1301_Physics ProblemsTechnical Physics

# 1301_Physics ProblemsTechnical Physics - 642 Particle...

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642 Particle Physics and Cosmology P46.59 Momentum of proton is qBr 1 60 10 0 250 1 33 19 ... C k g C s m ej bg a f pm s p 532 10 20 . k g cp p = −− 1 60 10 1 60 10 99 8 11 11 .. . kg m s J MeV 22 . Therefore, pc p = 99 8 . M eV . The total energy of the proton is EE c p p =+ = + = 0 2 2 983 3 99 8 944 a f a f M e V . For pion, the momentum qBr is the same (as it must be from conservation of momentum in a 2- particle decay). π = 99 8 eV E 0 139 6 = c p = + = 0 2 2 139 6 99 8 172 af a f M e V Thus, total after total before Rest energy == . Rest Energy of unknown particle =+= 944 172 1116 MeV MeV MeV (This is a Λ 0 particle!) Mass MeV = 2 c . P46.60 Σ→ + 00 Λ γ From Table 46.2, mc Σ = 11925 2 and Λ = 1115 6 2 . Conservation of energy in the decay requires K E o 0, , Σ + ΛΛ or 11156 2 2 MeV MeV F H G I K J + p m E Λ Λ . System momentum conservation gives pp Λ = , so the last result may be written as 2 2 MeV MeV F H G I K J + p m E Λ or 2 2 2 2 MeV MeV F H G I K J + E Λ . Recognizing that Λ 2 = . MeV and pc E γγ = we now have 211156 2 . MeV MeV MeV + E E . Solving this quadratic equation, E = 74 4 eV. P46.61 pp pn +→++
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