1281_Physics ProblemsTechnical Physics

# 1281_Physics ProblemsTechnical Physics - 622 P46.2 Particle...

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622 Particle Physics and Cosmology P46.2 The minimum energy is released, and hence the minimum frequency photons are produced, when the proton and antiproton are at rest when they annihilate. That is, EE = 0 and K = 0. To conserve momentum, each photon must carry away one-half the energy. Thus, E E Eh f min . == = = 2 2 938 3 0 0 MeV min . Thus, f min .. . . = × ×⋅ 938 3 1 60 10 6626 10 227 10 13 34 23 MeV J MeV Js Hz af ej λ × × c f min . . . 300 10 132 10 8 23 15 ms Hz m. P46.3 In γ →+ +− pp , we start with energy 2.09 GeV we end with energy 938.3 MeV + 938.3 MeV + 95.0 MeV + K 2 where K 2 is the kinetic energy of the second proton. Conservation of energy for the creation process gives K 2 118 = MeV . Section 46.3 Mesons and the Beginning of Particle Physics P46.4 The reaction is µν ν +→+ e muon-lepton number before reaction: −+ = 10 1 afa f electron-lepton number before reaction:
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## This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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