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Dual Nature of Light

Dual Nature of Light. Is light a wave or a particle?. Wave Properties. -Diffraction -Interference -Polarization. Diffraction. Constructive & Destructive Interference. Polarization. Energy. Wave E increases with A 2 /Intensity. Studies of Wave E. Planck – color ( f , l ) vs. T.

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Dual Nature of Light

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  1. Dual Nature of Light Is light a wave or a particle?

  2. Wave Properties • -Diffraction • -Interference • -Polarization

  3. Diffraction

  4. Constructive & Destructive Interference

  5. Polarization

  6. Energy

  7. Wave E increases with A2/Intensity.

  8. Studies of Wave E • Planck – color (f, l) vs. T. • As T inc. , f inc, (l decr)

  9. Radiation & TemperatureHot Objects Emit Waves

  10. Intensity/Brightness

  11. Classical physics could not accurately predict f vs. Temperature Problem:

  12. Max Planck related f to T. Light (EM) E, is quantized--it can only take on certain whole number values. E comes in little "chunks" of f x a constant now called Planck's constant, h: • EM radiation waves • chunks • quanta • photons

  13. Can calculate E in EM waves units quanta or photons based on frequency. E = hf. h is Plank’s constant 6.63 x 10-34 Js. E is energy in Joules f is frequency of radiation

  14. Show that if E = hf,E = hc.l • For waves, v = fl. • Rearrange f = v/l. • Vacuum/air EM v = c (3 x 108m/s). • f = c/l. • E = hff = c/l. • E = hc.l

  15. Ex 1. Each photon of a certain color light has an energy of 2.5 eV. What is the frequency of and color of the light?

  16. Solution:E = hff = E/h convert eV to Joules.(2.5 eV)(1.6 x 10-19J/eV) = 6.03 x 1014 Hz 6.626 x 10-34 J s Green Light

  17. 2. The energy of a certain photon is 2.9 eV. What type of wave is it? Be specific. • 2.9 eV x 1.6 x 10-19 J = 4.64 x 10-19 J. eV E = hf (4.64 x 10-19 J) = (6.63 x 10-34 Js) f Violet Light f = 7 x 10 14Hz

  18. Finish Ex Sheet • Hwk : Text Read 830 – 833 Do pg 833 #1-4 and 839 #2, & 856 # 2-4, 9.

  19. Do Now: A photon of light has energy = 2.072 eV. What color is it? • 2.072 eV (1.6 x 10-19 J/eV). • 3.3152 x 10-19 J • E = hf. • (3.3152 x 10-19 J) = (6.63 x 10-34 Js)f • f = 5.00 x 1014 Hz. • Orange

  20. So EM Radiation comes in chunks, maybe it’s not waves. • All objects above 0 K radiate EM waves as E. • Hotter = more E = higher freq. (different color) • E = hf (J).

  21. What is the evidence? • Photoelectric effect. http://phet.colorado.edu/en/simulation/photoelectric

  22. Photoelectric Effect When EM waves shine on a metal surface, the E in wave may be enough to kick out surface e-. Materials that emit e- in this way are called photoemissive. The ejected e- are called photoelectrons.

  23. Phet observations

  24. http://phet.colorado.edu/en/simulation/photoelectric • Use higher Amplitude/Intensity/brightness – • more e- fly off w same speed. • Current increases (A, C/s) • Increased f • e- fly off faster w higher KE.

  25. Classical Mechanics cannot explain why increasing A or exposure time does not increase photoemission. After all: • Boat would be tossed higher & faster with increased wave amplitude. • But not ejected e-.

  26. Einstein: Maybe EM waves quantized too – photons. • Increasing f, increases E of each photon. • Increasing Intensity (A) increases number of photons hitting more e- so more fly out. • Envision EM as little chucks. High f are heavier. • http://phet.colorado.edu/en/simulation/photoelectric

  27. Photoemission only works with metals with weakly bound e-.

  28. Each metal, has a threshold frequency fo = lowest f that will free an e-. Light frequencies below the fo eject no e-, no matter how intense or bright the light.

  29. Light frequencies above the fo eject e-, no matter how low the A (how dim).

  30. A metal has a threshold frequency fo in the blue light range. What will happen if very bright red light is shone upon the metal? No e- will be emitted more e- will be emitted The emitted e- will have less energy.

  31. Einstein confirmed EM waves come in quanta or photons, each has E =hf. Higher-frequency photons have more E, so they can knock out e- & make the e- fly out faster with greater KE.

  32. 2. A metal has a threshold frequency in the blue light range.What will happen if UV light is shone upon the metal? • nothing b) the emitted e- will have more energy (KE) c) more e- will be emitted with the same energy.

  33. Increasing the intensity/ampl, increases the rate of e- emission or the current; the number of e- ejected, but each e- won’t gain any extra energy. Only increased f increases E of photoelectrons.

  34. A metal has a threshold frequency in the blue light range. 3. What will happen if the blue light is made twice as bright? • nothing b) the emitted e- will have more energy (KE) c) more e- will be emitted with the same energy.

  35. All EM waves can be described as quanta or photons. The energy carried by photons is: Ephoton = hf or Ephoton = hc/l. (for photon traveling at speed of light).This energy is absorbed by photo-emissive materials.

  36. The min. frequency to free e- is fo.The min energy needed to free an e- is called work function Wo, or F.Metals have low Wo.

  37. 4. A certain metal has a work function (Wo) of 1.7 eV. If photons of energy 3.0 eV are absorbed by the metal: • a) No e- will be emitted at that energy. • b) More e- will be emitted than would be at the Wo. • c) Higher KE e- will be emitted than would be at Wo.

  38. Summary: • EM waves as chunks of energy/photons travel at c. • Calculate the Energy J E = hf, or • E = hc/l. • Evidence for photons – from Photoelectric Effect Experiment – • f not A responsible for KE of ejected e-. • High f = high E, photon. • High A = high number of photons. • Photo-emissive materials have: • fo = min f to eject e- (Hz) • Wo= min E to eject e- (J)

  39. Read Txt 830-835Intro to Modern Physics Packet #1-14

  40. Graph of Photoelectric Experiment • KE of photoelectron vs. frequency.

  41. max KE of photo e- vs. f for metal. As f of EM wave increases, KE increases, slope = h. F (work function), is minimum energy needed to eject e-. Work function

  42. State the work function & threshold frequency of this metal eV 0.4 0.0 0.4 0.8 1.0 2 3 4 5 6 7 8 x 1014 Hz

  43. Wo & fo. The threshold frequency fo, = min frequency needed to eject e- so the min energy needed, Wo in J is: Wo = hfo. Any photon E left over after the work function, goes into KE of e-.

  44. 5. A particular metal has a threshold frequency fo, of 5 x 1014 Hz. What is its work function Wo in J & eV? Wo = hfo. 3.3 x 10-19 J 2.07 eV

  45. Ephoton = hf is the total E available. • Absorbed photon E splits between Wo & KE photo e-, so total E of absorbed by e- is: • Epho = Wo + KE. • The maximum KE of ejected e- is: • KEelc = Epho – Wo. • Don’t forget Wo = hfo.

  46. 6: Photoelectric Effect:Light having f = 1 x 1015 hz shines on a sodium surface. The photoelectrons have a maximum KE of 3 x 10-19 J.Find the threshold frequency for sodium.

  47. Photon Photoelectron.Etot = Wo + KE.Etot – KE = Wo.hf – KE = hfo. fo = (hfphoton – KEmax) (h)

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