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

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Presentation on theme: "The photoelectric effect Contents: Einstein’s proposed experiment Solving photoelectric problems Example 1 | Example 2Example 1Example 2 Whiteboard Photon."— Presentation transcript:

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

2 Waves Photons Color Brightness Energy per photon changes E = hf Radio waves vs X-rays Wavelength Changes Amplitude Changes # of photons changes many = bright few = dim CCD Devices, High speed film big = red small = blue small = dim big = bright Light

3 Photo Electric Effect How it works Einstein’s idea +- V Reverse the voltage Photon Energy = Work + Kinetic Energy hf =  + E max  - Work function (Depends on material) hf – photon energy E max – max KE of photoelectrons (= eV s ) – where V s is the stopping potential

4 Photon Energy = Work + Kinetic Energy hf =  + E max 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? 117 nm

5 Photon Energy = Work + Kinetic Energy hf =  + E max 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? 12.4 V

6 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? Photon energy = Work + Kinetic Energy Photon Energy = 3.25 eV + 7.35 eV = 10.60 eV Convert to J: (10.60 eV)(1.602E-19 J/eV) = 1.69812E-18 J E = hf = hc/, = hc/E = (6.626E-34 Js)(3.00E8 m/s)/(1.69812E-18 J) = 1.17059E-07 m = 117 nm Photon Energy = Work + Kinetic Energy hf =  + E max hf = hf o + eV  - Work function (Depends on material) f o - Lowest frequency that ejects e - Electron charge V - The uh stopping potential

7 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? E = hf = hc/ = (6.626E-34 Js)(3.00E8 m/s)/(70.9E-9) = 2.80367E-18 J Convert to eV: (2.80367E-18 J)/(1.602E-19 J/eV) = 17.5 eV Photon Energy = Work + Kinetic Energy 17.5 eV = 5.10 + Kinetic Energy Kinetic Energy = 12.4 eV, so 12.4 V would stop the electrons. Photon Energy = Work + Kinetic Energy hf =  + E max hf = hf o + eV  - Work function (Depends on material) f o - Lowest frequency that ejects e - Electron charge V - The uh stopping potential

8 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 Chromium4.6 Nickel4.6 Metal Work Function Whiteboards: Photoelectric effect 11 | 2 | 3 | 4234

9 6.00 eV Photon energy = Work function + Kinetic energy of electron Photon energy = 2.15 eV + 3.85 eV = 6.00 eV 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?

10 147 nm 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 10 8 m/s)/(1.3553 x 10 -18 J) = 1.4667x 10 -07 m = 147 nm 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)

11 6.67 V 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 10 8 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 + eV s 11.079 eV - 4.41 eV = 6.6688 eV = eV s V s = 6.67 V 112 nm light strikes a metal with a work function of 4.41 eV. What is the stopping potential of the ejected electrons? (2)

12 3.70 eV 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 10 8 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 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)

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


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