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Nuclear Fission Energy in Nuclear Fissions

Nuclear Fission Energy in Nuclear Fissions. Nuclear reactions (fission).

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Nuclear Fission Energy in Nuclear Fissions

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  1. Nuclear FissionEnergy in Nuclear Fissions

  2. Nuclear reactions (fission) Otto Hahn and Frederic Strassmann managed to bombard a nucleus of with neutrons (1938). The product isotopes of this reaction were radioactive and they thought they were turning the uranium nuclei into heavier nuclei. Instead, when they analysed the product isotopes chemically, they discovered that the nuclei had split into lighter radioactive isotopes of barium and krypton. Incoming neutron Kr-90 nucleus U-235 nucleus Ba-144 nucleus

  3. Nuclear fission The word equation of this particular nuclear fission is represented underneath: • What is the total number of nucleons before and after the reaction? • Less than before • More than before • The same as before

  4. Wrong. Try again

  5. Nuclear fission The word equation of this particular nuclear fission is represented underneath: • What is the total number of nucleons before and after the reaction? • Less than before • More than before • The same as before

  6. Nuclear fission • Other experiments showed that: • Several neutrons are released in the fission • Different products of the fission reaction can occur, but the number of protons and neutrons and the total energy is conserved. • The products are radioactive. In fact, they normally have a greater neutron/proton ratio than the stable nuclei of the same elements • Slow neutrons achieve nuclear fission of U-235 more easily than fast ones. This is because of the potential well around a nucleon. • The energy released in nuclear fission is much higher than energy released by chemical processes.

  7. Nuclear fission Calculate the energy released Q in the following nuclear reaction. For your calculations click on the link below to find out the nuclear masses. Nuclear mass finder m(U-235) = 235.04438 u m(n) = 1.00866 u Dm = tot m (before) – tot mass (after) = 0.1921 u m(Ba-144) = 143.92279 u E equivalence of mass loss = 0.1921 x 931 MeV = = 178.85 MeV 89.92089 u m(Kr-90) = 2m(n) = 2.01732 u

  8. Energy in the nuclear fission So, one nucleus of U-235 gives out about 200 MeV of energy, because the energy released is the same as the energy equivalence of the mass loss in the reaction. If we consider 6 x 1023 nuclei of U-235, i.e. the number of nuclei in 235 g of Uranium 235, the energy release is 12 x 1023 MeV = 2 x 1013 J. Calculate the energy released from the fission of 1 kg of U-235 Work out the amount of U-235 needed to run a 100 MW power station for one year, if its efficiency is 35%.

  9. Energy in the nuclear fission t = 60 x 60 x 24 x 365 (s) = 3.1536 x 107 s Energy produced per year (100% eff) = Power x t = = 108 W x 3.1536 x 107 s ~ 3 x 1015 J m(U-235) required = 9 x 1015 : 8 x 1013 = 112.5 kg

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