Ch44Word - 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 =

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. Very-high-energy 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)....
View Full Document

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.

Page1 / 13

Ch44Word - 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 =

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online