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Topic 7.2 Extended B – Nuclear Stability NUCLEON POPULATIONS Stable isotopes exist for elements having atomic numbers Z = 1 to 83, excepting 43 and 61. Up to Z = 20, the neutronproton ratio is close to 1. Beyond Z = 20, the neutronproton ratio is bigger than 1, and grows with atomic number. Since decay decreases protons and neutrons at the same rate, the daughter nucleus would be no more stable than the parent, for the larger atomic numbers. Thus the larger unstable nuclei become stable by a combo decay and - decay (which turns neutrons into protons). FYI: Since there are 6 more neutrons than protons, sulfur-38 is susceptible to - decay. Question: Topic 7.2 Extended How many - decays would it undergo? B – Nuclear Stability whichOF element would it transmute? PQuestion: AIRING ETo FFECT STABLE NUCLEI There are 168 stable nuclei having an even number of protons and neutrons. There are 107 stable nuclei having either Z even and N odd, or Z odd and N even. There are only 4 stable nuclei having an odd number of each. 26 Thus, you would expect 27 13Al to be stable, but 13Al to be unstable, and this is the case. General Criteria for Nuclear Stability (1) Isotopes with Z < 83 are stable. (2) Most Even-Even, Even-Odd, Odd-Even isotopes are stable. (3) Isotopes with Z < 20 are stable if Z = N. Else N > Z. Is sulfur-38 stable, or is it unstable? Since Z = 16 < 83 for sulfur, (1) is satisfied. Since N = 38 - 16 = 22, and Z = 16, (2) is satisfied. Since Z < 20 and N Z (3) is not satisfied. We may conclude that sulfur-38 is NOT stable. FYI: 1H has one proton, and 1 electron for a total mass of 1.00727 u + 0.000548 u = 1.007818 u. This value matches the value in the table: Topic 7.2 Extended FYI: has two (which includes the 2 electrons) and 2 neutrons, B – Nuclear Stability for a total mass of 2(1.007825) + 2(1.008665) = 4.03298 u. This 4He 1H Bvalue INDING ENERGY DOES NOT match the value in the table: In the world of nuclear reactions we have to keep FYI: The DIFFERENCE between the constituent masses and the track of the mass of the nucleus4 if we are to resulting mass is called the mass He has a mass defect of determine the energy of a defect. reaction. 0.030377 u. unified atomic mass unit 4.03298 To this- 4.002603 end we =define the (u) using a neutral carbon-12 atom as our standard of precisely 12.000000 u. Thus 1 u = 1.660610-27 kg Atomic Mass Unit (u) Particle Masses and Energy Equivalents Particle Mass (u) Mass (kg) Energy (MeV) 1u 1.660610-27 931.5 Electron 0.000548 9.109510-31 0.511 Proton 1.00727 1.6726510-27 938.28 1H atom 1.007825 1.6735610-27 938.79 Neutron 1.008665 1.6750010-27 939.57 4He 4.002603 6.6467210-27 3738.8 atom FYI: This energy deficit is called the total binding energy (Eb) of the helium nucleus. In fact, this is the energy released in the nuclear Topic 7.2 Extended reaction which combines the nucleons to form helium. B – Nuclear Stability FYI: The sun generates its energy through the nuclear reaction Bcombining INDING ENERGY hydrogen into helium. Each reaction liberates 28.3 MeV. We can convert the mass defect m into equavalent energy using E = mc2: 0.030377 u 1.660610-27 kg ( 3108 ms-1 )2 E = 1 u 1 eV 4.540010-12 J E = 1.610-19 J E = 28.375 MeV To save time you can use the conversion from mass to MeV: 1 u = 931.5 MeV Atomic Mass Unit (u) Thus E = 0.030377 u Eb = mc2 931.5 MeV 1 u = 28.30 MeV Binding Energy (Eb) FYI: The energy to ASSEMBLE the 4He nucleus from hydrogen on the sun is from gravitational compression. Causing fusion here on earth, Topic 7.2 Extended we have to obtain this energy in a different way. B – Nuclear Stability FYI: The energy to DISASSEMBLE the 4He nucleus can come from BINDING ENERGY photons of very high energy, or particle beams such as electrons, Just as 28.3 MeV are protons, or anti-protons (p-bars). released when the 4He fuses... FUSION ... we can reverse the process if we somehow "inject" 28.3 MeV into the helium nucleus: FISSION FYI: This is why nuclear bombs7.2 pack aExtended real punch! Topic B – Nuclear Stability BINDING ENERGY We can calculate the average binding energy per nucleon for helium (or any stable isotope) using the following formula: binding energy per nucleon = Thus for 4He Eb A Average Binding Energy (A = 4) we have 28.3 MeV = 7.075 MeV Eb = 4 A To disassemble the nucleus we would need to supply 7.075 MeV / nucleon. If we could remove one nucleon, it would take about 7.075 MeV. FYI: Recall that it takes 13.6 eV to remove an electron from a hydrogen atom. Compare this CHEMICAL energy to the NUCLEAR energy of 7,075,000 eV to remove a nucleon. This gives us a rough comparison NUCLEAR ENERGY to CHEMICAL ENERGY of 7,075,000 eV / 13.6 eV = 520,221! FYI: The higher the binding energy per nucleon, the harder it is to break apart the nucleus. Thus, the bigger Eb/A the more stable the Topic 7.2 Extended nucleus. Fe (iron) is the most stable element. B – Nuclear Stability BINDING ENERGY We can calculate the average binding energy per nucleon for all of the elements: For the elements with A < 56 note that fusion results in more stable nuclei. For the elements with A > 56 note that fission results in more stable nuclei. Note that iron is the most stable element. Question: Why will it cease fusion when it reaches this point? Question: If the universe started7.2 out asExtended hydrogen and helium, and Topic stars can only breed nuclei up to iron, where do all the higher elements B – Nuclear Stability come from? BINDING ENERGY This brings us to an aside on stellar evolution: During the lifetime of a star there are two opposing forces maintaining an equilibrium. (1) Gravitation is trying to collapse the star. (2) Radiation pressure is opposing the gravitational collapse. If a star is sufficiency massive, Gravitational Collapse Nuclear Reactions Start it will maintain nuclear reactions Balanced by Radiation until it becomes IRON. Pressure When such a star becomes iron, it will have no more radiation pressure to oppose gravity. It will collapse (due to gravity), and its complex iron nuclei will decay into ALL NEUTRONS! FYI: This is called the IRON CATASTROPHE. Topic 7.2 Extended B – Nuclear Stability MAGIC NUMBERS Remember how Pauli's Exclusion Principle told us how many electrons could be at any energy level? Thus, the magic numbers were 2, 8, 18, 32, etc. for electrons about a nucleus. Theoretically, these numbers came from Schrödinger's equation. To make a long story short, Schrödinger's equation also reveals something about the shell structure of the nucleus! The shells of the nucleus fill up with nucleons just like electron shells fill up with electrons. A nuclear shell is said to be closed if it has the optimum number of protons OR neutrons in it. Here is the series of those optimum numbers: 2, 8, 20, 28, 50, 82, or 126. Magic Numbers FYI: Nuclei having this many protons (or neutrons) are more stable than the average nuclei (and usually have more stable isotopes).