CHAPTER 44 - 5

CHAPTER 44 - 5 - 25....

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25. The total kinetic energy after the decay of the stationary p +  is the  Q -value: Q   = [ m (p + ) –  m (e + ) –  m ( ν )] c 2   = [(139.6 MeV/ c 2 ) – (0.511 MeV/ c 2 ) – 0] c 2  = 139.1 MeV. For momentum conservation we have 0 =  p e  –  p ,   or   ( p c ) 2  = ( p e c ) 2  =  E e 2  – ( m e c 2 ) 2  =  K e 2  + 2 K e m e c 2 . For energy conservation we have Q  =  K e  +  p c  =  K e  +  p e c ,   or   ( p e c ) 2  = ( Q  –  K e ) 2  =  Q 2  – 2 K e Q   +  K e 2 . When we combine this with the result from momentum conservation, we get K e   =  Q 2 /2( Q  +  m e c 2 ) = (139.1 MeV) 2 /2(139.1 MeV + 0.511 MeV) =       69.3 MeV . 26. The minimum initial kinetic energy of the neutron and proton must provide the rest energy of the  K + K  pair: K p  +  K n  = 2 m K c 2  = 2(493.7 MeV) = 987.4 MeV. Because the neutron and proton have the same speed but different masses, they have slightly different kinetic energies:
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This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at Berkeley.

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CHAPTER 44 - 5 - 25....

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