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1302_Physics ProblemsTechnical Physics

1302_Physics ProblemsTechnical Physics - Chapter 46 P46.62...

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Chapter 46 643 P46.62 π µ ν µ + : From the conservation laws for the decay, m c E E π µ ν 2 139 6 = = + . MeV [1] and p p µ ν = , E p c ν ν = : E p c p c µ µ ν 2 2 2 2 2 105 7 105 7 = + = + d i a f b g a f . . MeV MeV or E E µ ν 2 2 2 105 7 = . MeV a f . [2] Since E E µ ν + = 139 6 . MeV [1] and E E E E µ ν µ ν + = d id i a f 105 7 2 . MeV [2] then E E µ ν = = 105 7 139 6 80 0 2 . . . MeV MeV a f . [3] Subtracting [3] from [1], 2 59 6 E ν = . MeV and E ν = 29 8 . MeV . P46.63 The expression e dE E k T B gives the fraction of the photons that have energy between E and E dE + . The fraction that have energy between E and infinity is e dE e dE e dE k T e dE k T e e e E k T E E k T E k T E E k T E k T E E k T E k T z z z z = = = B B B B B B B B B 0 0 0 b g b g . We require T when this fraction has a value of 0.0100 (i.e., 1.00%) and E = = × 1 00 1 60 10 19 . . eV J . Thus, 0 010 0 1.60 10 1.38 10 19 23 . = × × e T J J K e j e j or ln . . . 0 010 0
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