notice. have a wavelength (called the DeBroglie wavelength) associated with its motion. v• This wavelength is calculated from the momentum equation: An electron of mass 9.1 x 10-31 kg travelling λ = h/p = h/mv at 1.4 x 104 ms-1 has a DeBroglie wavelengthwhere λ = DeBroglie wavelength (m) of: h = Planck’s Constant λ = h/p = h/mv p = Particle Momentum = 6.63 x 10-34/(9.1 x 10-31)(1.4 x 104) (kgms ) -1 m = Mass (kg) = 5.2 x 10-8 m v = Velocity (ms ) -1 A measurable quantity, so electrons are able to display wave like properties like interference and diffraction.
  50. 50. Light & Matter RevisionQuestion Type: De Broglie WavelengthA sketch of a cathode ray tube (CRT) is Q26: Calculate the de Broglieshown in Figure 5. In this device, electrons wavelength of the electrons. Youof mass 9.10 × 10-31 kg are must show your working.accelerated to a velocity of 2.0 × 10 m s . A 7 -1 λ= h/mvfine wire mesh in which the gap between thewires is w = 0.50 mm the de Broglie wavelength washas been placed in the path of the electrons, found to be 3.64 x 10-11 m.and the pattern produced is observed on the Q27: Explain, with reasons,fluorescent screen. 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. 51. Light & Matter RevisionQuestion Type: De Broglie WavelengthNeutrons are subatomic particles and, like electrons, canexhibit both particle-like and wave-like behaviour.A nuclear reactor can be used to produce a beam ofneutrons, 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. λ= h/mv gives 2.0 × 103 ms–1 (or 1985 ms–1)Q29: The neutron beam is projected onto ametal crystal with interatomic spacing of 3.0 ×10-10 m.Would you expect to observe a diffractionpattern? Explain your answer.There would be a diffraction pattern becausethe wavelength is of the same order ofmagnitude as the interatomic spacing.
  52. 52. 3.11 Practical ApplicationsThe photoelectric effect haspractical applications in manyareas: Solar Cells 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 Night Vision Goggles 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. 53. 4.2 Electron Interference The now familiar Young’s Initially, with only small . . .. . double slit experiment is numbers, electrons arrive performed with the light at the screen in a random source replaced by an fashion. “electron gun” which fires Even as the numbers reach a single electron at a time. 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 time8 electrons 270 electrons from picture (a) to the picture (d) was 20 min. This result shows particles (in this case electrons) can display wave like properties.2000 electrons 6000. electrons
  54. 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., haveThe arrangement of the K.E.’s of 30 to 500 eV), have aspots is interpreted to Platinum Sample de Broglie wavelength betweenprovide information about 2.2 x 10-10 m and 0.5 x 10-10 m.the ordered arrangement of This fits nicely into the range ofatoms on the surface and distances between atoms inthe distances between the solids and can thereforespots gives information on strongly diffract from them.the distance between theatoms. e- energy = 65 eV
  55. 55. Light & Matter RevisionQuestion Type: Electron Diffraction A beam of electrons, pass through two very thin slits (approximately 10-10 m) Q30: Explain whether the and are detected on a electron detector particle model or the wave screen as shown in Figure 5. model best explains the expected observations from this experiment. 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. 56. 4.4 Electrons and X Rays Diffraction Pattern Al foilHigh energy X Ray Beamelectron beamA high energy electronbeam is fired at an The distance between the bright lines inAluminium target both patterns is the same, meaning theThe diffraction pattern wavelengths of both beams was the same.shown is produced. Hence if λX Ray is known, then the DeWhen the electron beam is Broglie wavelength of the electrons isreplaced with an X Ray also known.beam a somewhat similardiffraction is produced
  57. 57. Light & Matter RevisionQuestion Type: X ray EnergyA beam of X-rays, wavelength λ = 250pm (250 × 10-12 m), is directed onto athin aluminium foil as shown in Figure4a. The X-rays scatter from the foil ontothe 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
  58. 58. Light & Matter RevisionQuestion Type: X ray Diffraction After the X-rays pass through the foil, Q32: Explain why the electrons a diffraction pattern is formed as produce a diffraction pattern similar shown in Figure 4b. to that of the X-rays. In a later experiment, the X-rays are Electrons have a de Broglie replaced with a beam of energetic wavelength which was the same electrons. Again, a diffraction pattern (or very similar) to the wavelength is observed which is very similar to of the X-rays. the X-ray diffraction pattern. This is Q33: Assuming the two diffraction shown in Figure 4c. 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. 59. Light & Matter RevisionQuestion Type: X ray & electron Diffraction 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. Q35: If the X-rays have frequency ofQ34: How does this evidence support the 1.5 x 1019 Hz, what is the wavelengthconcept of matter waves? of the electrons?The circular pattern is a diffractionpattern. With X-rays, the bright lines λ = c/f = 3.0 x 108/1.5 x 1019represent regions of constructive = 2.0 x 10-11 minterference alternating with regions of Q36: What is the momentum of an X-destructive interference. ray photon?The corresponding electron patternshows similar diffraction properties p = h/λ = 6.63 x 10-34/2.0 x 10-11which suggests electrons have wave = 3.3 x 10-23 kgms-1properties, as diffraction is a wavephenomenon.
  60. 60. 4.5 Evidence for Photons as Waves It appears trying to predict the fate of individual photons is not possible,Young’s double slit experiment is but as their numbers build the wavewell known as evidence for the wave model allows the prediction of theirlike nature of light. average behaviour with greatThis experiment creates difficulties accuracy.for the particle (photon) model. Thus photons (en masse) can displayHow can individual photons, which wave like properties although theymust pass through one or other of are generally regarded as discrete,the slits, interact with themselves to individual particles.produce an interference pattern ?When the experiment is carried outat extremely low light intensities, itis found that, for a while, nointerference pattern is noticed.However, as time passes thenumber of photons arriving at thescreen builds and an interferencepattern does emerge with morephotons going to the areas ofbright bands and less to areas ofdark bands.
  61. 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. 62. Light & Matter RevisionQuestion Type: Photon energyA monochromatic violet light of wavelength390 nm is emitted by a light source with apower of 200 Watt.Q37: How many photons leave thislight source in a 10 second interval? 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 Q38: What is the momentum of each of these photons? p = h/λ = 1.7 x 10-27 kgms-1
  63. 63. Chapter 5• Topics covered:• Quantised Energy Levels for Atoms.• Emission Spectra• Absorption Spectra• Solar Spectrum.• Quantised Energy Levels• Wave Particle Duality.
  64. 64. 5.0 Quantised Energy Levels for Atoms• The Bohr model for the atom, restricting electrons to The 1st five certain regions or orbits with electron shells fixed energy levels, worked with their standing well in describing atomic structure and behaviour. waves• 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 Disallowed orbit 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
  65. 65. 5.1 Emission Line SpectraWhen atoms of substances are exposed to The energy absorbed correspondscertain kinds of energy sources, the to jumps between the fixed energyelectrons surrounding the nucleus can levels of the electron shells withingain energy and move up to higher energy the atomlevels, called “excited states”. The excited electrons return to their ground state by emitting a photon of a certain frequency (colour)The emitted photons from all electrons transitionsform “a picture” called an “Emission Spectrum”.These Spectra are unique to each element andare used to identify unknowns in a field calledspectroscopy. Helium Emission Spectra Sodium Emission Spectra
  66. 66. Light & Matter RevisionQuestion Type: Emission SpectraA class looks at spectra from twosources. Q40: State the electron mechanism,i. an incandescent light globe in each of the sources below, thatii. a mercury vapour lamp produces each spectrum.They observe that the spectra are i. Incandescent light globeof different types. ii. Mercury vapour lampQ39: State the type of spectrum i Thermal motion of free electronsseen from each source.i. Incandescent light globe ii Energy transition of bound electronsii. Mercury vapour lamp i Continuous (or broad) ii Discrete (or individual lines)
  67. 67. 5.2 Absorption Line Spectra If you look at the spectrum from a common incandescent light bulb, youll 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 The pattern of absorption lines indiscreet energy levels by absorbing a a spectrum will be unique to aphoton. chemical element so we can useSo certain atoms under certain absorption lines to detect theconditions will absorb very specific presence of specific elements inwavelengths of light dependent on the astronomical objects.configuration of their electrons.This will remove energy and leave blank An emission spectrum is simply thespaces (dark lines) in the spectrum. reverse of an absorption spectrum.
  68. 68. Light & Matter RevisionQuestion Type: Absorption SpectraFigure 5a shows part of the emissionspectrum of hydrogen in more detail.With a spectroscope, Val examines thespectrum of light from the sun. Thespectrum is continuous, with coloursranging from red to violet. Howeverthere were black lines in the spectrum,as shown in Figure 5b.Q41: Explain why these dark lines arepresent 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.
  69. 69. 5.3 The Solar Spectrum 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” It was later discovered that eachIn 1802 English chemist William Wollaston was chemical element wasthe first to note the appearance of a number of associated with a set of spectraldark lines in the solar spectrum. lines, and that the dark lines in the solar spectrum were causedIn 1814, Joseph Von Fraunhofer by absorption by those elementsindependently rediscovered the in the upper layers of the sun.lines and began a systematic Some of the observed featuresstudy and careful measurement of are also caused by absorption inthe wavelength of these features. oxygen molecules in Earth’s In all, he mapped over 570 lines. atmosphere.
  70. 70. 5.4 Quantised Energy Levels The energy is associated withElectrons surrounding the nucleus these “jumps” is given by E = hfcan “jump” up and down between whereallowed energy levels by absorbing E = Energy of the emitted photonor emitting a photon. for a downward jump OR theThe ABSORPTION and EMISSION energy absorbed from an incidentSPECTRA for atoms show the photon for an upward jump. (eV)energy values associated with the h = Planck’s Constant“jumps”. f = Frequency of Photon (Hz) Mercury Energy Levels Ionisation 10.4 eV n=5 8.8 eV n=4 6.7 eV n=3 Electronsabsorbing State5(n = 3 can of States in = 2 to n = represent The n excited state n 10.4 Electrons Ground more than = 1) eV 4.9 eV n=2 energy from a photon1.8 eV by: be represents thea state electron 2. By emitting lowest photon return to the groundcollision willfor allowed energy levels stripped completely4.9 eV photon Emitting a single 6. 7the Mercury electrons a from eV atoms. 1. followed byin Mercuryphoton energy level atom leaving it ionised OR 0 eV n=1 Ground State
  71. 71. Light & Matter RevisionQuestion Type: Energy LevelsThe spectrum of photons emitted by Q42: What is the lowest energyexcited atoms is being investigated. photon that could be emittedShown in Figure 6 is the atomic energy from the excited atoms?level diagram of the particular atom The smallest energy gap wasbeing studied. Although most of the that from n = 3 to n = 2, anatoms are in the ground state, some energy difference of 1.8 eV.atoms are known to be in n = 2 and n = Q43: Calculate the wavelength of3 excited states. 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. 72. Light & Matter RevisionQuestion Type: Energy LevelsFigure 2 shows the energy levelsof 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 HzFigure 2 shows that electrons in a Electrons have an equivalent or desodium atom can only occupy specific Broglie levels.Q45: Describe how the wave nature of Only orbits of an integral number ofelectrons can explain this. wavelengths are allowed as these will form a standing wave
  73. 73. Light & Matter RevisionQuestion Type: Energy Level DiagramsPart of the visible region of thespectrum of light emitted from excitedhydrogen gas has three lines as shownin Figure 3.The energy level diagram for thehydrogen 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.
  74. 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 BUT THIS IS WHAT PHYSICS IS ALL behaviour. ABOUT –• This is an unsatisfactory situation, PUSHING THE ENVELOPE – as it defies our need for simple all EXPLORING THE UNKNOWN – encompassing explanations for the TRYING TO FIT RATIONAL AND LOGICAL behaviours we observe around us and in the universe. EXPLANATIONS TO THAT WHICH HAS BEEN OBSERVED.• It would appear, at this stage, we have not yet unravelled all the details of the behaviour of light and matter.
  75. 75. Ollie Leitl 2005
  1. A particular slide catching your eye?

    Clipping is a handy way to collect important slides you want to go back to later.