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Unformatted text preview: PHYS2020, Fall 2006 CAPA Set 15 1. [1pt] Europium has a work function of 2.50 eV . What is the maximum kinetic energy of electrons if the metal is illuminated by light of wavelength 270 nm ? The energy of the photon first goes into liberating the electron from the atom. The remaining energy is given as kinetic energy to the liberated electron. Thus, KE = hf W where W is the work function. Since c = f , we can write this as E = hc/ W . For the numbers given here, we get KE = hc/ W = 4 . 14 10 15 eV s 3 10 8 m/s/ 270 10 9 m 2 . 50 eV = 4 . 60 eV 2 . 50 eV = 2 . 1 eV . 2. [1pt] What is their speed? Since the kinetic energy is (2 . 1 eV ) is much less than the rest mass energy (511000 eV ), this is a nonrelativistic problem. Therefore KE = 1 / 2 mv 2 and v = radicalBig 2 KE m . This can be solved by v = radicalBig 2 2 . 1 eV 511000 eV/c 2 = 8 . 22 10 6 c 2 = 0 . 00287 c = 860000 m/s . Or one can convert the kinetic energy to Joules and use the mass of the electron in kilograms: v = radicalbigg 2 2 . 1 eV 1 . 6 10 19 J/eV 9 . 11 10 31 kg = radicalBig 7 . 38 10 11 m 2 /s 2 = 860000 m/s . 3. [1pt] By what potential difference must a proton ( m = 1 . 67 10 27 kg ) be accelerated (from rest) to have a wavelength = 4 . 46 10 12 m ?...
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This note was uploaded on 02/11/2008 for the course PHYS 2020 taught by Professor Dubson during the Spring '06 term at Colorado.
 Spring '06
 DUBSON
 Physics, Energy, Kinetic Energy, Work, Photon, Light

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