1108_Physics ProblemsTechnical Physics

1108_Physics ProblemsTechnical Physics - Chapter 39...

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

View Full Document Right Arrow Icon
Chapter 39 449 P39.45 (a) Em c == γ 2 20 0 . G eV with mc 2 0511 = . M e V for electrons. Thus, = × × 20 0 10 391 10 9 4 . . eV 0.511 10 eV 6 . (b) = 1 1 2 4 uc bg . from which = 0 999 999 999 7 . (c) LL u c L p p =− F H G I K J × × = 1 300 10 767 10 767 2 3 2 . .. m 3.91 10 m c m 4 *P39.46 (a) P × energy J 100 10 s W t 2 200 10 15 13 . (b) The kinetic energy of one electron with vc = 09999 . is −= F H G G I K J J ××= × −− 1 1 10 9 9 9 9 1 9 11 10 69 7 8 20 10 572 10 2 2 31 2 14 12 ej e j mc . . . kg 3 10 m s J J 8 Then we require 001 100 2 5 72 10 12 . . J J N N = × × 21 0 350 10 4 12 7 J J . P39.47 Conserving total momentum of the decaying particle system, pp before decay after decay 0 m u m u ve = µµ γγ 207 . Conservation of mass-energy for the system gives EEE v µπ += : mc pc mc v 22 207 273 m p c m e v e . Substituting from the momentum equation above, 207
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online