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The photoelectric effect Contents: Einstein’s proposed experiment Solving photoelectric problems

The photoelectric effect Contents: Einstein’s proposed experiment Solving photoelectric problems Example 1 | Example 2 Whiteboard Photon vs wave theory. small  = blue. big  = red. small = dim. big = bright. Light. Waves Photons. Wavelength Changes.

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The photoelectric effect Contents: Einstein’s proposed experiment Solving photoelectric problems

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  1. The photoelectric effect • Contents: • Einstein’s proposed experiment • Solving photoelectric problems • Example 1 | Example 2 • Whiteboard • Photon vs wave theory

  2. small  = blue big  = red small = dim big = bright Light Waves Photons Wavelength Changes Energy per photon changes E = hf X-Rays, UV, Gamma Color Amplitude Changes # of photons changes many = bright few = dim CCD Devices, High speed film Brightness

  3. V - + Reverse the voltage • Einstein • Proposes Photon theory • Experiment: Light ejects electrons Ammeter detects • Some Potential V stops all electrons • This is called the “Stopping potential” • (Pretty tough, huh?) TOC

  4. - + Electric Force Consider an ejected electron hurtling toward the negative plate: Remember V = W/q, so W = Vq Definition of electron volt Ekin turns to potential energy If 5.12 V is the stopping potential, then Ekin = 5.12 eV

  5. V - + Einstein’s Photon theory predicts: Photon energy = Work function + Kinetic energy of electron hf =  + Ekmax hf = hfo + eVs  - Work function (Depends on material) fo - Lowest frequency that ejects e - Electron charge Vs - The uh stopping potential TOC

  6. Ag (silver) 4.26 Au (gold) 5.1 Cs (cesium) 2.14 Cu (copper) 4.65 Li (lithium) 2.9 Pb (lead) 4.25 Sn (tin) 4.42 Chromium 4.6 Nickel 4.6 Work Function Metal

  7. Photon energy = Work function + Kinetic energy of electron hf =  + Ekmax hf = hfo + eVs  - Work function fo - Lowest frequency that ejects e - Electron charge Vs - The uh stopping potential e = 1.602 x 10-19 C V = W/q E = hf = hc/ Example 1: A certain metal has a work function of 3.25 eV. When light of an unknown wavelength strikes it, the electrons have a stopping potential of 7.35 V. What is the wavelength of the light? TOC

  8. Photon energy = Work function + Kinetic energy of electron hf =  + Ekmax hf = hfo + eVs  - Work function fo - Lowest frequency that ejects e - Electron charge Vs - The uh stopping potential e = 1.602 x 10-19 C V = W/q E = hf = hc/ Example 2: 70.9 nm light strikes a metal with a work function of 5.10 eV. What is the maximum kinetic energy of the ejected photons in eV? What is the stopping potential? TOC

  9. Whiteboards: Photoelectric effect 1 | 2 | 3 | 4 TOC

  10. Photons of a certain energy strike a metal with a work function of 2.15 eV. The ejected electrons have a kinetic energy of 3.85 eV. (A stopping potential of 3.85 V) What is the energy of the incoming photons in eV? Photon energy = Work function + Kinetic energy of electron Photon energy = 2.15 eV + 3.85 eV = 6.00 eV W 6.00 eV

  11. Another metal has a work function of 3.46 eV. What is the wavelength of light that ejects electrons with a stopping potential of 5.00 V? (2) E = hf = hc/, Photon energy = Work function + Kinetic energy of electron Photon energy = 3.46 eV + 5.00 eV = 8.46 eV E = (8.46 eV)(1.602 x 10-19 J/eV) = 1.3553 x 10-18 J E = hf = hc/,  = hc/E = (6.626 x 10-34 Js)(3.00 x 108 m/s)/(1.3553 x 10-18 J)  = 1.4667x 10-07 m = 147 nm W 147 nm

  12. 112 nm light strikes a metal with a work function of 4.41 eV. What is the stopping potential of the ejected electrons? (2) E = hf = hc/, 1 eV = 1.602 x 10-19 J Photon energy = Work function + Kinetic energy of electron E = hf = hc/ = (6.626 x 10-34 Js)(3.00 x 108 m/s)/(112 x 10-9 m) E = 1.7748 x 10-18 J E = (1.7748 x 10-18 J)/(1.602 x 10-19 J/eV) = 11.079 eV Photon energy = Work function + Kinetic energy of electron 11.079 eV = 4.41 eV + eVs 11.079 eV - 4.41 eV = 6.6688 eV = eVs Vs = 6.67 V W 6.67 V

  13. 256 nm light strikes a metal and the ejected electrons have a stopping potential of 1.15 V. What is the work function of the metal in eV? (2) E = hf = hc/, 1 eV = 1.602 x 10-19 J Photon energy = Work function + Kinetic energy of electron E = hf = hc/ = (6.626 x 10-34 Js)(3.00 x 108 m/s)/(256 x 10-9 m) E = 7.7648 x 10-19 J E = (7.7648 x 10-19 J)/(1.602 x 10-19 J/eV) = 4.847 eV Photon energy = Work function + Kinetic energy of electron 4.847 eV = Work function + 1.15 eV 11.079 eV - 1.15 eV = 3.70 eV W 3.70 eV

  14. Einstein’s Photon theory predicts: Photon energy = work function + Kinetic energy of electron hf =  + Ekmax Ekmax = hf -  • Photon Theory predicts: • Ekmax rises with frequency • Intensity of light ejects more • Wave Theory predicts: • Ekmax rises with Amplitude (Intensity) • Frequency should not matter • Survey says Millikan does the experiment 1915 conclusion 1930 conclusion TOC

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