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Unformatted text preview: Solutions for HandIn Set #8 PHYS 212  Spring 2007 A54 Phtotoelectric Effect Application of simple classical theories of the photo electric effect suggests the following: The energy of emitted electrons should depend on the intensity of the light; this is contradicted by experiment. There should be an observable time delay for an electron to absorb enough energy to be emitted by lowintensity light; electrons are observed to be emitted instantaneously. A55 Kinetic energies of emitted electrons This is a twopart problem; both parts use the equation governing the photoelectic effect: K max = hf . The information about the results of the experiment with 700 nm light allows us to determine the work function . Rearranging the equation above gives = hf K max = hc  K max = 1240 eV nm 700 nm . 25 eV = 1 . 77 eV . 25 eV = 1 . 52 eV . For the 400 nm light we have K max = hf = hc  = 1240 eV nm 400 nm 1 . 52 eV = 3 . 1 eV 1 . 52 eV = 1 . 58 eV . A59 Electron in a box (a) The relationship between kinetic energy and momen tum is the same as it is in classical mechanics: K = p 2 2 m , so p = 2 mK = q 2 5 . 11 10 5 eV/ c 2 75 eV = 8 . 75 keV/ c, or equivalently p = 2 mK = p 2 9 . 1 10 31 kg 75 eV 1 . 6 10 19 J/eV = 4 . 67 10 24 kg m/s . (b) Using the de Broglie relationship gives = h p = hc pc = 1240 eV nm 8 . 75 10 3 eV = . 142 nm or equivalently = h p = 6 . 63 10 34 J s 4 . 67 10 24 kg m/s = 1 . 42 10 10 m (c) Second Excited State b Low probability (d) From the sketch of the wavefunction it is clear that the box is 1.5 wavelengths long, orthe box is 1....
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This homework help was uploaded on 04/07/2008 for the course PHYS 212 taught by Professor Ladd during the Spring '08 term at Bucknell.
 Spring '08
 Ladd
 Energy

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