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# Ppt djy 2011 2 topic 7 and 13 nuclear reactions

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### Ppt djy 2011 2 topic 7 and 13 nuclear reactions

1. 1. IB Physics Power Points Topic 07 and 13 Atomic and Nuclear Physicswww.pedagogics.ca Nuclear Reactions
2. 2. Nuclear TransmutationsDefinition: A nuclear reaction where one nuclide ischanged into another. Examples of nuclearreactions include fission, fusion, and radioactivedecay
3. 3. Artificial (induced) Transmutations• target nucleus is bombarded with another particlesuch as a nucleon, an alpha particle or another smallnucleus• if the target nucleus „captures‟ an incoming particle atransmutation reaction occurs
4. 4. Artificial (induced) TransmutationsThis reaction was first observed by Rutherford in 1919. 4 2+ 14 17 1 + 2 He 7 N 8 O+ p 1This nuclear reaction equation is balanced. The sumsof the mass and atomic numbers are equal for bothsides of the reaction arrow (don‟t worry about theelectrons ie the ionic charges).
5. 5. PracticeA neutron is observed to strike a 16O nucleus and adeuteron (2H) is given off. What is the nuclide thatresults? 1 16 2 0 n 8 O 1 H ? 1 16 2 15 0 n 8 O 1 H 7 N
6. 6. Nuclear Reactions – common notations 4 alpha particle 2 He or 1 neutron 0 n or n electron 0 1 e or 1 proton p or p 1
7. 7. Unified atomic mass unit 1.660538782(83)×10−27 kg /12 1 amu = 931.46 MeV/c2 = 0.93146 GeV/c2
8. 8. Einstein’s mass – energy equivalence E=mc2 Big Idea mass and energy are interconvertible Einstein‟s equation relates rest mass to an equivalent energy 1 kg = c2 J of energy (a lot!) 1 amu = 931.46 MeV = 0.93146 GeV of energy
9. 9. Now convert the energy value (in Joules) to electron volts 10 19 1.494 10 1.6 10 934 MeV
10. 10. mass defect Consider a helium nucleus: 2 protons and 2 neutronsTotal mass of mass 4.001504 uprotons andneutrons? 4.031882 u mass defect = 4.031882 – 4.001504 = 0.030378 u (Δm or σ)
11. 11. Practice1.The mass of an atom of Ne-20 is 19.992435 u.Determine the mass defect for this nuclide. m 10 (0.000549 1.007276 1.008665) 19.992435 m 0.172465 u
12. 12. binding energy – explaining the missing mass “assembling” a nucleus binding energy Individual protons and neutrons
13. 13. Practice1. Determine the binding energy in O-16 (MeV and J).The mass of an O-16 atom is 15.994915 u. m 8 (0.000549 1.007276 1.008665) 15.994915 m 0.137005 u E 0.137005 u 931.5 127.6 MeV
14. 14. binding energy per nucleon depends on nuclide iron has highest value – mental note
15. 15. PracticeUse the graph to estimate the binding energy per nucleon for Fe-56. Verify your estimate mathematically. Mass of Fe-56 ATOMis 55.934940 u m (26 0.000549) (26 1.007276) (30 1.008665) 55.934940 m 0.52846 amu0.52846 amu 931.5 MeV c 2 492.26049 MeV492.26049 MeV 8.79 MeV / nucleon 56
16. 16. How much energy is required to remove a neutron from a C-13nucleus? Write a balanced nuclear equation for this reaction.C-13 atom mass = 13.003355 amu, and of course C-12 has amass of ? 13 12 1 6 C 6 C 0 n m (12.000 1.008665) 13.003355 m 0.00531 amu 2 0.00531 amu 931.5 MeV c 4.946 MeV
17. 17. STOP HERE and Complete Worksheet 1
18. 18. fusion and fission reactions where does the energy come from?
19. 19. fusion and fission reactionsFusion 2 light nuclides (low binding energy per nucleon) combineto make one heavy nuclide (higher binding energy pernucleon)Fission One heavy nuclide (lower binding energy per nucleon) splitsto form lighter nuclides (higher binding energy per nucleon)
20. 20. (235.043924 1.008665) (90.910187 141.929630 3 1.008665) 0.186777 931.5 173.99MeV
21. 21. 1H + 1H1 1 2 H+ 1 0 +1 e + ν + 0.4 MeV1 2 3H+ H1 1 2 He + 5.5 MeV3 3 4 12 He + He2 2 He + 2 H + 12.9 MeV 1
22. 22. Big Fusion Problem Number 1: Confinement of plasmaPossibility 1 – inertial confinement: beams of laserlight or ions compress a fuel pellet from all sides whileheating it.Possibility 2 – magnetic confinement: using a magneticfield to cause plasma (charged gas particles) tocirculate endlessly within a confined space..
23. 23. Big Fusion Problem Number 2: Extracting energyOnce a fusion reaction is initiated and confined, theenergy must be extracted in order to be any use (andto stop the whole thing melting down).One possibility is to use a molten lithium blanketsurrounding the fusion reaction to transfer the heat toa water based heat transport system.