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Early Quantum Mechanics

This chapter discusses key discoveries in early quantum mechanics, focusing on Wein's Law and its implications for blackbody radiation. It explores how the wavelength of emitted light varies with temperature, illustrating the sun's emission characteristics. The chapter also delves into Planck's Quantum Hypothesis, introducing photons as energy packets and providing calculations of photon energies for various wavelengths. Additionally, it presents the photoelectric effect, detailing how light interacts with metals to emit electrons and highlights the principles behind scattering phenomena, including the Compton Effect.

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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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