p44_051 - 51. (a) Q = (5m2 H m3 He m4 He m1 H 2mn )c2 =...

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51. (a) Q =(5 m 2 H m 3 He m 4 He m 1 H 2 m n ) c 2 =[5(2 . 014102 u) 3 . 016029 u 4 . 002603 u 1 . 007825 u 2(1 . 008665 u)](931 . 5MeV / u) = 24 . 9MeV. (b) Assuming 30 . 0% of the deuterium undergoes fusion, the total energy released is E = NQ = µ 0 . 300 M 5 m 2 H Q. Thus, the rating is
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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