CHAPTER 44 - 13 - is . Thus we can write the reaction as +...

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  u u   d d   d d   u u 56. For energy conservation of the particle’s decay, we have m P c 2  =  E D1  +  E D2  =  K D1  +  m D1 c 2  +  K D2  +  m D2 c 2 . The two decay fragments must move in opposite directions.  For momentum conservation we have 0 =  p D1  –  p D2 ,   or   ( p D1 c ) 2  = ( p D2 c ) 2  =  E D1 2  – ( m D1 c 2 ) 2  =  E D2 2  – ( m D2 c 2 ) 2 . When we use the result from energy conservation, we get  E D1 2  – ( m D1 c 2 ) 2  = ( m P c 2  –  E D1 ) 2  – ( m D2 c 2 ) 2 , which gives E D1 = [( m P c 2 ) 2  + ( m D1 c 2 ) 2  – ( m D2 c 2 ) 2 ]/2 m P c 2 . Thus the kinetic energy is K D1 E D1  –  m D1 c 2  = {[( m P c 2 ) 2  + ( m D1 c 2 ) 2  – ( m D2 c 2 ) 2 ]/2 m P c 2 } – (2 m P c 2 m D1 c 2 /2 m P c 2 = [( m P c 2  –  m D1 c 2 ) 2  – ( m D2 c 2 ) 2 ]/2 m P c 2 . 57. From Table 44–2 we see that the p 0  has charge = 0,  B  = 0,  L  = 0, and  S  = 0.  From Table 44–3 we see that the p 0  could be either   or  .  Actually it is a combination of the two, but we will simplify the analysis by assuming the p
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Unformatted text preview: is . Thus we can write the reaction as + n p + p ; . Thus we see the annihilation of a pair. 58. We find the speed of the proton from its kinetic energy: K = {[1/(1 v 2 / c 2 ) 1/2 ] 1} mc 2 ; 7.0 10 6 MeV = {[1/(1 v 2 / c 2 ) 1/2 ] 1}(938 MeV), which gives = 1/(1 v 2 / c 2 ) 1/2 = 7.46 10 3 . This means the speed is very close to the speed of light. If we rearrange and use the approximation (1 x ) 1/2 1 x /2, we get v / c = [1 (1/ 2 )] 1/2 1 (1/2 2 ) = 1 [1/2(7.46 10 3 ) 2 ], which gives v / c = 1 9 10 9 ....
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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 University of California, Berkeley.

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