ph135cS8 - Ph135c. Solution set #8, 5/27/10 1. (a)...

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Unformatted text preview: Ph135c. Solution set #8, 5/27/10 1. (a) Proceeding as in last weeks HW, we can write: = G F 2 ( g V 2 + 3 g A 2 ) 2 2 Z m e dE e E e q E e 2- m e 2 (- E e ) 2 = G F 2 ( g V 2 + 3 g A 2 ) 2 2 m e 5 Z x 1 dxx p x 2- 1( x- x ) 2 where x = m e , which in this case is about 1 . 036. Numerically, the integral is 1 . 94 10- 6 . This gives: 8 . 6 10- 31 MeV 22 . 8 yr (b) From Bertulain, the Coulomb factor is: F ( Z,E e ) = 2 1- e- 2 where = Ze 2 v e , where v e = E e E e 2- m e 2 is the electron velocity. For 3 H decay, Z = 2, and so including this factor gives: = G F 2 ( g V 2 + 3 g A 2 ) 2 2 Z m e dE e E e q E e 2- m e 2 F (2 ,E e )(- E e ) 2 = G F 2 ( g V 2 + 3 g A 2 ) 2 2 m e 5 Z x 1 dxx p x 2- 1 4 e 2 x x 2- 1 1- exp(- 4 e 2 x x 2- 1 ) ( x- x ) 2 Numerically, this integral is 2 . 78 10- 6 , which modifies our estimate of the lifetime to: 15 . 9 yr This is closer to the measured value....
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ph135cS8 - Ph135c. Solution set #8, 5/27/10 1. (a)...

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