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Early Quantum Mechanics

Early Quantum Mechanics. Chapter 27. Three Major Discoveries. Wein’s Law. Treat’s light solely as a wave Hot “blackbodies” radiate EM The hotter the object, the shorter the peak wavelength Sun (~6000 K) emits in blue and UV A 3000 K object emits in IR 2.90 X 10 -3 mK = l peak T.

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Early Quantum Mechanics

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  1. Early Quantum Mechanics Chapter 27

  2. Three Major Discoveries

  3. Wein’s Law • Treat’s light solely as a wave • Hot “blackbodies” radiate EM • The hotter the object, the shorter the peak wavelength • Sun (~6000 K) emits in blue and UV • A 3000 K object emits in IR 2.90 X 10-3 mK = lpeakT

  4. Wein’s Law: Ex 1 Estimate the temperature of the sun. The Suns light peak at about 500 nm (blue) 500 nm = 500 X 10-9 m T = 2.90 X 10-3 mK lpeak T = 2.90 X 10-3 mK = 6000 K 500 X 10-9 m

  5. Wein’s Law: Ex 2 Suppose a star has a surface temperature of about 3500 K. Estimate the wavelength of light produced. 2.90 X 10-3 mK = lpeakT l peak = 2.90 X 10-3 mK T l peak = 2.90 X 10-3 mK = 8.29 X 10-7 m = 830 nm 3500 K

  6. Planck’s Quantum Hypothesis • Energy of any atomic or molecular vibration is a whole number • Photon – the light particle • Photons emitted come in “packets” • E = hf • h = 6.626 X 10-34 J s (Planck’s constant)

  7. Photons: Ex 1 Calculate the energy of a photon of wavelength 600 nm. 600 nm X 1 X 10-9m = 6 X10-7 m 1 nm c = lf f = c/l = (3X108 m/s)/(6 X10-7 m) = 5 X 1014 s-1 E = hf E = (6.626 X 10-34 J s)(5 X 1014 s-1) = 3.3 X 10-19 J

  8. Photons: Ex 1a Convert your answer from the previous problem to electron Volts (1 eV = 1.6 X 10-19 J)

  9. Photons: Ex 2 Calculate the energy of a photon of wavelength 450 nm (blue light).

  10. Photons: Ex 2a Convert your answer from the previous problem to electron Volts (1 eV = 1.6 X 10-19 J)

  11. Photons: Ex 3 Estimate the number of photons emitted by a 100 Watt lightbulb per second. Assume each photon has a wavelength of 500 nm. 500 nm X 1 X 10-9m = 5 X10-7 m 1 nm c = lf f = c/l = (3X108 m/s)/(5 X10-7 m) = 6 X 1014 s-1

  12. 100 Watts = 100 J/s (we are looking at 1 second) E = nhf n = E/hf n = 100 J (6.626 X 10-34 J s)(6 X 1014 s-1) n = 2.5 X 1020

  13. Photoelectric Effect (Einstein) • When light shines on a metal, electrons are emitted • Can detect a current from the electrons • Used in light meter, scanners, digital cameras (photodiodes rather than tubes)

  14. Three Key Points • Below a certain frequency, no electrons are emitted • Greater intensity light produces more electrons • Greater Frequency light produces no more electrons, but the come off with greater speed

  15. More intensity • More photons • More electrons ejected with same KE

  16. Greater Frequency • No more electrons ejected • Electrons come off with greater speed (KE)

  17. hf = KE + W hf = energy of the photon KE = Maximum KE of the emitted electron W = Work function to eject electron

  18. hf = KE + W: Ex 1 What is the maximum kinetic energy of an electron emitted from a sodium atom whose work function (Wo) is 2.28 eV when illuminated with 410 nm light? 410 nm = 410 X 10-9 m or 4.10 X 10-7 m 2.28 eV X 1.60 X 10-19J = 3.65 X 10-19 J 1 eV

  19. c = lf f = c/l f = c/(4.10 X 10-7 m) = 7.32 X 1014 s-1 hf = KE + W KE = hf – W KE = (6.626 X 10-34 J s)(7.32 X 1014 s-1) - 3.65 X 10-19 J KE = 1.20 X 10-19 J or 0.75 eV

  20. hf = KE + W: Ex 2 What is the maximum kinetic energy of an electron emitted from a sodium atom whose work function (Wo) is 2.28 eV when illuminated with 550 nm light? ANS: 2.25 eV

  21. hf = KE + W: Ex 3 What is the maximum wavelength of light (cutoff wavelength) that will eject an electron from an Aluminum sample? Aluminum’s work function (Wo) is 4.08 eV? 4.08 eV X 1.60 X 10-19J = 6.53 X 10-19 J 1 eV

  22. hf = KE + W hf = 0 + W (looking for bare minimum) c = lf f = c/l hf = W hc = W l l = hc = (6.626 X 10-34 J s)(3.0 X 108 m/s) W 6.53 X 10-19 J l = 3.04 X 10-7 m = 304 nm (UV)

  23. Photon/Matter Interactions • Electron excitation (photon disappears) • Ionization/photoelectric effect (photon disappears) • Scattering by nucleus or electron • Pair production (photon disappears)

  24. Electron Excitation • Photon is absorbed (disappears) • Electron jumps to an excited state • Ionization/Photoelectric Effect • Photon is absorbed (disappears) • Electron is propelled out of the atom

  25. Pair Production • Photon closely approaches a nucleus • Photon disappears • An electron and positron are created. • Scattering • Photon collides with a nucleus or electron • Photon loses some energy • Speed does not change, but the wavelength increases

  26. 3. Scattering: Compton Effect • Electrons and nuclie can scatter photons • Scattered photon is at a lower frequency than incident photon • Some of the energy is transferred to the electron or nucleus

  27. l’ = l + h (1 – cos q) moc l’ = wavelength of scattered photon l = wavelength of incident photon mo = rest mass of particle q = angle of incidence

  28. Compton Effect: Ex 1 X-rays of wavelength 0.140 nm are scatterd from a block of carbon. What will be the wavelength of the X-rays scattered at 0o? l’ = l + h (1 – cosq) moc l’ = 140 X10-9m + (6.626 X 10-34 J s)(1 – cos0) (9.11 X 10-31 kg)(3 X 108m/s) l’ = 140 X10-9m + 0 l’ = 140 nm

  29. Compton Effect: Ex 2 What will be the wavelength of the X-rays scattered at 90o? l’ = l + h (1 – cos q) moc l’ = 140 X10-9m + (6.626 X 10-34 J s)(1 – cos 90) (9.11 X 10-31 kg)(3 X 108m/s) l’ = 140 X10-9m + 2.4 X 10-12 m l’ = 142 nm

  30. Compton Effect: Ex 3 What will be the wavelength of the X-rays scattered at 180o? l’ = l + h (1 – cos q) moc l’ = 140 X10-9m + (6.626 X 10-34 J s)(1 – cos 180) (9.11 X 10-31 kg)(3 X 108m/s) l’ = 140 X10-9m + 4.8 X 10-12 m l’ = 145 nm (straight back)

  31. 4. Pair Production • Photon Disappears • e- and e+ are produced • They have opposite direction (law of conservation of momentum) • When e- and e+ collide they annihilate each other  a new photon appears

  32. Principle of Complimentarity • Neils Bohr • Any experiment can only observe light’s wave or particle properties, not both • Different “faces” that light shows

  33. The Discovery of the Electron (Thomson) • Cathode Ray Tube • Charged particles produced (affected by magnetic field)

  34. Concluded that atom must have positive and negative parts • Electron – negative part of the atom • Only knew the e/m ratio • Plum Pudding Model

  35. Charge and Mass of the Electron (Millikan) • Oil drop experiment • Determines charge on electron (uses electric field to counteract gravity) • Quantized • e = 1.602 X 10-19 C • m = 9.11 X 10-31 kg

  36. The Nucleus (Rutherford) • Gold Foil Experiment • Discovers nucleus (disproves Plum Pudding Model) • Planetary Model

  37. Wave Nature of Matter • Everything has both wave and particle properties • DeBroglie Wavelength E2 = p2c2 + m2c4 (consider a photon) E2 = p2c2 (photon has no mass) E = pc E = hf hf = pc c = lf

  38. hf = plf p = mv (for a particle) hf = mvlf h = mvl l = h mv Everything has a wavelength Diffraction pattern of electrons scattered off aluminum foil

  39. DeBroglie Wavelength: Ex 1 Calculate the wavelength of a baseball of mass 0.20 kg moving at 15 m/s l = h mv l = (6.626 X 10-34 J s) (0.20 kg)(15 m/s) l = 2.2 X 10-34 m

  40. DeBroglie Wavelength: Ex 2 Calculate the wavelength of an electron moving at 2.2 X 106 m/s l = h mv l = (6.626 X 10-34 J s) (9.11 X 10-31 kg)(2.2 X 106 m/s) l = 3.3 X 10-10 m or 0.33 nm

  41. DeBroglie Wavelength: Ex 3 Calculate the wavelength of an electron that has been accelerated through a potential difference of 100 V V = PE (PE =KE) q V = 1 mv2 2 q v2 = 2qV/m

  42. v = (2qV/m)1/2 v=[(2)(1.602 X 10-19 C)(100V)/(9.11 X 10-31 kg)]1/2 v = 5.9 X 106 m/s l = h mv l = (6.626 X 10-34 J s) (9.11 X 10-31 kg)(5.9 X 106 m/s) l = 3.3 X 10-10 m or 0.33 nm

  43. Electron Microscope • Electron’s wavelength is smaller than light • Magnetic focusing

  44. DNA

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