Fission and Fusion Book pg 286-287 cgrahamphysics.com 2016
Review BE is the energy that holds a nucleus together. This is equal to the mass defect of the nucleus. Also called separation energy. The energy required to decompose an atom, or nucleus into its constituent particles, equal to the energy equivalent of the mass defect. Elements with a high binding energy per nucleon are very difficult to break up. Iron 56 is close to the peak of the curve and has one of the highest binding energies per nucleon of any isotope.
Nuclear fission Nuclear fission and nuclear fusion Nuclear fission is the splitting of a large nucleus into two smaller (daughter) nuclei. An example of fission is 235 U + 1 n ( 236 U*) 140 Xe + 94 Sr + 2( 1 n) 92 0 92 54 38 0 In the animation, 235 U is hit by a neutron, and capturing it, becomes excited and unstable: It quickly splits into two smaller daughter nuclei, and two neutrons. cgrahamphysics.com 2016 94 Sr 140 Xe
Chain reaction Nuclear fission and nuclear fusion Note that the splitting was triggered by a single neutron that had just the right energy to excite the nucleus. 235 U + 1 n ( 236 U*) 140 Xe + 94 Sr + 2( 1 n) 92 0 92 54 38 0 Note also that during the split, two more neutrons were released. If each of these neutrons splits subsequent nuclei, we have what is called a chain reaction. 1 2 4 8 Exponential Growth Primary Secondary Tertiary
Critical Mass The smallest possible amount of fissionable material that will sustain a chain reaction Too few neutrons released - reaction will slow down Too many neutrons released - reaction becomes uncontrolled - causes explosion - nuclear meltdown
Atomic bomb nuclear fision cgrahamphysics.com 2016
Fission Nucleon loses energy when split mass decreases high BE p.n. High nucleon number elements undergo fission Low nucleon number nuclides may undergo fusion they all try to reach the nuclide that is most stable: iron - 56
Stars start to shine - fission cgrahamphysics.com 2016
Fission Fission brought on by absorption of neutrons is called induced fission Fission that happens if a nucleus splits is called spontaneous fission Induced fission is called neutron activation
Neutron Activation Radioactive decay is spontaneously If atoms are bombarded with different particles, these change into new elements Neutrons are particularly effective in inducing artificial transmutation to produce radioactive isotopes
Neutrons are produced Beryllium is bombarded with α - particles 4 2 12 6 α + Be C + n 4 9 0 1 Identify the bombarding particle: 14 1 1 14 7N + 0n 1H + 6 C 6 1 1 4 4 3Li + 0n + 1 p 2 α + 2 α 205 81Tl + 1 1 206 1 1 p + 0n 82Pb + 0 n
Liquid drop model The balance of electrostatic forces b/w protons and the short range nuclear force hold the nucleus together Absorbing a slow neutron the unstable atom U-236 is formed and the balance b/w forces is disturbed The long range Coulomb force of repulsion dominates now over short ranged nuclear force of attraction
A possible fission reaction 1 0 235 92 144 56 36 90 n + U Ba + Kr + 2 n n = 1.009u U - 235 = 235.044u Ba 144 = 143.923u Kr 90 = 89.920u Find the mass defect and energy equivalent m = 235.044 + 1.009 143.923 + 89.920 + 2 x 1.009 = 236.053 235.861 = 0.192u E = 0.192 x 931.5 = 178.85 ~ 179 MeV 0 1
Conservation laws govern fission Momentum is conserved (relativistically) Total charge is conserved (charge of products is charge of reactants) The number of nucleons remains constant Mass energy is conserved ( E = mc 2 ) The KE after a reaction is greater than the KE of the initial neutron the mass defect m has been changed into energy E. KE is due to energy of fired neutron
Fusion Two lighter nuclei fuse together Very high pressure and temperature are needed Nucleus must overcome Coulomb repulsion Nucleus must come under the influence of the strong nuclear force If two nuclei with low atomic numbers could join together they would have - higher BE p.n - overall mass decrease - lose energy
Example For this reaction find the mass defect and the energy released Mass of 2 H= 2.014102u Mass of 3 H= 3.016049u Mass of 4 He= 4.002604u Mass of n = 1.008665u m = 2.014102 + 3.016049 4.002604 + 1.008665 = 5.030151 5.011269 = 0.01882u BE = 0.01882 x 931.5 = 17.6MeV The energy released = KE of helium nucleus and neutron Advantage: no radioactive elements are produced Disadvantage: obtaining and maintaining high temperature and pressure to initiate fusion
Revisit BE curve - Nuclides with nucleon number ~ 60 are most stable - High nucleon number nuclides undergo fission - Low nucleon number nuclides undergo fusion
To reach most stable nuclide Fission: - loosing energy - increases BE of fission nuclei - becomes more stable Fusion - PE of new nucleus is less than that of individual atoms - increases BE of fused nucleus - becomes more stable
Example 238 1 92U + 0 1 n 38 90 Sr + 146 54Xe + 3 0 n Use the graph to compare the BE 238-U: 7.6 x 238 =1800MeV 90 Sr: 8.7 x 90 = 780MeV 146 Xe: 8.2 x 146 = 1200MeV 1200 + 780 = 1980MeV The BE of fission nuclei is greater than BE of U 238 losing energy increases BE nucleus more stable
Fusion in the Sun p : + p : 2 1 H 2 1 H + p : 3 2 He 3 2 He + 3 2 He 4 2 He +2 p : For a complete cycle the first two reactions must occur twice Result: - 1 helium nucleus - 2 positrones - 2 protons, which are available for future fusions - 2 neutrinos
Example H 2 4 e= 4.00260u H 1 2 = 2.01410u n 0 1 = 1.00867u p 1 1 = 1.00728u e 1 0 = 0.00055u a) What is the mass of the H a) m He = 4.00260 931.5 = 3728.4MeVc ;2 4 b) 2p + 2n 2He m = 2 1.00728 + 2 1.00867 931.5 3728.4 = 3755.7 3728.4 = 27.3 MeVc ;2 BE = 27.3 MeV c) BE = E A = 27.3 4 = 6.8MeV 2 2 e) 1 H + 1 H 4 2 He m = 2 x 2.01410 4.00260 = 0.0256u BE = 0.0256 x 931.5 = 23.8 ~ 24MeV 4 2 e nucleus? b) What is the BE of the 4 2 He nucleus? c) What is the BE p.n. for a 4 2 He nucleus? 2 d) What is the BE p.n. for a 1 H nucleus? e) When two deuterium nuclei fuse to form a helium nucleus how much energy is released? d) 1p + 1n H m = 1.00728 + 1.00867 2.01410 = 2.01595 2.01410 = 0.00185u BE= 0.00185 x 931.5 = 1.723MeV and BE p.n. = E = 1.273 = 0.86MeV A 2 1 2
Example Determine the number x of neutrons produced and calculate the energy released in the following fission reaction: 235 1 144 U + n Ba + 90 Kr + X n 92 235 U = 235.043929u 144 Ba = 143.922952u 90 Kr = 89.919516u 1 n = 1.008665u Solution Mass left side: 236 Mass right side: 144 + 90 = 234 Hence x = 2 0 m = 235.043929 + 1.008665 143.922952 + 89.919516 + 2 1.008665 = 236.052594 235.859798 = 0.192796u E = 0.192796 x 931.5 = 179.6MeV ~ 180 MeV 56 36 0 1