p44_019

# P44_019 - 238 Pu nucleus then N = M m = 1 00 kg(238 u(1 661 × 10 − 27 kg u = 2 53 × 10 24 Thus P = NQ ln 2 T 1 2 =(2 53 × 10 24(5 50 × 10 6

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19. If R is the decay rate then the power output is P = RQ ,where Q is the energy produced by each alpha decay. Now R = λN = N ln 2 /T 1 / 2 ,where λ is the disintegration constant and T 1 / 2 is the half-life. The relationship λ =( ln2) /T 1 / 2 is used. If M is the total mass of material and m
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Unformatted text preview: 238 Pu nucleus, then N = M m = 1 . 00 kg (238 u)(1 . 661 × 10 − 27 kg / u) = 2 . 53 × 10 24 . Thus, P = NQ ln 2 T 1 / 2 = (2 . 53 × 10 24 )(5 . 50 × 10 6 eV)(1 . 60 × 10 − 19 J / eV)(ln 2) (87 . 7 y)(3 . 156 × 10 7 s / y) = 558 W ....
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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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