CHAPTER 44 - 8

# CHAPTER 44 - 8 - either or. u u d d thetwo,...

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u u   d d   u u   d d d _ d u _ d u u u _ p p š   0 š   0 u _ u _ d _ s d u d s u u u _ p K u _ u p K either   or  .  Actually it is a combination of  the two, but we will simplify the diagram by  assuming that one p 0  is   and the other is  .   We see that the exchanged particle is p + . We see that the exchanged particle is p. Note that high-energy  p    collisions  usually produce many particles.

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40. We see that the exchanged particle is p 0 . 41. The total energy of the proton is E  =  K  +  mc 2  = 25 GeV + 0.938 GeV =       26 GeV . We find the momentum from ( pc ) 2  =  E 2  – ( mc 2 ) 2  = (26 GeV) 2  – (0.938 GeV) 2 , which gives  pc  = 26 GeV. We find the wavelength from λ  =  h / p  =  hc / pc = (1.24 × 10 –12  MeV     m)/(26 GeV)(10 3  MeV/GeV) =
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CHAPTER 44 - 8 - either or. u u d d thetwo,...

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