Particles and Waves

1. Particles and Waves

2. Early Atom Models • Democritus (2500 years ago)—tiny indivisible particles, atmos • Dalton (1803) • Each element composed of extremely small particles • Atoms of any element are identical and different from others • Atoms not created or destroyed • Compounds always have same relative numbers of kinds of atoms

3. The Plum Pudding Model of the Atom • The atom was once thought to be a solid ball of positive material of pudding-like consistency in which were embedded discrete negatively charged objects called electrons. • In this (incorrect) model of the atom, the atom is a ball of uniform density.

4. Ernest Rutherford • Ernest Rutherford was the first to show thatthe atom does nothave uniform density • Most of the itsmass is located at itscenter. Ernest Rutherford won1908 Nobel Prize forstudies in radioactivity;discovered nucleus in 1911.

5. The Nucleus

6. Rutherford Model of the Atom Based on his alpha-particlescattering experiment on gold, Rutherford concludedthat the atom consisted ofa hard central core wheremost of the mass of theatom rested.

7. Nuclear Structure • Facts about the nucleus: • Protons and neutrons have about thesame mass, and each is about 2000times as massive as the electron. • The number of protons is the sameas the number of electrons (not shown)which orbit the nucleus. • For any given element, all nuclei have the same number of protons,but the number of neutrons willvary.

8. Nuclei, Nucleon, and Quarks

9. Structure of Sub-Atomic Particles

10. Quarks Murray Gell-Mann tookthe name quark from "Three quarks for musterMark", in James Joyce'sbook Finnegan's Wake.(1963) Whimsical names--called "flavors"--for the quarks

11. There Are Two Families of Elementary Particles

12. Families of Particles

13. Sample Fermionic HadronsBaryons (           ) and Anti-baryon (           ) Symbol Name QuarkContent ElectricCharge Mass(GeV/c2) Spin proton 1 0.938 1/2 anti-proton -1 0.938 1/2 neutron 0 0.940 1/2 lambda 0 1.116 1/2 omega -1 1.672 3/2 Hadrons

14. Ion Neutral atom Neutral Atoms and Ions • For neutral atoms # electrons = # of protons • For ions # electrons ≠ # protons

15. Isotopes • Atoms can have the same number of protons but different numbers of neutrons • These are the isotopes of an element

16. Isotopes Can Be Common or Rare

17. Physics of the Early 20th Century

18. The Radiation Spectrum from Hot Objects This filament is emitting all of the colors of the rainbow, which makes it "white" hot. This filament is not yet hot enough  to emit significant amounts of blue  light.

19. Blackbody Radiation Hot objects emit red and orange, while very hot objects emit red, orange, and blue light. Prism disperses electromagneticenergy into its component parts.

20. Predicting Black-Body Spectra • 19th century concern--predict the intensity of radiation emitted by a black body at a specific wavelength. • Wilhelm Wien’s theory predicted the overall form of the curve but at long wavelengths his theory disagreed with experimental data. • Rayleigh and Jeans theory fitted the experimental data at long wavelengths but it had major problems at shorter wavelengths. • The problem was a l term in the denominator. It meant that as the wavelength tended to zero, the curve would tend to infinity. • However, experimental data shows there is a peak wavelength for each temperature, and the energy emitted at either side of this peak drops. The Rayleigh-Jeans Law predicted no peak wavelength.

21. Problems of Classical Physics Classical physics predicted a continual increase in radiated energy with increasing frequency, the “ultraviolet catastrophe,” which did not occur.

22. Planck's Quantum • Planck explained the blackbody radiationspectrum by postulating that the radiationwas emitted by oscillating atoms, and furthermore that the energy was quantized. • The energy of these "atomic oscillators"had to be an integer multiple of hf, wheref is the frequency of vibration of the atoms and h = 6.63 x 10-34 J-s.E = nhf       n = 0, 1, 2, 3, ......... • Albert Einstein later applied a similar quantum concept to light. Max Planck (1858-1947)German physicist1918 Nobel Prize

23. The Photoelectric Effect—The Question

24. The Photoelectric Effect—The Answer • Einstein found an answer to the question by assuming that the energy of light was different for light of different frequencies • Unless the light was of high enough frequency to have the energy to release an electron, nothing happened • Einstein won the Nobel Prize for this discovery, NOT for the theory of relativity To cause the release of an electron light has to have energy equal at least to the Work Function W

25. Light Comes in Packets • A “packet of light is called a photon • A photon is a “quantum” of light—the smallest unit • Photons have no “rest energy” • They move at the speed of light but have no mass • The energy of a photon is given by: E = hf Lower energy Higher energy

26. Example KEmax = hf - W0-----------------------------------------------Problem:  A metal has a workfunction of 8 x 10-19 J. What is the maximum kinetic energy of electrons emitted from the metal when light of frequency f = 2 x 1015 Hz is shone on the surface? Solution:KEmax = 5.26 x 10-19 J

27. Photon Energy Example How many photons stream forth in onehour from a light bulb radiating 100 wattsof light energy?  Assume l = 500 nm.-------------------------------------------------------- c = l ff = c / l  = (3 x 108 m/s) / (500 x10-9 m)  = 6 x 1014 Hz E = h f    = (6.63 x 10-34 J-s) (6 x 1014 s-1)    = 3.98 x 10-19 J     (Solution continued) 100 watts = 100 J/s One hour = 3600 s Total energy radiated = 3.6 x 105 J  N = 3.6 x 105 J / 3.98 x 10-19 J        = 9 x 1023

28. The ElectronVolt (eV) and the Rule of 1240 The kinetic energy of an electronaccelerated across a potentialdifference of one volt is one electronvolt (eV).The eV is not a unit of charge, or a unit of voltage; it is a unitof energy. The energy E in electronvolts (eV) of a photon is related to its wavelength l in nanometers (nm) through the followingrelationship:E = (1240 eV-nm) / l

29. Photoelectric Effect ProblemLight of wavelength  l = 400 nm shineson a metal surface whose work functionW0 is 2.0 ev. What is the maximum kinetic energyof the photoemitted electrons? SolutionE = 1240 / 400    = 3.1 eVKE = 3.1 - 2.0      = 1.1 eV

30. Photocells in Garage Door Openers

31. Photocells in Movie Projection

32. Compton Scattering A.H. Compton discovered that x-ray photons when collided with electrons in a graphite crystal displayed particle properties of momentum equal to h / l. • When the x-rays emerged from the crystal they had a wavelength longer than their original value. • The lengthening of their wavelength showed that they had lost energy. The lost energy was accounted for in the KE of an ejected electron from the crystal.

33. Compton Scattering--explanation l' = (lo) + [(h / mc)(1-cosq)] where q is the angular deviation of the x-rays from their original path. Eo = Ef + KEelectronKEelectron = Eo - EfKEelectron = hfo - hffKEelectron = h ( fo - ff )KEelectron = hc ( 1/ lo - 1/ lf )where f = c / l

34. Deriving the Momentum of a Photon from special relativityDE = Dmc2 momentum = p = (E / c2)(v) but v = c, therefore p = E / c since E = hf = h (c / l), p = (hc / l) / c p = h / l In 1927, Arthur H. Compton was awarded the Nobel Prize in Physics for his discovery of the particle properties of x-rays.

35. Photon Momentum What power (in Watts) must the beam of a 0.5 kg flashlight deliver if the recoil momentum of the outgoing photons is to suspend the flashlight against the earth's gravity? (Assume that all photons from the bulk are visible photons with wavelength l =5000Å.) A certain flashlight bulb consumes 10 Watts of electrical power. If the conversion efficiency from electrical power to light energy is 10%, what mass could the outgoing beam of light be able to suspend against the earth's gravity? By what percentage is this force, when averaged over a period of one second, expected to fluctuate?

36. Fundamental Particle Relationships

37. Why Do Excited Atoms Emit Light In Narrow Bands (lines)

38. Atomic Spectra • Atoms in heated gases emit and absorb light of certain wavelengths. • Shown above are three emission spectra and one absorption spectrum.

39. Line Spectrum of Atomic Hydrogen In 1885 Johann Balmer discovered an equation which describes the emission-absorption spectrum of atomic hydrogen:1 / l = 1.097 x 107 (1 / 4 - 1 / n2)        where n = 3, 4, 5, 6, ...Balmer found this by trial and error, andhad no understanding of the physicsunderlying his equation. Why are there lines in the first place? To answer this question we have to understand a few things

40. Neils Bohr Explains the Hydrogen Atom and Balmer’s Results Neils Bohr, a Danish physicist, treated thehydrogen atom as if it were an electron ofcharge -e orbiting in a circular path abouta proton of charge +e.

41. Bohr Model

42. Energy Levels in Hydrogen Hydrogen atom explained

43. Energy Levels in Hydrogen En = -13.6 eV /n2-----------------------  E1 = - 13.6   eV E2 = -   3.40 eV E3 = -   1.51 eV E4 = -   0.85 eV E5 = -   0.54 eV E6 = -   0.38 eV

44. Energy Transitions in Atoms  Energy of photon = Energy lost by electron                            hf  = Ei - Ef

45. Calculating Wavelengths of Emitted Light hf  = Ei - Ef E3  --->  E2:Ei  =  - 1.51 eVEf  =  - 3.40 eV------------------------------ hf   = - 1.51 - (-3.40)      = 1.89 eV------------------------------ l = (1240 eV-nm) / E   = 1240 / 1.89   = 656 nm

46. Wavelength(nm) Relative Intensity Transition Color 383.5384 5 9 -> 2 Violet 388.9049 6 8 -> 2 Violet 397.0072 8 7 -> 2 Violet 410.174 15 6 -> 2 Violet 434.047 30 5 -> 2 Violet 486.133 80 4 -> 2 Bluegreen (cyan) 656.272 120 3 -> 2 Red 656.2852 180 3 -> 2 Red Balmer Series

47. Other Energy Transitions The final state in the energytransitions is n = 3 for thePaschen series, n = 2 for the Balmer series, and n = 1 for the Lyman series.Recalling that the range ofvisible wavelengths isapproximately 300-700 nm,one can see that only transitions ending at n = 2 emit light in the visible range

49. The De Broglie Wavelength Louis deBroglie     (1892-1987)  1929 Nobel Prize p = momentum l = h/p Electron beam produces a pattern similar to the one produced by light 

50. Electron's de Broglie Wavelength What is the wavelength of an electronmoving at a speed v = 2 x 104 m/s?----------------------------------------------------- m = 9.1 x 10-31 kgp = 18.2 x 10-27 kg-m/sh = 6.63 x 10-34 J-sl = h / p   = 3.6 x 10-8 m   = 360 x 10-10 m   = 36 x 10-9 m   = 36 nm