Lecture Notes on Nuclear Physics - Great Ideas in Science | PHYS 2018, Study notes of Physics

Download Lecture Notes on Nuclear Physics - Great Ideas in Science | PHYS 2018 and more Study notes Physics in PDF only on Docsity! Physics 2018: Great Ideas in Science: The Physics Module Nuclear Physics Lecture Notes Dr. Donald G. Luttermoser East Tennessee State University Edition 1.0 Abstract These class notes are designed for use of the instructor and students of the course Physics 2018: Great Ideas in Science. This edition was last modified for the Fall 2007 semester. ii) Water: H2O (2 hydrogen atoms + 1 oxygen atom). iii) Methane: CH4 (1 carbon atom + 4 hydrogen atoms). f) Molecules can adhere to each other through chemical bonds making a structured lattice =⇒ solids. B. An Overview of Nuclear Reactions. 1. In order to get nuclear particles to interact with each other, one needs to collide these particles with a very high kinetic energy (i.e., high speed). a) Since the nucleus is composed of both neutrons (zero charge) and protons (positive charge), two nuclei close together will experience a strong repulsive force due to Coulomb’s electric force law. b) Remember that forces are derivatives of potential energy. Hence if one wishes to bring two nuclei closer and closer together, one must increase the kinetic energy of said nu- clei to high enough values to overcome the electric poten- tial field. c) From thermodynamics heat is nothing more than the av- erage kinetic energy of the particles that make up a system (i.e., matter). d) Hence, one way to get nuclei moving fast is to heat them up. 2. Reactions in chemistry and physics can be one of two types: a) Endothermic reactions absorb energy. III–3 b) Exothermic reactions release energy. 3. The element iron (Fe) has a nucleus that is the most stable of all the atomic nuclei. a) Elements lighter (i.e., less massive) than iron will produce exothermic nuclear reactions when they fuse together to make heavier elements =⇒ nuclear fusion. b) Elements heavier (i.e., more massive) than iron will pro- duce exothermic nuclear reactions when they break apart to make lighter elements =⇒ nuclear fission. 4. The amount of energy released duing these exothermic reactions is proportional to the amount of the mass difference between the parent (original) and daughter (offspring) particles through Einstein’s famous equation E = mc2 . (III-1) 5. Conservation Laws. a) If we assign a baryon number B of +1 to each baryon (nucleon or hyperon) and –1 to an antinucleon or antihy- peron, then in a closed system ∑ B = constant. (III-2) b) Similarly, if we assign a lepton number L of +1 to each lepton (i.e., e−, µ, ν, etc.) and of –1 to antileptons (i.e., e+, µ, ν, etc.), then in a closed system ∑ L = constant. (III-3) c) A similar conservation law does not exist for bosons — the mesons and field particles (i.e., photons). III–4 d) Charge must be conserved in a nuclear reaction. e) Mass-energy, via E = m c2, must be conserved in a nuclear reaction. f) Momentum must be conserved. Hence a matter-antimatter reaction must create two photons to conserve momentum (e.g., e− + e+ → 2γ). C. Thermonuclear Reactions. 1. In 1938, it became clear that the long-term energy source for stars must be thermonuclear fusion reactions. In these reactions, lighter elements burn to form heavier elements =⇒ nucleosyn- thesis. 2. Two nuclei will fuse to form one nuclei if they come within 10−13 cm of each other — but they must be moving fast enough to overcome the Coulomb repulsion that exists between like charged particles. a) Particles must be at a high temperature to be moving fast. b) This high temp completely ionizes all of the nuclei. c) Temps must build even more to get the kinetic energy to overcome the Coulomb barrier. 3. In main sequence stars, H is fused into He. Since H is composed of 1 baryon and He, 4 baryons (2p + 2n), 4 H nuclei must be used to construct one He nuclei. 4×mH = 4 × 1.0078 amu = 4.0312 amu −mHe = 4.0026 amu ∆m = 0.0286 amu III–5 d) We can now turn to quantum mechanics to solve the prob- lem. As discussed in the subsection on particle spin, in reality elementary particles are not little billiard balls col- liding with each other as a result of following trajectories. Instead, they follow probability distributions described by their wave functions. In quantum mechanics, there is a small probability that wave functions can penetrate en- ergy barriers that are higher than the energy of the wave function. This effect is known as quantum tunneling. e) Using quantum mechanics, we can describe a tempera- ture needed to produce a sufficient number of tunneling events to sustain a nuclear reaction (see page 335 of the textbook) as Tquantum = 4µmZ 2 1Z 2 2e 4 3kBh2 , (III-6) where µm is the reduced mass of the colliding “particles” and h is Planck’s constant. f) In this equation, two protons can come together (i.e., fuse) at a temperature of 107 K, which is consistent with the central temperature deduced for the Sun. g) A more detailed calculation from statistical mechanics shows that the bulk of the energy is being liberated by reactions involving particles in the high energy tail of a Maxwellian distribution. i) Particles with energies at the Gamow Peak will be the ones that supply most of the energy through thermonuclear reactions. ii) The Gamow Peak corresponds to a local maxi- mum in the two probability functions: the e−E/kBT III–8 Maxwell-Boltzmann distribution term and the e−bE −1/2 quantum tunneling penetration term, where, b ≡ 23/2π2µ1/2m Z1Z2e 2 h . iii) The Gamow Peak for a given temperature will occur at the energy of E◦ = ( bkBT 2 )2/3 , (III-7) for the Sun, the Gamow Peak is at 6 keV. 5. By making use of statistical mechanics in conjunction with quan- tum mechanics, stellar interior modelers set up power laws that describe the energy production rate per unit mass of the form ε = ε◦XiXxρ αT β , (III-8) where the X ’s are the mass fractions of the fusing particles, and ε◦, α. and β are constants that depend upon the reactions in- volved (more to come on this). D. Various Reaction Chains 1. Two different fusion processes convert H into He, the first is important for stars with Tc < ∼ 1.8 × 107 K (M <∼ 1.3M , ∼F5 V star) and is called the proton-proton chain. a) The first of this reaction chain is called the PP I chain: Energy Reaction Reaction Released Time (MeV) 1H + 1H −→ 2H + e+ + νe 1.442 1.4 × 109 yr 1H + 2H −→ 3He + γ 5.493 6 sec 3He + 3He −→ 4He + 1H + 1H 12.859 106 yr III–9 i) 1H = hydrogen atom (1 proton). ii) 2H = heavy hydrogen (1 proton + 1 neutron) = deuterium. iii) 3He = light helium (2 protons + 1 neutron). iv) 4He = helium (2 protons + 2 neutron) = alpha particle. v) γ = Gamma ray photon. vi) e+ = positron (positive charge) = anti-electron (antimatter). This positron interacts with the free electrons in the core virtually immediately which produces 2 additional gamma ray photons. vii) νe = electron neutrino (neutral particle). The neutrino’s absorption cross section is negligible and leaves the stellar core (and star) immediately with- out further interaction. The energy loss from the neutrino is 0.263 MeV which has not been included in the Energy Released column. viii) In this PP I change, please note that the follow- ing reaction can take place 1.4% of the time that the first reaction takes place 1H + e− + 1H −→ 2H + νe the so-called “pep” (proton-electron-proton) reac- tion which releases 1.4 MeV and loses an additional 1.4 MeV in energy loss from the neutrino. III–10 i) 8B = boron-8 atom (5 protons + 3 neutrons). ii) 8Be = beryllium atom (4 protons + 4 neutrons). iii) The neutrino energy loss in this PP chain is 7.2 MeV. The Davis solar neutrino experiment, which de- tected only 1/3-rd of the predicted solar neutrinos was most sensitive to these 8B beta decay neutrinos. Recently, neutrinos have been found to oscillate be- tween the 3 known neutrino states which accounts for the low detection rate of the Davis experiment. iv) This PP III chain dominates the other PP chains in stars with central temperatures of T >∼ 2.5×107 K in the production of helium. v) The Energy Released and the Reaction Time have the same meaning as they did for the PP I chain. 2. For more massive stars (Tc > ∼ 1.8 × 107 K, M >∼ 1.3M , ∼F5 V star), the CNO cycle is the dominant reaction chain. a) This reaction chain uses carbon as a catalyst: Energy Reaction Reaction Released Time (MeV) 12C + 1H −→ 13N + γ 1.95 1.3 × 107 yr 13N −→ 13C + e+ + νe 2.22 7 min 13C + 1H −→ 14N + γ 7.54 2.7 × 106 yr 14N + 1H −→ 15O + γ 7.35 3.2 × 108 yr 15O −→ 15N + e+ + νe 2.71 82 sec 15N + 1H −→ 12C + 4He 4.96 1.1 × 105 yr i) 12C = carbon-12 (6 protons + 6 neutrons). III–13 ii) 13C = carbon-13 (6 protons + 7 neutrons). iii) 13N = nitrogen-13 (7 protons + 6 neutrons) [ra- dioactive]. iv) 14N = nitrogen-14 (7 protons + 7 neutrons). v) 15N = nitrogen-15 (7 protons + 8 neutrons). vi) 15O = oxygen-15 (8 protons + 7 neutrons) [ra- dioactive]. vii) 16O = oxygen-16 (8 protons + 8 neutrons). viii) The neutrino energy loss in the 13N beta de- cay is 0.710 MeV and the neutrino energy loss is 1.000 MeV for the second 15O beta decay. ix) The Energy Released and the Reaction Time have the same meaning as they did for the PP I chain. b) Note that this reaction sequence does not make any new elements other than He! c) For the last step in the CNO cycle, an additional set of reactions can take place: III–14 Energy Reaction Reaction Released Time (MeV) 15N + 1H −→ 16O + γ 12.126 1.0 × 107 yr 16O + 1H −→ 17F + γ 0.601 3.0 × 1010 yr 17F −→ 17O + e+ + ν 2.762 3 min 17O + 1H −→ 14N + 4He 1.193 2.0 × 1011 yr i) 17F = fluorine-17 (9 protons + 8 neutrons). ii) 18O = oxygen-16 (8 protons + 10 neutrons). iii) The neutrino energy loss in the 17F beta decay is 0.94 MeV. iv) The resulting 14N isotope can then be used back in the 14N + 1H −→ 15O + γ reaction in the primary CNO cycle. 3. The thermonuclear reaction rate, ε (in erg/gm/s), is very sen- sitive to temperature. For the two hydrogen to helium reaction chains (i.e., the proton-proton chain and CNO cycle), we can write Eq. (IV-48) as two separate equations: εpp = ε◦ ρ X 2 ( T 106 )β (III-9) εcc = ε◦ ρ XXCN ( T 106 )β , (III-10) where X is the mass fraction of hydrogen (as defined on page IV-8 of the course notes), XCN is the weighted average of the combined mass fraction of carbon and nitrogen (since these species are the III–15 ancient red giant star that no longer exists. To quote Carl Sagan, we are star stuff! d) Should the core of a red giant obtain temperatures that exceed a few hundred million Kelvins, another reaction can take place via an alpha (α) capture: 12C + 4He −→ 16O + γ which releases 7.161 MeV of energy. e) Most of the 16O in the Universe is made in this fashion. 5. Finally, if somewhat higher temperatures are ever encountered inside a star, which happens during stellar evolution of massive stars, even heavier elements can be created from the fusion of additional α-particles and α-particle by-products: Energy Minimum Reaction Released Temperature (MeV) Required (106 K) 14O + 4He −→ 20Ne + γ 4.730 700 20Ne + 4He −→ 24Mg + γ 9.317 1500 24Mg + 4He −→ 28Si + γ 9.981 1800 28Si + 4He −→ 32S + γ 6.948 2500 32S + 4He −→ 36Ar + γ 6.645 3500 12C + 12C −→ 24Mg + γ 13.930 800 16O + 16O −→ 32S + γ 16.539 2000 a) Various silicon burning reactions can occur at tempera- tures exceeding 3 × 109 K. Silicon burning produces the iron (Fe) group elements. b) Once Fe is formed, reactions that produce heavier ele- ments are all endothemic and have a tough time forming via the standard thermonuclear burning. Such elements, III–18 and the elements not built upon α-particles, are created via the r- (for rapid neutron capture) and s- (for slow neutron capture) processes. These processes will be dis- cussed in the supernovae section of the course notes (i.e., §VII). c) The reaction times of this heavy element nucleosynthesis will be discussed in the stellar evolution sections of the notes. III–19