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Unformatted text preview: Final Examination December 10, 2007 Physics 115A, Fall 2007 E. Abers Final Examination Solutions and Comments Question 1 [20 Points] Relativistic Kinematics Suppose a  particle (mass m ) at rest decays into an electron (mass m e ) and a photon. 1 Let be the angular frequency of the photon. Let p be the (magnitude of the) momentum of the electron. Part a) Write the energy conservation law in terms of m , m e , , p , and possibly planckover2pi1 and c . Part b) Write the momentum conservation law in terms of these quantities. Part c) What is p in terms of m , m e , and possibly planckover2pi1 and c (but not )? Dont bother putting in the numbers in this problem. GO ON TO THE NEXT PAGE SOLUTION Part a) m c 2 = radicalbig m 2 e c 4 + p 2 c 2 + planckover2pi1 Part b) p = planckover2pi1 c Part c) m c 2 = radicalbig m 2 e c 4 + p 2 c 2 + cp m c 2 cp = radicalbig m 2 e c 4 + p 2 c 2 m 2 c 4 2 m c 3 p = m 2 e c 4 p = c m 2  m 2 e 2 m 1 This decay has been looked for but never observed. 2 Final Examination Question 2 [20 Points] The Hydrogen Atom A hydrogen atom at rest decays from the n = 3 level to the n = 1 level by emitting a photon. (This is the Lyman line.) Usually we ignore the recoil energy of the atom, the kinetic energy of the atom after the decay. Write a formula (not a numerical value) for the ratio of the recoil kinetic energy of the final atom to the energy of the emitted photon. Make any approximations you want that introduce errors of 1% or smaller. Use non relativistic kinematics for all particles except the photon the recoil energy is small. Note: The ratio of these two energies is dimensionless . Write your answer in the form Numerical constant other dimensionless factors some power parenleftbigg electron mass atomic mass parenrightbigg some power where is the fine structure constant. GO ON TO THE NEXT PAGE SOLUTION The energy of the transition is E 3 E 1 = 2 mc 2 2 parenleftbigg 1 1 9 parenrightbigg = 4 9 2 mc 2 so energy conservation requires planckover2pi1 = 4 9 2 mc 2 planckover2pi1 2 2 2 Mc 2 (1) The last term is much smaller than planckover2pi1 , and so to a good approximation planckover2pi1 = 4 9 2 mc 2 and p = planckover2pi1 c = 4 9 2 mc The momentum of the photon is p = planckover2pi1 c By momentum conservation this is the momentum of the recoiling atom also. Its kinetic energy is therefore T = p 2 2 M = planckover2pi1 2 2 2 Mc 2 where M is the atomic (not the electron) mass. The ratio of this energy to the photon energy is planckover2pi1 2 Mc 2 = 1 2 Mc 2 4 9 2 mc 2 = 2 9 2 m M Physics 115A, Fall 2007 3 Note: If you want to do better, you can solve the quadratic equation (1) for planckover2pi1 exactly, or find a power series for the frequency of the photon, for example by by successive approximations: planckover2pi1 = 4 9 2 mc 2 1 M 8 81 4 m 2 c 2 + = 4 9 2 mc 2 1 + 2 9 2 m M...
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This test prep was uploaded on 04/18/2008 for the course PHYS 115a taught by Professor Abers during the Winter '08 term at UCLA.
 Winter '08
 Abers
 Mass, Photon

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