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# ph135cS8 - Ph135c Solution set#8 1(a Proceeding as in last...

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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: Γ = G F 2 ( g V 2 + 3 g A 2 ) 2 π 2 Z Δ m e dE e E e q E e 2- m e 2 (Δ- E e ) 2 = G F 2 ( g V 2 + 3 g A 2 ) 2 π 2 m e 5 Z x 1 dxx p x 2- 1( x- x ) 2 where x = Δ m e , which in this case is about 1 . 036. Numerically, the integral is 1 . 94 × 10- 6 . This gives: Γ ≈ 8 . 6 × 10- 31 MeV ⇒ τ ≈ 22 . 8 yr (b) From Bertulain, the Coulomb factor is: F ( Z,E e ) = 2 πη 1- e- 2 πη where η = 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: Γ = G F 2 ( g V 2 + 3 g A 2 ) 2 π 2 Z Δ m e dE e E e q E e 2- m e 2 F (2 ,E e )(Δ- E e ) 2 = G F 2 ( g V 2 + 3 g A 2 ) 2 π 2 m e 5 Z x 1 dxx p x 2- 1 4 πe 2 x √ x 2- 1 1- exp(- 4 πe 2 x √ x 2- 1 ) ( x- x ) 2 Numerically, this integral is 2 . 78 × 10- 6 , which modifies our estimate of the lifetime to: τ ≈ 15 . 9 yr This is closer to the measured value....
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ph135cS8 - Ph135c Solution set#8 1(a Proceeding as in last...

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