ch43-p018 - × 10 27(3.90 × 10 – 25 kg = 462 kg 18 If P...

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At 200 MeV per event, this means (1.18 × 10 29 )/200 = 5.90 × 10 26 fission events occurred. This must be half the number of fissionable nuclei originally available. Thus, there 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
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Unformatted text preview: × 10 27 )(3.90 × 10 – 25 kg) = 462 kg. 18. 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 = (1.89 × 10 16 J)/(1.60 × 10 – 19 J/eV) = 1.18 × 10 35 eV = 1.18 × 10 29 MeV....
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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