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Unformatted text preview: Ch. 44 Page 1 CHAPTER 44 Elementary Particles Note: A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc / = (6.63 10 34 J s)(3 10 8 m/s)(10 9 nm/m)/(1.60 10 19 J/eV) ; E = (1.24 10 3 eV nm)/ = (1.24 10 12 MeV m)/ . 1. The total energy of the proton is E = K + m p c 2 = 6.35 GeV + 0.938 GeV = 7.29 GeV . 2. The total energy of the electron is E = K + m e c 2 = 35 GeV + 0.511 MeV = 35 GeV. Because the mass is negligible, the momentum is p = E / c . We find the wavelength from = h / p = hc / E = (1.24 10 12 MeV m)/(35 GeV)(10 3 MeV/GeV) = 3.5 10 17 m . 3. We find the magnetic field from the cyclotron frequency: f = qB /2p m ; 2.8 10 7 Hz = (1.60 10 19 C) B /2p(1.67 10 27 kg), which gives B = 1.8 T . 4. Veryhighenergy protons will have a speed v c . Thus the time for one revolution is t = 2p r / v = 2p(1.0 10 3 m)/(3.00 10 8 m/s) = 2.1 10 5 s = 21 s . 5. The cyclotron frequency is f = qB /2p m ; If we form the ratio for the two particles, we get f 2 / f 1 = ( q 2 / q 1 )( m 1 / m 2 ); f 2 /(26 MHz) = (2)(1/4), which gives f 2 = 13 MHz . 6. ( a ) The maximum kinetic energy is K = ! mv 2 = q 2 B 2 R 2 /2 m . If we form the ratio for the two particles, we get K / K p = ( q / q p ) 2 ( m p / m ); K /(8.7 MeV) = (2) 2 (1/4) = 1, which gives K = 8.7 MeV . We find the speed from K = ! mv 2 ; (8.7 MeV)(1.60 10 13 J/MeV) = ! (4)(1.67 10 27 kg) v 2 , which gives v = 2.0 10 7 m/s . ( b ) For deuterons we get K d / K p = ( q d / q p ) 2 ( m p / m d ); K d /(8.7 MeV) = (1) 2 (1/2) = 1/2, which gives K d = 4.3 MeV . We find the speed from K d = ! mv d 2 ; (4.3 MeV)(1.60 10 13 J/MeV) = ! (2)(1.67 10 27 kg) v d 2 , which gives v d = 2.0 10 7 m/s . Note that the particle and the deuteron have the same q / m . ( c ) The cyclotron frequency is f = qB /2p m ; If we form the ratio for the two particles, we get f d / f = ( q d / q )( m / m d ) = (1/2)(4/2) = 1, so they require the same frequency. We find the frequency from Ch. 44 Page 2 f d / f p = ( q d / q p )( m p / m d ); f d /(26 MHz) = (1)(1/2), which gives f d = f = 13 MHz ....
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This homework help was uploaded on 03/26/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.
 Spring '08
 ROSS
 Energy, Photon

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