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

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

Photo Electric Effect Reverse the voltage How it works Einstein’s idea Photon Energy = Work + Kinetic Energy hf =  + Emax hf = hfo + eV  - Work function (Depends on material) fo - Lowest frequency that ejects e - Electron charge V - The uh stopping potential + - V Reverse the voltage Ex. 1:  = 3.25 eV, V = 7.35 V,  = for light? Ex. 2:  = 70.9 nm,  = 5.10 eV, V = for electrons?

Photon Energy = Work + Kinetic Energy hf =  + Emax hf = hfo + eV  - Work function (Depends on material) fo - Lowest frequency that ejects e - Electron charge V - The uh stopping potential 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 =  + Emax hf = hfo + eV  - Work function (Depends on material) fo - Lowest frequency that ejects e - Electron charge V - The uh stopping potential 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.

Whiteboards: Photoelectric effect 1 | 2 | 3 | 4 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 Metal Work Function Whiteboards: Photoelectric effect 1 | 2 | 3 | 4

Photons of a certain energy strike a metal with a work function of 2 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 6.00 eV

Another metal has a work function of 3. 46 eV 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  = 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 147 nm

112 nm light strikes a metal with a work function of 4. 41 eV 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 11.079 eV = 4.41 eV + eVs 11.079 eV - 4.41 eV = 6.6688 eV = eVs Vs = 6.67 V 6.67 V

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 4.847 eV = Work function + 1.15 eV 11.079 eV - 1.15 eV = 3.70 eV 3.70 eV

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