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Topic 7. 3 Nuclear Reactions, Fission and Fusion

Topic 7. 3 Nuclear Reactions, Fission and Fusion. These notes were typed in association with Physics for use with the IB Diploma Programme by Michael Dickinson For further reading and explanation see: Physics, Tsokos (purple): Ch 6.3 Physics, Giancoli (mountain): Ch 30.

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Topic 7. 3 Nuclear Reactions, Fission and Fusion

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  1. Topic 7. 3 Nuclear Reactions, Fission and Fusion These notes were typed in association with Physics for use with the IB Diploma Programme by Michael Dickinson For further reading and explanation see: Physics, Tsokos (purple): Ch6.3 Physics, Giancoli (mountain): Ch 30

  2. Nuclear Reactions

  3. 7.3.1 Describe and give an example of an artificial (induced) transmutation.7.3.2 Construct and complete nuclear equations. • We already stated that natural transmutation is when during alpha and beta decay, the radioactive nuclide changes into one of a different atomic number or become a different element. • Artificial transmutation is when the transmutation process is induced by an external means. • This can happen when high speed particles like neutrons, alpha particles, other nuclides and even gamma radiation bombards the nucleus with enough force to cause it to split. • The parent nuclide undergoes transmutation and the two smaller, daughter, nuclides are formed.

  4. 7.3.1 Describe and give an example of an artificial (induced) transmutation.7.3.2 Construct and complete nuclear equations. • 1919 – Rutherford • Alpha particles into nuclei of nitrogen-14 gas • Some alpha particles absorbed and proton was emitted. • Oxygen–17 was formed • *** see board for equation*** • Lithium-6 bombarded with neutrons will yield an alpha particle and Hydrogen-3 • *** see board for equation*** • The following link will give you so practice writing equations. • http://tb014.k12.sd.us/Chemistry/Neclear%20Reactions/pr1.html

  5. 7.3.3 Define the term unified atomic mass unit. IB Definition • Unified Mass Unit(u) – is defined as 1/12 of the mass of a neutral carbon-12 atom. • u = 1.661 x 10-27kg = 931.5 MeV/c2 • Using this system the mass of carbon-12 is exactly 12.000000u. • The masses of nuclear particles are given in terms of “u” • Take a look at your data book. That is the extra number after “kg” • http://schools.bibb.k12.ga.us/cms/lib01/GA01000598/Centricity/Domain/292/Data%20Book.pdf

  6. 7.3.4 Apply the Einstein mass-energy equivalence relationship.7.3.5 Define the concepts of mass defect, binding energy and binding energy per nucleon. Lets take a minute to really examine this “unified mass units” • 6 x proton masses = 6 x 1.007276 u = 6.043656 u • 6 x neutron masses = 6 x 1.008665 u = 6.051990 u • 6 x electron masses = 6 x 0.000549 u = 0.003294 u • Total = 12.098940 u • Mass of Carbon-12 = 12.000000 u • Difference = 0.098949 u

  7. 7.3.4 Apply the Einstein mass-energy equivalence relationship.7.3.5 Define the concepts of mass defect, binding energy and binding energy per nucleon. • The difference is known as the mass defect. • The mass defect occurs because energy is needed to bind the nucleons together in the nucleus. • This results in a reduction in the energy of the system and a corresponding reduction in the mass of the bound atom

  8. 7.3.4 Apply the Einstein mass-energy equivalence relationship.7.3.5 Define the concepts of mass defect, binding energy and binding energy per nucleon. IB Definition • Mass Defect – the difference between the mass of a combined atom and the sum of the mass of the constituent parts. • Binding Energy – the energy needed to bind the constituents of an atom together. It corresponds to the mass defect of the atom.

  9. 7.3.4 Apply the Einstein mass-energy equivalence relationship.7.3.5 Define the concepts of mass defect, binding energy and binding energy per nucleon. • The mass defect of the carbon-12 atom corresponds to the binding energy of the atom. • The energy (MeV), corresponding to a mass defect of 0.09894u can be calculated or converted. E = 0.09894u x 931.5 MeV = 92.2MeV 1u E = 92.2 x 106eV x 1.6 x 10-19J = 1.479 x 10-11J 1eV

  10. 7.3.4 Apply the Einstein mass-energy equivalence relationship.7.3.5 Define the concepts of mass defect, binding energy and binding energy per nucleon. • This is where Einstein got his ideas. • He suggest that mass is simply another form of energy. • The relationship is as follows IB Formula – E = mc2 Where E is energy, m is mass, and c is the speed of light.

  11. 7.3.4 Apply the Einstein mass-energy equivalence relationship.7.3.5 Define the concepts of mass defect, binding energy and binding energy per nucleon. • The equation predicts how much energy corresponds to a particular loss in mass. • We can use this equation to calculate the binding energy of the carbon-12 atom. (The energy that is needed to separate a nucleus into its constituents).

  12. First a conversion • Mass=0.098940u = 0.098940u x 1.661x10-27kg = 1.643x10-28kg 1u Now the formula • E = mc2 • E = 1.643x10-28kg x (3x108m/s)2 • E = 1.479x10-11J Convert from J to eV • E = 1.479x10-11J x 1eV = 92.2MeV 1.6 x 10-19J

  13. Fission and fusion

  14. 7.3.6 Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon.7.3.8 Describe the processes of nuclear fission and nuclear fusion.7.3.9 Apply the graph in 7.3.6 to account for the energy release in the processes of fission and fusion. Learn about how a nuclear power plant works. http://science.howstuffworks.com/nuclear-power.htm

  15. 7.3.8 Describe the processes of nuclear fission and nuclear fusion. • Fusion • The process that powers the sun and the stars. • Two atoms of hydrogen combine together, or fuse, to form an atom of helium. • Some of the mass of the hydrogen is converted into energy. • Easiest fusion reaction is combining deuterium (or “heavy hydrogen) with tritium (or “heavy-heavy hydrogen”) to make helium and a neutron. • Both of which are easy to get making it potentially the “perfect” power source • http://fusioned.gat.com/what_is_fusion.html

  16. 7.3.8 Describe the processes of nuclear fission and nuclear fusion. • Fission • Fission is a nuclear reaction in which an atomic nucleus splits, or fissions, into fragments (ie. certain forms of Uranium) • Usually two fragments of comparable mass. • The split releases kinetic energy, as heat, and radiation and neutrons. • These neutrons can be used to “fission” other Uranium nuclei and start a chain reaction. • If the mass of the parts is added up, it is less than the original, or starting mass. • This missing mass, also known as the mass defect, is changed into energy. • http://reactor.engr.wisc.edu/tour/reactor.htm

  17. 7.3.6 Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon.7.3.9 Apply the graph in 7.3.6 to account for the energy release in the processes of fission and fusion.

  18. 7.3.6 Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon.7.3.9 Apply the graph in 7.3.6 to account for the energy release in the processes of fission and fusion.

  19. 7.3.6 Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon.7.3.9 Apply the graph in 7.3.6 to account for the energy release in the processes of fission and fusion. • IB Definition • Binding Energy per Nucleon – A nuclide’s binding energy is equal to the total binding energy for an atom divided by the number of constituents nucleons. • The graph shows several things: • How the binding energy per nucleon changes with respect to the nuclide’s atomic number, Z. • The binding energy per nucleon rises sharply from atomic number, (Z=1) until it reaches a maximum at Z=26 (Iron). • The trend then falls away more gradually towards the heavier nuclides.

  20. 7.3.6 Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon.7.3.9 Apply the graph in 7.3.6 to account for the energy release in the processes of fission and fusion. • ***Remember*** The higher the binding energy between the nucleons, the more stable the nucleus will be. • Unstable nuclides below atomic number 26(iron) undergo nuclear fusion in an attempt to increase their stability. • Unstable nuclides above atomic number 26 undergo nuclear fission. • The steeper the slope the more energy is releases. • Fusion – very steep – much more energy released per nucleon • Fission – shallow – much less energy released per nucleon

  21. 7.3.6 Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon.7.3.9 Apply the graph in 7.3.6 to account for the energy release in the processes of fission and fusion. • The graph explains/predicts: • Relative stability of elements around atomic number 26 • Large abundance of iron in universe • The energy released in nuclear reactions.

  22. 7.3.10 State that nuclear fusion is the main source of the Sun’s energy • There are several different possibilities of this sequence. We’ll look at path 1.

  23. 7.3.10 State that nuclear fusion is the main source of the Sun’s energy • Here are two sites that go into more detail and explanation. • http://burro.astr.cwru.edu/Academics/Astr221/StarPhys/ppchain.html • http://hyperphysics.phy-astr.gsu.edu/hbase/astro/procyc.html • The sequence of fusion reactions that occur in the Sun is known as the proton-proton cycle. ****See the board for equations or the web sites**** • Part 1: • Part 2: • Part 3: • The first two parts have to happen twice in order to produce enough He–3 for the third part.

  24. 7.3.10 State that nuclear fusion is the main source of the Sun’s energy • The total energy released in this fusion reaction is: (0.42 x 2) + (5.49 x 2) + 12.86 = 24.68 MeV • The 2 positrons produced in the first part of the sequence also releases 1.02 MeV of energy which has to be included in the final sum. 24.68 + (1.02 x 2) = 26.72 MeV

  25. 7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions. Example Problem 3 • Estimate the energy released in the fission reaction of Uranium-235 as it transmutes into Barium-141 and Krypton-92 • Answer: about 176 MeV

  26. Example Problem 3 • Estimate the energy released in the fission reaction of Uranium-235 as it transmutes into Barium-141 and Krypton-92 • Answer: about 176 MeV • Solution: The difference between the binding energies of the original Uranium and the new Barium and Krypton is a good estimate of the total energy released in this reaction. • Using the graph from 7.3.6 we can find the binding energy per nucleon • U-235 = 7.6MeV • Ba-141 = 8.3MeV • Kr-8.6 = 8.6MeV

  27. Example Problem 3 • Estimate the energy released in the fission reaction of Uranium-235 as it transmutes into Barium-141 and Krypton-92 • Answer: about 176 MeV • Solution: Calculating the total binding energies: • U-235 = 7.6MeV x 235 = 1786MeV • Ba-141 = 8.3MeV x 141 = 1170.3MeV • Kr-8.6 = 8.6MeV x 92 = 791.2MeV Total = 1961.5MeV • Difference = 1961.5MeV – 1786MeV = about 176MeV

  28. 7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions. Example Problem 4 • Repeat the previous example using the following information and find the exact answer. Atomic mass of U-235 235.043924u Atomic mass of Kr-92 91.926300u Atomic mass of Ba-141 140.914400u • Answer:173 MeV

  29. Example Problem 4 • Repeat the previous example using the following information and find the exact answer. Atomic mass of U-235 235.043924u Atomic mass of Kr-92 91.926300u Atomic mass of Ba-141 140.914400u • Solution: The difference between the binding energies of the original Uranium and the new Barium and Krypton is a good estimate of the total energy released in this reaction. • Calculate the binding energy using the mass defects

  30. Example Problem 4 • Calculate the binding energy using the mass defects Atomic mass of U-235 235.043924u Constituent mass U-235 236.958995u Mass defect U-235 1.915071u = 1783.9MeV Atomic mass of Kr-92 91.926300u Constituent mass Kr-92 92.766940u Mass defect Kr-92 0.840640u = 783.1MeV Atomic mass of Ba-141 140.914400u Constituent mass Ba-141 142.174557u Mass defect Ba-141 1.260157u = 1173.8MeV

  31. Example Problem 4 • Difference in binding energies = (1173.8 + 783.1) – 1783.9 = 173MeV

  32. 7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions. Example Problem 5 • Using the binding energy per nucleon curve, estimate the energy released in the fission reaction of Uranium-235 to Strontium-88 and Xenon-136 shown below. For this calculation, assume that the initial kinetic energy of the neutron is negligible. • Answer: about 112MeV

  33. Example Problem 5 • Using the binding energy per nucleon curve, calculate the energy released in the fission reaction of Uranium-235 to Strontium-88 and Xenon-136 shown below. For this calculation, assume that the initial kinetic energy of the neutron is negligible. • Solution: The difference between the binding energies of the original Uranium and the new Strontium and Xenon is a good estimate of the total energy released in this reaction. Uranium-235 7.6MeV x 235 = 1786MeV Strontium-88 8.6MeV x 88 = 756MeV Xenon-136 8.4MeV x 136 = 1142MeV Total = 1898MeV

  34. Example Problem 5 • Using the binding energy per nucleon curve, calculate the energy released in the fission reaction of Uranium-235 to Strontium-88 and Xenon-136 shown below. For this calculation, assume that the initial kinetic energy of the neutron is negligible. • Solution: The difference between the binding energies of the original Uranium and the new Strontium and Xenon is a good estimate of the total energy released in this reaction. • Difference = 1898MeV – 1786MeV = 112MeV

  35. 7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions. Example Problem 6 • Repeat the previous example, using the following information. Atomic mass of U-235 235.043924u Atomic mass of Sr-88 87.905618u Atomic mass of Xe-136 135.907210u • Answer: 126.5MeV

  36. 7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions. Example Problem 7 • Calculate the energy released in the fusion reaction between Deuterium, H-2, and Tritium, H-3, as shown in the equation below, using the following information. Atomic mass of H-2 2.014102u Atomic mass of H-3 3.016049u Atomic mass of He-4 4.002602u • Answer: 17.59MeV

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