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### Full chapter

1. 1. Online HW help available in Science 242 Computer lab: Tues 9-10:30 (MK) and 3:30-4:20 (TS) MWF 11:30-12:20 (TS) Talk this afternoon: Mission to find ice on the Moon 3:00-4:15 Sci 258.
2. 2. Chapter 38: Photons, electrons and atoms <ul><li>Light behaves like a wave: </li></ul><ul><li>Interference </li></ul><ul><li>Diffraction </li></ul><ul><li>Refraction </li></ul><ul><li>Polarization </li></ul>
3. 3. Interference Diffraction
4. 4. Refraction Polarization
5. 5. 38-1: Emission and absorption of light Emission line spectrum Continuous spectrum Classical optics cannot explain these observations!
6. 6. c = f 
7. 7. 38-2: Photoelectric effect Einstein won the Nobel Prize for this work in 1922!
8. 8. Work function of a metal (  ) : The minimum amount of energy an electron must gain so that it can be ejected from the metal surface. No e-’s are ejected if f < threshold frequency. 1 eV = energy given to an e- by accelerating it through 1 V
9. 9. Measure the e-’s maximum KE by varying the voltage, find what “stopping potential” V 0 makes the e-’s stop (KE = 0). W tot = -eV 0 =  K = K final - K initial = 0 - K max K max = eV 0
10. 11. Planck’s constant h = 6.626 x 10 -34 J-s = 4.136 x 10 -15 eV-s
11. 12. <ul><li>When UV light of  = 254 nm falls on a clean copper surface, the potential difference required to stop the photoelectrons is 0.181 V. </li></ul><ul><li>What is the work function for the surface? Compare to the known value. </li></ul><ul><li>What is the photoelectric threshold  for this surface? </li></ul>
12. 13. Photoelectric effect question <ul><li>Suppose a metal has a work function of 4.5 eV. You shine light with an energy of 5.2 eV on the metal. What is the potential difference required to stop the current from flowing? </li></ul><ul><li>0.7 V </li></ul><ul><li>4.5 V </li></ul><ul><li>5.2 V </li></ul><ul><li>9.7 V </li></ul><ul><li>None of the above </li></ul>
13. 14. Photoelectric effect question <ul><li>You shine light with an energy of 4.3 eV on an unknown metal. You find that a potential difference of 1.2 V is required to reduce the current to zero. How much kinetic energy do the photoelectrons have? </li></ul><ul><li>4.3 V </li></ul><ul><li>1.2 V </li></ul><ul><li>5.5 V </li></ul><ul><li>3.1 V </li></ul><ul><li>You need to know the work function to solve the problem </li></ul>
14. 15. 38.3 Atomic Line Spectra and Energy Levels What causes the different colors?
15. 16. Each elements has its own unique spectrum
16. 17. Atoms have internal energy
17. 18. The Balmer series for hydrogen: Visible light electrons falling to n=2 Balmer’s formula (n f = 2) : 1 /  = R (1/n f 2 - 1/n i 2 ) R = Rydberg constant = 1.097 x 10 7 m -1
18. 20. Hydrogen emission spectrum
19. 21. The hydrogen atom
20. 22. Energies are < 0 because we set E = 0 when atom is ionized. All other energies must be < 0 (like potential energy)
21. 23. <ul><li>You can use a simple formula like Balmer’s only for atoms with one electron </li></ul><ul><li>Hydrogen </li></ul><ul><li>Positronium </li></ul><ul><li>Singly ionized helium (He + ) </li></ul><ul><li>Doubly ionized lithium (Li 2+ ) </li></ul>For other atoms, things get weird!
22. 24. Energy levels and transitions of the many-electron atom: Sodium Quantum states of the valence electron
23. 25. An atom will absorb a photon only if the photon has exactly the right energy to “excite” a transition
24. 26. Absorption spectrum
25. 27. Very high resolution absorption spectrum tells us what the sun’s atmosphere is made of
26. 28. 38.4 The Nuclear Atom
27. 29. Rutherford’s experiment
28. 30. The nucleus is not just small, it is REALLY small! Radius = 7x10 -15 m Radius = 7x10 -14 m
29. 31. 38.5 The Bohr Model <ul><li>Classical physics predicts that the electron should spiral into the nucleus </li></ul><ul><li>Cannot explain emission spectra </li></ul>
30. 32. <ul><li>The e- stays in certain stable orbits, emits no radiation unless it jumps to a lower level </li></ul><ul><li>The angular momentum of the e- is quantized </li></ul><ul><li>the attaction between p and e- provides the centripetal acceleration </li></ul>n = principal quantum number The Bohr model:
31. 33. From Coulomb’s law, the force between the proton and electron is This is the centripetal force, mv 2 / r Where q 1 = q 2 = e for the hydrogen atom F = 1 4  0 q 1 q 2 r 2
32. 34. Bohr radius a 0 =  0 h 2 /  me 2 = 5.29 x 10 -11 m So when the electron is in any energy level n: KE of the electron in the nth level: K n = 1/2 mv 2 = _____ Total energy E n = K n + U n = ??? -1 4  0 e 2 r PE of the electron in the nth level: U n =
33. 35. Compare this with
34. 36. Reduced mass: the nucleus is not infinite in mass, Bohr model is off by 0.1% m r = m 1 + m 2 m 1 m 2
35. 37. Positronium
36. 38. <ul><li>The Bohr approximation works well for most atoms, because the nucleus is so massive compared to the electron. Not so for positronium. </li></ul><ul><li>What is the reduced mass in the positronium “atom”? </li></ul><ul><li>The mass of the electron </li></ul><ul><li>2x the mass of the electron </li></ul><ul><li>4x the mass of the electron </li></ul><ul><li>1/2 the mass of the electron </li></ul><ul><li>1/4 the mass of the electron </li></ul><ul><li>No clue! </li></ul>
37. 39. What does this do to the energy levels (and resulting frequencies of light) of positronium? <ul><li>They are half of that for hydrogen </li></ul><ul><li>They are 2x that for hydrogen </li></ul><ul><li>They are 1/4 that for hydrogen </li></ul><ul><li>They are 4x that for hydrogen </li></ul>
38. 40. Ionized Helium is also a 1-electron atom
39. 41. <ul><li>Why is the emission spectrum of ionized helium similar to that of hydrogen? </li></ul><ul><li>Because hydrogen and helium are similar chemically </li></ul><ul><li>Because several of the energy levels of hydrogen and helium are the same </li></ul><ul><li>Because hydrogen and helium have similar atomic masses </li></ul><ul><li>It is a total coincidence </li></ul>
40. 42. 38.6 The Laser Light Amplification through Stimulated Emission of Radiation absorption emission <ul><li>A photon can be emitted </li></ul><ul><li>Spontaneously </li></ul><ul><li>Stimulated by a collision </li></ul><ul><li>Stimulated by another photon </li></ul><ul><li>An atom can be excited </li></ul><ul><li>Due to a collision </li></ul><ul><li>Due to absorbing a photon </li></ul>Nobel prize given to Charlie Townes!
41. 43. Population inversion: more atoms in high energy state than low, without the high temperatures. Non equilibrium state! Ratio of states: Maxwell-Boltzmann distribution = exp(-(E ex -E gr )/kT n ex n gr = exp(-E ex /kT) exp(-E gr /kT)
42. 44. How a laser works
43. 46. The electromagnetic spectrum
44. 47. Maser: Naturally occurring!
45. 48. 38-7. X-ray production and scattering <ul><li>Roentgen discovered X rays in 1895. </li></ul><ul><ul><li>Not affected by direction of E field: No charge </li></ul></ul><ul><ul><li>No mass either </li></ul></ul><ul><ul><li>Must be a photon! </li></ul></ul>=10 3 -10 6 V
46. 49. <ul><li>As the e- are “braked” to rest, two things happen: </li></ul><ul><li>their KE becomes photons of all energies: Bremsstrahlung (braking radiation). Does not depend on target material! </li></ul><ul><li>Some e- of the target material become excited and produce x rays as they decay to lower states </li></ul>Classical physics cannot explain this! =10 3 -10 6 V
47. 50. There should be an equal sign here! Relationship between voltage and max photon energy: =
48. 51. The process by which x-rays are generated in this way can be compared to the photoelectric effect. How? <ul><li>It is basically the same as the photoelectric effect </li></ul><ul><li>it is the opposite of the photoelectric effect </li></ul>
49. 52. Compton scattering: direct confirmation of the quantum nature of x rays Photons lose energy as a result of scattering off of an electron in the target material. Here m = electron mass. The missing energy goes into the recoil of a loosely bound electron
50. 53. This peak is due to Compton scattering
51. 54. 38.8 Continuous spectra: Blackbody radiation Spectral emittance Intensity distribution I(  ) = # of photons at some wavelength 
52. 55. The temperature of a blackbody uniquely determines the wavelength at which most of the photons are emitted: = total intensity emitted over all wavelengths I = total number of photons at all energies = the area under the curve = the integral of I(  ) over all wavelengths 0--> infinity
53. 56. 38-9 Wave-particle duality Particles: :waves