p44_015 - were 2(5 . 90 10 26 ) = 1 . 18 10 27 nuclei. The...

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15. If P is the power output, then the energy E produced in the time interval ∆ t (= 3 y) is E = P t = (200 × 10 6 W)(3 y)(3 . 156 × 10 7 s / y) = 1 . 89 × 10 16 J, or (1 . 89 × 10 16 J) / (1 . 60 × 10 19 J / eV) = 1 . 18 × 10 35 eV = 1 . 18 × 10 29 MeV. At 200 MeV per event, this means (1 . 18 × 10 29 ) / 200 = 5 . 90 × 10 26 Fssion events occurred. This must be half the number of Fssionable nuclei originally available. Thus, there
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Unformatted text preview: were 2(5 . 90 10 26 ) = 1 . 18 10 27 nuclei. The mass of a 235 U nucleus is (235 u)(1 . 661 10 27 kg / u) = 3 . 90 10 25 kg, so the total mass of 235 U originally present was (1 . 18 10 27 )(3 . 90 10 25 kg) = 462 kg....
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