1285_Physics ProblemsTechnical Physics

1285_Physics ProblemsTechnical Physics - 626 P46.18...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
626 Particle Physics and Cosmology P46.18 (a) p e →+ + γ Baryon number: +→ + 100 B 0, so baryon number conservation is violated. (b) From conservation of momentum for the decay: pp e = . Then, for the positron, Ep cE ee e 2 2 0 2 =+ bg , becomes E E e 2 2 0 22 0 2 = + γγ di ,, . From conservation of energy for the system: EE E pe 0, or EE E ep =− so E p 2 0 2 0 2 2 + . Equating this to the result from above gives EEE p 2 0 2 0 2 0 2 2 +=− + , or E E p = = = 0 2 0 2 0 2 9383 0511 29383 469 , .. . MeV MeV MeV MeV af . Thus, = = 0 938 3 469 469 , . MeV MeV MeV . Also, p E cc == 469 MeV and c e 469 MeV . (c) The total energy of the positron is E e = 469 MeV . But, E vc e 0 0 2 1 , , so 1 109 10 2 0 3 F H G I K J = × v c E E e e , . . MeV 469 MeV which yields: = 09999994 . . P46.19
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online