p44_023 - . 58 10 24 ) = 1 . 72 10 25 . (c) One neutron...

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23. (a) The energy yield of the bomb is E =(66 × 10 3 megaton)(2 . 6 × 10 28 MeV / megaton) = 1 . 72 × 10 27 MeV. At 200 MeV per Fssion event, (1 . 72 × 10 27 MeV) / (200 MeV) = 8 . 58 × 10 24 Fssion events take place. Since only 4 . 0% of the 235 U nuclei originally present undergo Fssion, there must have been (8 . 58 × 10 24 ) / (0 . 040) = 2 . 14 × 10 26 nuclei originally present. The mass of 235 U originally present was (2 . 14 × 10 26 )(235 u)(1 . 661 × 10 27 kg / u) = 83 . 7kg. (b) Two fragments are produced in each Fssion event, so the total number of fragments is 2(8
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Unformatted text preview: . 58 10 24 ) = 1 . 72 10 25 . (c) One neutron produced in a Fssion event is used to trigger the next Fssion event, so the average number of neutrons released to the environment in each event is 1 . 5. The total number released is (8 . 58 10 24 )(1 . 5) = 1 . 29 10 25 ....
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