2. Topic 7.2 Extended B – Nuclear Stability  Stable isotopes exist for elements having atomic numbers Z = 1 to 83, excepting 43 and 61.  Up to Z = 20, the neutron- proton ratio is close to 1.  Beyond Z = 20, the neutron- proton 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). N UCLEON P OPULATIONS

3. Topic 7.2 Extended B – Nuclear Stability  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.  Thus, you would expect 27 Al to be stable, but 26 Al to be unstable, and this is the case. P AIRING E FFECT OF S TABLE N UCLEI 13 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 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: Since there are 6 more neutrons than protons, sulfur-38 is susceptible to  - decay. Question: How many  - decays would it undergo? Question: To which element would it transmute?

4. Topic 7.2 Extended B – Nuclear Stability  In the world of nuclear reactions we have to keep track of the mass of the nucleus if we are to determine the energy of a reaction.  To this end we define the unified atomic mass unit (u) using a neutral carbon-12 atom as our standard of precisely 12.000000 u. B INDING E NERGY  Thus 1 u = 1.6606  10 -27 kg Atomic Mass Unit (u) Particle Masses and Energy Equivalents ParticleMass (u)Mass (kg)Energy (MeV) 1 u 1.6606  10 -27 931.5 Electron0.000548 9.1095  10 -31 0.511 Proton1.00727 1.67265  10 -27 938.28 1 H atom1.007825 1.67356  10 -27 938.79 Neutron1.008665 1.67500  10 -27 939.57 4 He atom4.002603 6.64672  10 -27 3738.8 FYI: 1 H has one proton, and 1 electron for a total mass of 1.00727 u + 0.000548 u = 1.007818 u. FYI: 4 He has two 1 H (which includes the 2 electrons) and 2 neutrons, for a total mass of 2(1.007825) + 2(1.008665) = 4.03298 u. This value matches the value in the table: This value DOES NOT match the value in the table: FYI: The DIFFERENCE between the constituent masses and the resulting mass is called the mass defect. 4 He has a mass defect of 4.03298 - 4.002603 = 0.030377 u.

5. Topic 7.2 Extended B – Nuclear Stability  We can convert the mass defect  m into equavalent energy using E =  mc 2 : 0.030377 u B INDING E NERGY 1.6606  10 -27 kg 1 u ( 3  10 8 ms -1 ) 2 E = 4.5400  10 -12 J 1 eV 1.6  10 -19 J E = 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 0.030377 u 931.5 MeV 1 u = 28.30 MeV E = FYI: This energy deficit is called the total binding energy (E b ) of the helium nucleus. In fact, this is the energy released in the nuclear reaction which combines the nucleons to form helium. FYI: The sun generates its energy through the nuclear reaction combining hydrogen into helium. Each reaction liberates 28.3 MeV. E b =  mc 2 Binding Energy (E b )

6. Topic 7.2 Extended B – Nuclear Stability  Just as 28.3 MeV are released when the 4 He fuses... B INDING E NERGY ... we can reverse the process if we somehow "inject" 28.3 MeV into the helium nucleus: FUSION FISSION FYI: The energy to ASSEMBLE the 4 He nucleus from hydrogen on the sun is from gravitational compression. Causing fusion here on earth, we have to obtain this energy in a different way. FYI: The energy to DISASSEMBLE the 4 He nucleus can come from photons of very high energy, or particle beams such as electrons, protons, or anti-protons (p-bars).

7. Topic 7.2 Extended B – Nuclear Stability  We can calculate the average binding energy per nucleon for helium (or any stable isotope) using the following formula: B INDING E NERGY binding energy per nucleon = EbAEbA Average Binding Energy  Thus for 4 He (A = 4) we have EbAEbA = 28.3 MeV 4 = 7.075 MeV  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: This is why nuclear bombs pack a real punch!

8. Topic 7.2 Extended B – Nuclear Stability B INDING E NERGY  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. FYI: The higher the binding energy per nucleon, the harder it is to break apart the nucleus. Thus, the bigger E b /A the more stable the nucleus. Fe (iron) is the most stable element.  Note that iron is the most stable element.

9. Topic 7.2 Extended B – Nuclear Stability B INDING E NERGY  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. Gravitational Collapse Nuclear Reactions Start Gravitational Collapse Balanced by Radiation Pressure  If a star is sufficiency massive, it will maintain nuclear reactions until it becomes IRON. Question: Why will it cease fusion when it reaches this point?  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. Question: If the universe started out as hydrogen and helium, and stars can only breed nuclei up to iron, where do all the higher elements come from?

10. Topic 7.2 Extended B – Nuclear Stability M AGIC N UMBERS  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).