44_InstructorSolutions

# 44_InstructorSolutions - 44 PARTICLE PHYSICS AND COSMOLOGY...

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44-1 P ARTICLE P HYSICS AND C OSMOLOGY 44.1. (a) IDENTIFY and SET UP: Use Eq.(37.36) to calculate the kinetic energy K . EXECUTE: 22 1 10 . 1 5 4 7 1/ Km c m c vc ⎛⎞ =− = ⎜⎟ ⎝⎠ 31 14 9.109 10 kg, so 1.27 10 J mK −− = × (b) IDENTIFY and SET UP: The total energy of the particles equals the sum of the energies of the two photons. Linear momentum must also be conserved. EXECUTE: The total energy of each electron or positron is 221 3 1.1547 9.46 10 J. EKm c m c =+ = = × The total energy of the electron and positron is converted into the total energy of the two photons. The initial momentum of the system in the lab frame is zero (since the equal-mass particles have equal speeds in opposite directions), so the final momentum must also be zero. The photons must have equal wavelengths and must be traveling in opposite directions. Equal λ means equal energy, so each photon has energy 14 9.46 10 J. × (c) IDENTIFY and SET UP: Use Eq. (38.2) to relate the photon energy to the photon wavelength. EXECUTE: / Eh c = so 14 / /(9.46 10 J) 2.10 pm hc E hc == × = EVALUATE: The wavelength calculated in Example 44.1 is 2.43 pm. When the particles also have kinetic energy, the energy of each photon is greater, so its wavelength is less. 44.2. The total energy of the positron is 2 5.00 MeV 0.511MeV 5.51MeV. c = + = We can calculate the speed of the positron from Eq.(37.38): 2 2 2 2 0.511 MeV 1 1 0.996. 5.51 MeV 1 mc v mc E cE v c = = 44.3. IDENTIFY and SET UP: By momentum conservation the two photons must have equal and opposite momenta. Then Ep c = says the photons must have equal energies. Their total energy must equal the rest mass energy 2 Em c = of the pion. Once we have found the photon energy we can use f = to calculate the photon frequency and use / cf = to calculate the wavelength. EXECUTE: The mass of the pion is e 270 , m so the rest energy of the pion is 270(0.511 MeV) 138 MeV. = Each photon has half this energy, or 69 MeV. 61 9 22 34 (69 10 eV)(1.602 10 J/eV) s o 1 . 71 0 H z 6.626 10 J s E f f h ×× = = × ×⋅ 8 14 22 2.998 10 m/s 1.8 10 m 18 fm. 1.7 10 Hz c f × = × = × EVALUATE: These photons are in the gamma ray part of the electromagnetic spectrum. 44.4. (a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of 23 2.27 10 Hz × and a wavelength of 15 1.32 10 m. × (b) The energy of each photon will be 938.3 MeV 830 MeV 1768 MeV, += with frequency 22 42.8 10 Hz × and wavelength 16 7.02 10 m. × 44.5. (a) ee e 270 207 63 mm m m m m πμ ++ Δ= = = 63(0.511MeV) 32 MeV. E (b) A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you could always find a frame where energy was not conserved. This cannot occur. 44

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44-2 Chapter 44 44.6. (a) 34 14 23 1 8 (6.626 10 J s) 1.17 10 m 0.0117 pm (207)(9.11 10 kg)(3.00 10 m s) μμ hc hc h Em c m c λ ×⋅ == = = = × = ×× In this case, the muons are created at rest (no kinetic energy). (b) Shorter wavelengths would mean higher photon energy, and the muons would be created with non-zero kinetic energy.
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## This note was uploaded on 06/03/2010 for the course PHYS 227 taught by Professor Rabe during the Fall '08 term at Rutgers.

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44_InstructorSolutions - 44 PARTICLE PHYSICS AND COSMOLOGY...

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