problem43_53

# problem43_53 - 3 u V Me 5 931 u 10 495 3 3 2 = × = ∆ =...

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43.53: a) Only the heavier one Al) ( 25 13 can decay into the lighter one Mg). ( 25 12 b) X Z A A Z + = = + 1 , 0 X ) Mg ( ) Al ( 25 12 25 13 is a positron + β decay or X Z A - = = + 1 , 0 Mg X Al) ( 25 12 A Z 25 13 is an electron electron capture c) Using the nuclear masses, we calculate the mass defect for + decay: u 10 495 . 3 ) u 00054858 . 0 ( 2 u 985837 . 24 u 990429 . 24 ) 12 ) Mg ( ( ) 13 ) Al ( ( 3 e e 25 12 e 25 13 - × = - - = - - - - = m m M m M m . MeV 255
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Unformatted text preview: . 3 ) u V Me 5 . 931 ( ) u 10 495 . 3 ( ) ( 3 2 = × = ∆ = ⇒-c m Q For electron capture: u 10 4.592 u 985837 . 24 u 990429 . 24 Mg) ( ) Al ( 3 25 12 25 13-× =-=-= ∆ M M m MeV. 277 . 4 u) V Me (931.5 u) 10 592 . 4 ( ) ( 3 2 = × = ∆ = ⇒-c m Q...
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## This note was uploaded on 05/12/2008 for the course PHYS 262 taught by Professor Dougherty during the Spring '08 term at Lafayette.

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