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Shall We Play a Game of Thermonuclear War?

Shall We Play a Game of Thermonuclear War?. Several Mev. E. c. r. r. 10. cm. -13. 0. E. n. 30 Mev. Thermonuclear Energy Sources. Electrostatic Potential E c ~ e 2 /r 0  1.7(10 -6 ) ergs  1 Mev

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Shall We Play a Game of Thermonuclear War?

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  1. Shall We Play a Game of Thermonuclear War?

  2. Several Mev E c r r 10 cm -13 0 E n 30 Mev Thermonuclear Energy Sources • Electrostatic Potential • Ec ~ e2/r0 1.7(10-6) ergs  1 Mev • In a star pressure and high temperature cause penetration of the Coulomb barrier. Or do they? ~ ~ Thermonuclear Energy

  3. The Solar Core • The velocities are Maxwellian: • f(E) = (2/π1/2) (1/kT)3/2 e-E/kT E1/2 • E is the average thermal energy of a proton • <E> = 3/2 kT which is several keV at 2(107) K or 10-3 Ec! • The Sun does not have sufficient numbers of protons at high temperatures to fuel its energy output! Or does it? Thermonuclear Energy

  4. Barrier Penetration • Consider 84Po212: it is an α emitter with a laboratory half-life of 3(10-7) sec. • The particle has an energy of ~ 9 MeV. This particle should not be able to escape from a classical potential! • But atoms are NOT classical systems. • They are quantum mechanical entities • The particles in a nucleus are described by a spatial probability distribution. • The probability density is highest where the classical theory says the particle should be but the probability is non-zero at every other point. • Sometimes the α particle will find itself at r > r0! Thermonuclear Energy

  5. The Inverse Problem • We are interested in the inverse of the Po case. • We want to break the barrier coming in. • Classically there is no opportunity for the penetration but in QM there is a finite probability of the event; that is, the incoming particle will sometimes find itself at r < r0. • The closer you can get the incoming particle to the nuclei to be penetrated the better the probability. • Thus penetration is favored in a fast moving particle. Thermonuclear Energy

  6. Miniumum Mass for Stars Barrier Potential Z1Z2e2 • “Low Temperatures” favor penetration of low Z species • Low mass objects have lower values of T • Only H can be burnt • Core Temperature Tc > TI (ignition) for a particular fuel • If we have a uniform density: ρ = M/4πR3 • Tc (GmμmH) / (5kR) (Hydrostatic Ideal Gas) Thermonuclear Energy

  7. The Condition for Burning • At high density matter becomes degenerate and the thermal energy of a photon must be less than the Fermi energy: • Charge neutrality must be maintained: ne=np= ρ/mH from which one may obtain: Thermonuclear Energy

  8. The Minimum Mass • The Inter-nuclear Separation is: • Note that 2mekT is the thermal wavelength of an electron • One can then obtain: • For H ( μ = ½) at T = 107: M/M > 0.14 • A more detailed argument says: M/M > 0.05 to 0.08 Thermonuclear Energy

  9. Energy Release The Process is characterized by • Units: ergs gm-1 sec-1 • ε0: Constant depending on the reaction • a: A constant which is ~ 1 • n: 4 for H burning (pp cycle) and 30 for C burning Thermonuclear Energy

  10. Reaction Networks • Energy release: photons (hν) + neutrinos • Only the photons contribute to the energy generation • Neutrinos are lost to the system • Note that the photon generation can be after the reaction: specifically in ee+ annihilation • A reaction network is a specific set of nuclear reactions leading to either energy production and/or nucleosynthesis. • pp hydrogen burning is one such network • CNO hydrogen burning is another • 3α (He burning) is another Thermonuclear Energy

  11. The pp Chain Process: PPI • Initial Reaction: p + P → D + e+ + νe • Energy = 1.44207 MeV • Only D remains as the position undergoes immediate pair annihilation producing 511 KeV. The e is lost energy to the system. • This can be written: H1 (p,β+νe) D • Next: p + D → He3 + γ • Produces 5.49 MeV • At T < 107 K this terminates the chain Thermonuclear Energy

  12. PPI continued • Lastly (at T > 107 K): He3 + He3→ He4 + 2p • This produces 12.86 MeV • Total Energy for PPI: 2(1.18 + 5.49) + 12.86 = 26.2 MeV • Reaction 1 & 2 must occur twice • The neutrino in reaction 1 represents 0.26 MeV Thermonuclear Energy

  13. PPII He4 must be present and T > 2(107) K • He3 + He4→ Be7 + γ 1.59 MeV • Be7 + p → B8 + γ 0.13 MeV • B8→ Be8 + e+ + νe 10.78 MeV (7.2 in νe) • Be8→ 2He4 0.095 MeV • The total energy yield is 19.27 MeV (including the first two reactions of PPI. • Note that when Be or B are made they end up as He. Thermonuclear Energy

  14. PPIII A Minor Chain as far as probability is concerned. • Start with the Be7 from PPII • First reaction is Be7 + e-→ Li7 + ν • Next Li7 + p → 2He4 • This is the infamous neutrino that the Davis experiment tries to find. • Note that we cannot make Li either! Thermonuclear Energy

  15. Timescales for the Reactions Thermonuclear Energy

  16. Nucleur Reaction Rates • Energy Generation Depends on the Atomic Physics only (On A Per Reaction Basis) • Basics: 4H → He4 • 4H = 4(1.007825 AMU) = 6.690594(10-24) gm • He4 = 4.00260 AMU = 6.645917(10-24) gm • Mass = 4.4677(10-26) gm • E = mc2 = 4.01537(10-5) ergs = 25.1 MeV (vs 26.2 MeV ?!) • Mass Excess ≡ M = M - AmH • Reactions can be either exothermic or endothermic • Burning Reactions that produce more tightly bound nuclei are exothermic; ie, the reactions producing up to the iron peak. Thermonuclear Energy

  17. Conservation Laws • Electric Charge • Baryon Number • p = n = 1 • e = ν = 0 • Spin Thermonuclear Energy

  18. The Units in ε = ε0ρaTn • ε = ergs g-1 s-1 = g cm2 s-2 g-1 s-1 = cm2 s-3 • Assume a = 1 ρa = ρ : g cm-3 • So cm2 s-3 = ε0 g cm-3 Kn • ε0 = (cm2 s-3)/(g cm-3 Kn) = (cm2/g) cm3 s-3 K-n • So how do we get ε0? • Bowers & Deming Section 7.3 • Clayton Thermonuclear Energy

  19. Reaction Cross Section • Cross Section σ • Let X = Stationary nucleus • Let a = Moving nucleus • Probability of a reaction per unit path length is nXσ where nX is the density of particle X • nXσ = # cm-3 cm2 = # cm-1 (This is 1 / MFP) • MFP = (1 / nXσ) = vt where t = time between collisions and v = speed of particles (the one(s) that are moving). • t = 1 / nXσv or Number of Reactions (per Stationary particle/s) = nXσv • Total number of reactions: r = nanXvσ(v) • r = cm-3 cm-1 cm s-1 = # / (cm3 s) • The cross section must be a function of v Thermonuclear Energy

  20. Integrate Over Velocity • There actually exist a range of velocities f(v) so we must integrate over all velocities: • The total energy is then rQ/ρ. • r = reaction rate in # / (cm3 s) • Q = energy produced per reaction • ρ = density g cm-3 • rQ/ρ = cm-3 s-1 g-1 cm3 ergs = ergs s-1 g-1 = ε = ε0ρaTn Thermonuclear Energy

  21. Penetration Factor • The penetration factor is defined as: • m is the reduced mass of the system Thermonuclear Energy

  22. The Cross Section • We write the cross section thus: • S(E) is the “area” of the reaction - set by the deBroglie wavelength of the particle • S(E) varies slowly with E • Now change variable to E and rewrite <v> Thermonuclear Energy

  23. Gamow Peak • <σv> is the combination of two pieces: • e-E/kT (Maxwellian) decreases with increasing E and represents the decreasing number of particles with increasing E. • e-b/SQRT(E) (Penetration) increases with E and represents the increasing probability of a reaction with E [speed]. • E0 is the peak energy for <σV> and can be thought of as the most effective energy for that T Thermonuclear Energy

  24. Analytically • S(E) varies slowly with E and its primary contribution is at E0 ==> want S(E0) • Next approximate the Peak by a Gaussian and you get: • See Clayton For the Approximation formulae Thermonuclear Energy

  25. Nuclear Burning Stages and Processes • Main Sequence Energy Generation • pp Hydrogen Burning • pp cycle: H(H,e+νe) D (H,γ) He3 then He3 (He3,2p) He4 • Slowest Reaction is: H(H,e+νe) D • The cross section for the above has never been measured. It also depends on the weak interaction which governs β decay : p → n + e+ + νe • For pp T ~ 107 K so this is the energy source for the lower main sequence (mid F - 2 M and later) • ε = ε0ρT4 for the pp cycle Thermonuclear Energy

  26. Reactions of the PP Chains † Computed for X = Y = 0.5, ρ = 100, T6 = 15 Thermonuclear Energy

  27. The CNO Cycle Thermonuclear Energy

  28. The CNO Cycle • Effective at T > 2(107) K • Mass > 2 M (O through early F stars) • The main energy generation is through the CN Cycle • The first two reactions are low temperature reactions. They produce 13C in zones outside the main burning regions. This material is mixed to the surface on the ascent to the first red-giant branch: 12C/13C is an indicator of the depth of mixing - not a signature that the CNO process has been the main energy generator. • 14N is produced by incomplete CN burning (and is the main source of that element). • The cycle when complete is catalytic. • ε = ε0ρT16 for the CNO cycle Thermonuclear Energy

  29. CNO Reactions † Computed for T6 = 25 Thermonuclear Energy

  30. 6Li (H,4He) 3He 7Li (H,4He) 4He 10B (2H,4He) 24He 9Be (2H,24He) 3He D (H,γ) 3He These are parts of the pp cycle and destroy any of these elements that are present. But these species still exist! How do they survive? Pre-Main Sequence Nuclear Sources Thermonuclear Energy

  31. Post-Main Sequence Processes The Triple α Process • Mass Criteria: M/M > 0.5 • Tc≈ 108 K • Summary: 3 4He →12C + γ Q = 7.27 MeV • Details: 4He (4He, ) 8Be (4He,) 12C • 4He (4He, ) 8Be is endothermic • The third 4He must be present when 4He (4He, ) 8Be occurs or Be8 → 2He4 occurs (Timescale is 2.6(10-16) s) • This is a three body collision! It happens because 12C has an energy level at the combined energy of 8Be + 4He. This is a resonance reaction. • ε = ε0ρ2T30 for 3α Thermonuclear Energy

  32. Other Common Reactions These happen along with 3α • A Very Important Reaction: 12C(4He,)16O • This reaction has a very high rate at the temperatures and densities of 3α • After 3α one has C and O. • 14N (4He,e+νe) 18O (4He,) 22Ne • This burns to completion under 3α conditions so all N is destroyed. Thermonuclear Energy

  33. Carbon Burning • C is the next fuel - it is the lowest Z nuclei left after 3α • Tc > 6(108) K • ε = ε0ρT32 for carbon burning Thermonuclear Energy

  34. Oxygen Burning Thermonuclear Energy

  35. Neutrino Energy Loss • Cross Section is σ ~ 10-44 cm2 • Interaction Probability σnbR • nb = baryon number • R = radius of star • Neutrinos are a source of energy loss but do not contribute to the pressure Thermonuclear Energy

  36. Davis Experiment • 17Cl37 + νe→ 18Ar37 + e- • The threshold energy is 0.81 MeV for the reaction. • This is a “superallowed” transition as the quantum numbers of excited 18Ar37 at 5.1 MeV are the same as the ground state in 17Cl37. • We would expect to be able to detect a single event at E > 5.1 MeV • Primary νe sources in the Sun are the pp and CN cycles. Thermonuclear Energy

  37. Sources of Solar Neutinos Thermonuclear Energy

  38. Experimental Results • The only neutrino the Davis experiment can see is the B8 neutrino - the least significant neutrino source in the chains. • Original Predicted Rate was 7.8 SNU with a detected rate of 2.1 ± 0.3 SNU • GALLEX tests the p + p neutrino production: its rate is about 65% of the predicted values. • Better solar models are cutting the predicted rates • Diffusion • Convection • Better opacities Thermonuclear Energy

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