p43_032 - 32. (a) We adapt Eq. 43-20: NPu = 0.002 g 239...

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32. (a) We adapt Eq. 43-20: N Pu = µ 0 . 002 g 239 g / mol ( 6 . 02 × 10 23 nuclei / mol ) 5 × 10 18 nuclei . (b) Eq. 43-19 leads to R = N ln 2 T 1 / 2 = 5 × 10 18 ln 2 2 . 41 × 10 4 y =1 . 4 × 10 14 / y which is equivalent to 4
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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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