1300_Physics ProblemsTechnical Physics

1300_Physics ProblemsTechnical Physics - 641 Chapter 46...

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Chapter 46 641 P46.56 (a) ∆∆≈ Et = , and ∆= = × × t r c 14 10 31 0 47 10 15 8 24 . . m ms s ∆≈ = ×⋅ × × F H G I K J E t = 1055 10 23 10 160 10 34 24 11 13 2 . . . . . Js s J 1 MeV J MeV ej m E c cc = ≈× 2 22 1 4 10 10 .~ MeV MeV (b) From Table 46.2, mc π 2 139 6 = . M eV a pi-meson . P46.57 Λ 2 1115 6 = Λ 0 →+ p p 2 938 3 = 2 139 6 = The difference between starting rest energy and final rest energy is the kinetic energy of the products. KK p += 37 7 . MeV and ppp p == Applying conservation of relativistic energy to the decay process, we have 938 3 938 3 139 6 139 6 37 7 2 2 .. . af +− L N M O Q P ++ L N M O Q P = pc MeV. Solving the algebra yields pc pc p 100 4 . Then, Km c m c pp p =+ = 2 2 2 2 100 4 5 35 M e V K = 139 6 100 4 139 6 32 3 . . MeV . P46.58 By relativistic energy conservation in the 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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