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7 QUANTUM BEHAVIOUR The quantum nature of light. Nein , light is made of particles!. Observe phenomena where light displays a particle-like nature Observe and explain the photoelectric effect in terms of the quantum theory of light. Light is a wave motion!.

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## 7 QUANTUM BEHAVIOUR The quantum nature of light

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**7 QUANTUM BEHAVIOURThe quantum nature of light**Nein, light is made of particles! • Observe phenomena where light displays a particle-like nature • Observe and explain the photoelectric effect in terms of the quantum theory of light Light is a wave motion!**What about evidence for the quantised nature of radiation?**Gamma source**What about evidence for the quantised nature of**radiation….**The Electron Volt….**Energy in Joules = energy in eV x 1.6x10-19 JeV-1 This is the work done when an electron is moved through a potential difference of 1 volt. Since the charge of the electron is 1.6x10-19 C, then 1 eV = 1.6x10-19J.**The Photoelectric Effect….**Note: Energy to extract an electron = Φ (the work function)**Note:**GRAD is Planck's constant!**Four key observations of the Photoelectric Effect**1.The number of photoelectrons emitted per second is proportional to the intensity of the incident radiation. 2. For a given metal, there is a certain minimum frequency of light below which no emission occurs no matter how intense the radiation. 3.Photoelectrons are emitted with a range of kinetic energies up to a certain maximum value. This maximum value increases if light of higher frequency (shorter wavelength) is used. The maximum kinetic energy is completely independent of the radiation intensity. 4.Even with a source of low intensity, there is no time delay between starting to illuminate the metal and the production of photoelectrons.**Note:**gradient is Planck's constant!**Planck’s Equation**E = hf = hc/λ Einstein’s Photoelectric effect equation Photon energy = Work function + Kinetic Energy hf = Φ + Ek Key: h – Planck’s constant (6.63x10-34Js) f – frequency of photon(Hz) Φ – work function (min energy required to eject a photon (J)) c – speed of light (3x108ms-1) λ – wavelength (m)**Starter:**Annotate the graph with the answers to the following questions. Q1. Add a line to the graph for a metal with a lower work function than the one depicted. Q2. Show how you would determine the work function for a metal from the graph. Q3. Highlight the point on the graph corresponding to irradiation of the metal with the shortest wavelength light. What colour is the light in this case? Q4. Highlight the point on the graph corresponding to irradiation of the metal with the longest wavelength light. What colour is the light in this case? Q5. Light of wavelength 488 nm irradiates the metal surface. Use the graph to determine the maximum kinetic energy electrons could have when the surface is irradiated in this way.**Example questions….**h is Planck’s constant (6.63x10-34Js) Blue light has a frequency of 7.7x1014Hz. What energy does a blue photon have? (in both J and eV) What is the energy of a red photon (in J and eV) in red light with a wavelength of 7x10-7m. Light of frequency 6.7x1014Hz shines on to clean caesium metal. What is max kinetic energy of electron emitted? (Φ for caesium = 3.43x10-19J) Light of wavelength 4.2x10-7m is incident on potassium metal. If the kinetic energy of the emitted electron is 0.78eV what is the work function?**Two lasers have identical output powers.**Q1. What can you say about the amount of energy emitted by each laser every second? Q2. Laser A emits red light, laser B emits blue light. Which laser emits more photons every second? Justify your answer.**Estimating the photons emitted from a reading lamp…**The power of reading lamp is 40W Average wavelength of light emitted from the lamp is 5x10-7m. 1. Calculate the energy transferred by each photon. 2. Calculate the number of photons emitted by the lamp in each second.**Quantum calculations**• Learn about the “Try all paths” algorithm for quantum calculations • Explain light phenomena in terms of phasor addition for all possible paths Starter: Find the length of the resultant on adding the 3 phasor arrows shown. Assume that each one has length 1 unit.**“How do the photons know which path to take?”**“Try all paths” is a quantum rule obeyed by all photons and electrons. Richard Feynman argued that instead of just taking one route between the source and the detector, a photon will take all of the possible paths to the detector in one go. You can keep track of this photon going on along every possible route using phasors. Source Detector**Quanta and phasors:**E = hf and c = f can be used to work out the number of phasor rotations for a photon taking a particular path. What is the rate of rotation of a phasor associated a photon with energy 5.1x10-19J? What is the rate of rotation of a phasor with light of wavelength of 7x10-7m? A photon of wavelength 5.2x10-7m explores a path 4m long. What is its travel time and how many times does the phasor rotate? Now try questions 3 and 4 on page 157**Calculating the probability of arrival**P1 P2 Resultant Probability α (resultant phasor)2**Question…**Question: The resultant phasors for a photon reaching points A and B have magnitudes of 6.3 and 4.5 respectively. How many times more likely is it that a photon will arrive at point A than at point B?**Does the rule ‘explore all paths’ really work in all**situations. On computers explore the following software activities to check out this rule in three crucial situations…. reflection with Activity 110S, 120S, 140S, 170S; refraction with 190S and 200S diffraction with 210S.**Wave properties of electrons**• Describe the evidence for wave behaviour of entities we normally regard as particles* (* electrons, neutrons, large molecules, even bigger stuff...........) Starter: Describe a physical situation/observation where electrons are best described as particles. How does the particle description of electrons help to explain it?**ANIMATION**Copy and complete... Evidence for light as particles Evidence for light as waves Evidence for electrons as particles Evidence for electrons as waves**Kinetic energy (J) = electronic charge × accelerating**voltage KE = e × V • If the accelerating voltage in a computer monitor is 400V what is the total kinetic energy acquired by a single electron? • An electron acquires a Kinetic Energy of 600 x 10-17J in a cathode ray tube. What voltage was required to provide this energy? • 3. A kinetic energy of 0.02 mJ per second was transferred to a screen of a television during a TV programme. If the voltage was 500V how many electrons per second would be hitting the screen. • 4. (a) How many electrons would hit the screen in 5 seconds in an oscilloscope if a current of charge of 0.01 A is flowing? (Hint I = Q/t) • If the voltage across the same oscilloscope was 100V then what would be the total Kinetic Energy transferred to the screen by the electrons? Electronic charge = -1.6 x 10 -19 C**Kinetic energy = electronic charge × accelerating voltage**KE = e × V**To diffract electrons you need very small slits – the gaps**between atoms**Electron diffraction**This eerie green glow is caused by electrons in a cathode ray tube striking the phosphorescent coating on the inside of the glass bulb just behind the ruler. • In this case the diffraction is caused by the electrons passing through a thin layer of polycrystalline graphite (pencil "lead"). The regular array of carbon atoms in the crystals is responsible for the diffraction effects.**Prince Louis-Victor-Pierre-Raymond, 7th duc de Broglie**λ = ____h____ momentum Everything acts with quantum behaviour – if we give particles enough energy they will have a wavelength**For electrons only**λ = ____h____ momentum The distance it takes to make one complete turn of the phasor momentum = mass x velocity**The de Broglie wavelength can be thought of as the distance**travelled by the photon for one turn of the phasor associated with it. The de Broglie wavelength is given by λ = h/p, where p is the momentum of the particle (=mass x velocity, mv) If you know the particle’s kinetic energy, the speed can be worked out using KE = ½ mv2, and hence momentum (=mv) can be found. The de Broglie wavelength can then be calculated. Note: The energy expression used for photons (E=hc/ λ) can also be used to find the de Broglie wavelength of particles, but this method is normally reserved for very high energy electrons (with energies of 100s of MeV) where the relationship E = pc holds. NOW DO Q1-4 FROM P167**Solutions:**Q1. KE = e x V = 1.6 x 10-19 C x 100 V = 1.6 x 10-17 J Q2. KE = ½ mv2 = 1.6 x 10-17 = ½ x 9 x 10-31 x v2 v = 5.9 x 106 ms-1 Q3. momentum = mv = 9 x 10-31 x 5.9 x 106 = 5.3 x 10-24 kg ms-1 Q4. λ = h/p = 6.6 x 10-34 / 5.3 x 10-24 = 1.2 x 10-10 m**Two slit interference with electrons**..\doubleslite-n.wmv The greatest physics experiment… …in the world.**JJ and GP Thomson: father and son Nobel physics prize**winners Yeah whatever Dad. I got it for showing that the electron is a wave! Listen sonny, I got the Nobel prize for showing that the electron is a particle!

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