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Final Exam Key

# Final Exam Key - PHY 320 Spring 2004 Final Exam May 10 4:00...

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PHY 320 Spring 2004 Final Exam May 10 @ 4:00 pm (Ave. = 81.3%) 1 E k (eV) Slope = 1 E photon (eV) 2 -2 n = 1 n = 2 n = 3 n = Energy (eV) 0 –Z 2 E 0 –Z 2 E 0 /4 –Z 2 E 0 /9 1. Complete the table below by entering the particles (photon or electron) that initiate and result from the processes listed, and indicate if the process is elastic, inelastic, or not applicable. Below the table, list one process that demonstrates the particle-like nature of light and another process that shows the wave-like behavior of light. (Ave = 82.8%) In Out Compton Scattering Photon Photon Inelastic LEED Electron Electron Elastic Photoelectric Photon Electron n/a Xray Diffraction Photon Photon Elastic Particle-like light : photoelectric effect, Compton scattering Wave-like light : Xray diffraction 2. Carefully draw an energy-level diagram for a one-electron atom of nuclear charge Z with the levels n = 1, 2, 3, shown and energies labeled. Mark on the diagram the longest wavelength Balmer series transition. What is E 0 physically? (Ave = 85.0%) Using the relationship for one-electron energy levels: 2 2 o n Z E E n = Lowest Energy/ Longest λ of Balmer series: n i = 3 to n f = 2. The constant E 0 is the ionization energy of hydrogen, which is the energy necessary to remove the one electron from the ground state energy level. 3. For the photoelectric effect , sketch the kinetic energy E K of the electrons vs. the incoming photon energy E photon for a metal with ϕ = 2 eV. Indicate the slope and intercept values. In addition, find the stopping voltage for an incoming photon with 5 eV energy. (Ave = 93.3%) The photoelectric equation states that the incoming photon energy E photon equals the energy necessary to eject an electron (work function φ ) plus the electron’s kinetic energy E K . stop photon photon kinetic Graph: , where slope = 1, , intercepts , 5 eV 2 eV 3 eV E E x y eV E ϕ ϕ ϕ ϕ = = = = = 4. If a beam of protons travels through a very narrow single slit, sketch the intensity pattern that is formed on a detector past the slit. What happens to the pattern if the slit is made wider and why? What happens if the beam of protons is replaced by a beam of electrons with equal kinetic energy and why? (Ave = 76.8%) The intensity pattern is that for single-slit diffraction, which is the function (sinx/x) 2 , where minima occur for the condition n λ = asin θ .

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