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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 / λ = (6.63 × 10 –34 J · s)(3 × 10 8 m/s)(10 9 nm/m)/(1.60 × 10 –19 J/eV) λ ; E = (1.24 × 10 3 eV · nm)/ λ = (1.24 × 10 –12 MeV · m)/ λ . 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 λ = 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 µ 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 α / K p = ( q α / q p ) 2 ( m p / m α ); K α /(8.7 MeV) = (2) 2 (1/4) = 1, which gives K α = 8.7 MeV . We find the speed from K α = ! mv α 2 ; (8.7 MeV)(1.60 × 10 –13 J/MeV) = ! (4)(1.67 × 10 –27 kg) v α 2 , which gives v α = 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 α 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 α = ( q d / q α )( m α / m d ) = (1/2)(4/2) = 1, so they require the same frequency. We find the frequency from f d / f p = ( q d / q p )( m p / m d ); Ch. 44 Page 2 f d /(26 MHz) = (1)(1/2), which gives f d = f α = 13 MHz . 7. The size of a nucleon is d ˜ 2(1.2 × 10 –15 m) = 2.4 × 10 –15 m. Because 30 MeV « mc 2 , we find the momentum from K = p 2 /2 m , so the wavelength is λ = h / p = h /(2 mK ) 1/2 . For the α particle we have λ α = (6.63 × 10 –34 J · s)/[2(4)(1.67 × 10 –27 kg)(30 MeV)(1.60 × 10 –13 J/MeV)] 1/2 = 2.6 × 10 –15 m ˜ size of nucleon . For the proton we have λ p = (6.63 × 10 –34 J · s)/[2(1.67 × 10 –27 kg)(30 MeV)(1.60 × 10 –13 J/MeV)] 1/2 = 5.2 × 10 –15 m ˜ 2(size of nucleon)....
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics, Energy, Photon

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