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He Burning Overview

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He Burning Overview He burning overview • Lasts about 10% of H-burning phase • Temperatures: ~0.3 GK • Densities ~ 104 g/cm3 Reactions: 4He + 4He + 4He 12C (triple α process) 12C + 4He 16O (12C(α,γ)) Main products: carbon and oxygen (main source of these elements in the universe) 1 Helium burning 1 – the 3α process First step: α + α 8Be unbound by ~92 keV – decays back to 2 α within 2.6E-16 s ! but small equilibrium abundance is established Second step: 8Be + α 12C* would create 12C at excitation energy of ~7.7 MeV 1954 Fred Hoyle (now Sir Fred Hoyle) realized that the fact that there is carbon in the universe requires a resonance in 12C at ~7.7 MeV excitation energy Experimental Nuclear Astrophysics was born 2 How did they do the experiment ? • Used a deuterium beam on a 11B target to produce 12B via a (d,p) reaction. • 12B β-decays within 20 ms into the second excited state in 12C • This state then immediately decays under alpha emission into 8Be • Which immediately decays into 2 alpha particles So they saw after the delay of the β-decay 3 alpha particles coming from their target after a few ms of irradiation This proved that the state can also be formed by the 3 alpha process … removed the major roadblock for the theory that elements are made in stars Nobel Prize in Physics 1983 for Willy Fowler (alone !) Fred Hoyle Willy Fowler 3 How 12C is produced? Note: 8Be ground state is a 92 keV resonance for the α+α reaction 12 γ decay of C Note: 3 into its ground state Γα/Γγ > 10 so γ-decay is very rare ! 4 Calculation of the 3α rate in stellar He burning Under stellar He-burning conditions, production and destruction reactions for 12C*(7.6 MeV) are very fast (as state mainly α-decays !) Equilibrium: α+α 8Be 8Be + α 12C*(7.7 MeV) Equilibrium concentration: 8Be/4He ≈ 10-9! The 12C*(7.6 MeV) abundance is therefore given by the Saha Equation: 3/ 2 9/ 2 2π2 2π2 = 3 ρ 2 2 −Q/kT Y12C(7.6 MeV) Y4He N A e m12CkT m4HekT 2 with Q / c = m12C(7.7) − 3mα using m12C ≈ 3m4He 3 2π2 = 3/ 2 3 ρ 2 2 −Q/kT Y12C(7.6 MeV) 3 Y4He N A e m4HekT The total 3α reaction rate (per s and cm3) is then the total gamma decay rate (per s and cm3) from the 7.6 MeV state. This reaction represents the leakage out of the equilibrium ! 5 3α rate The total 3α rate r: Γ r = Y ρN γ 12C(7.6 MeV) A And with the definition 1 2 2 2 λ α = Yα ρ N < ααα > 3 6 A one obtains: note 1/6 because 3 identical particles ! 3 2 2π Γγ 2 < ααα >= ⋅ 3/ 2 ρ 2 2 −Q/kT N A 6 3 N A e m4HekT (Nomoto et al. A&A 149 (1985) 239) With the exception of masses, the only information needed is the gamma width of the 7.6 MeV state in 12C. This is well known experimentally by now. 6 Energy generation in triple-alpha process Strong temperature dependence 41 At T6=100: ε(3α) ~ T ! ε~T17 ε~T41 ε~T4 3α process: • Occurs mainly in the highest temperature regions of a star • Predominantly responsible for the luminosity of red giant stars 7 How to produce 16O via 12C(α,γ)16O? Too narrow! tail of the 1- resonance tails of subthreshold Relevant resonances temperature: T9~0.1-0.2 E0=0.3 MeV 12Cg.s.: 0+ 4He g.s.:0+ s-wave 0+ no states ! p-wave 1- d-wave 2+ Rolfs&Rodney 8 Production of 16O some tails of resonances just make the reaction strong enough … resonance (high lying) resonance (sub threshold) resonance (sub threshold) E2 DC Experimental complications: E1 E1 E2 • very low cross section makes direct measurement impossible • subthreshold resonances cannot be measured at resonance energy • Interference between the E1 and the E2 components 9 Uncertainty in the 12C(α,γ) rate The single most important nuclear physics uncertainty in astrophysics Affects: • C/O ratio~0.6 further stellar evolution (C-burning or O-burning ?) • iron (and other) core sizes (outcome of SN explosion) • Nucleosynthesis (see next slide) Some current results for S(300 keV): SE2=53+13-18 keV b (Tischhauser et al. PRL88(2002)2501 SE1=79+21-21 keV b (Azuma et al. PRC50 (1994) 1194) Ongoing experimental studies, e.g. via triple-alpha decays studies at IGISOL, Jyväskylä 10 Massive star nucleosynthesis model as a function of 12C(α,γ) rate Weaver and Woosley Phys Rep 227 (1993) 65 • This demonstrates the sensitivity • One could deduce a preference for a total S(300) of ~120-220 (But of course we cannot be sure that the astrophysical model is right) 11 Blocking of helium burning: 16O(α,γ)20Ne Unnatural parity! At higher T, resonances at higher energies (3-,1-,0+) start to contribute and enhance the rate too far from the threshold Only direct capture possible at relevant stellar temperatures low cross section 12 Ratio of α capture rates on 12C and 16O Rolfs&Rodney Relevant region 13 Helium burning: summary Resonance in Gamow window - C is made ! No resonance in Gamow window – C survives ! 14 But some C is converted into O … Carbon burning Burning conditions: for stars > 8 Mo (solar masses) (ZAMS) T~ 0.6-0.7 GK ρ ~ 105-106 g/cm3 Major reaction sequences: dominates by far of course p’s, n’s, and α’s are recaptured … 23Mg can β-decay into 23Na Composition at the end of burning: mainly 20Ne, 24Mg, with some 21,22Ne, 23Na, 24,25,26Mg, 26,27Al 16 20 O is still present in quantities comparable with Ne (not burning … yet) 15 Carbon burning High level density cross section and S-factor vary smoothly Rolfs&Rodney 16 Carbon burning Iliadis 17 Carbon burning Iliadis 18 Neon burning Burning conditions: for stars > 12 Mo (solar masses) (ZAMS=zero age main sequence star) T~ 1.3-1.7 GK ρ ~ 106 g/cm3 Why would neon burn before oxygen ??? Answer: Temperatures are sufficiently high to initiate photodisintegration of 20Ne 20Ne+γ 16O + α equilibrium is established 16O+α 20Ne + γ this is followed by (using the liberated helium) 20Ne+α 24Mg + γ The net effect: 2 20Ne 16O + 24Mg + 4.59 MeV 19 Neon burning Iliadis 20 Photodisintegration (Rolfs, Fig. 8.5.) 21 Calculations of inverse reaction rates A reaction rate for a process like: 20Ne + γ 16O+α can be calculated from the inverse reaction rate: 16O+α 20Ne + γ A simple relationship between the rates of a reaction rate and its inverse process (if all particles are thermalized) Derivation of “detailed balance principle”: Consider the reaction A+B C with Q-value Q in thermal equilibrium. Then the abundance ratios are given by the Saha equation: 3/ 2 3/ 2 n n g g m m kT A B = A B A B −Q / kT 2 e nC gC mC 2π 22 A+B C with Q-value Q in thermal equilibrium In equilibrium the abundances are constant per definition. Therefore in addition dn C = n n < συ > −λ n = 0 dt A B C C λC nAnB or = < συ > nC If <σv> is the A+B C reaction rate, and λC is the C A+B decay rate Therefore the rate ratio is defined by the Saha equation as well ! Using both results one finds 3/ 2 3/ 2 λ g g m m kT C = A B A B −Q / kT 2 e < σv > gC mC 2π µ or using mC~ mA+mB and introducing the reduced mass 23 Detailed balance and partition functions Detailed balance: 3/ 2 λ g g µkT C = A B −Q / kT 2 e < σv > gC 2π Need to know: - partition functions g - mass m of all participating particles can calculate the rate for the inverse process for every reaction Partition functions g: For a particle in a given state i this is just gi = 2Ji +1 However, in an astrophysical environment some fraction of the particles can be in thermally excited states with different spins. The partition function is then given by: −Ei / kT g = ∑ gie i 24 Photodisintegration ( , ) X(α,γ)Y exp ( , ) Y(γ,α)X ∝ − • Strong Q-value dependence • Q-value = mass difference = energy needed for the photodisintegration • Photodisintegration increases the relative abundance of even-even nuclei over that of even-odd nuclei • Reduces nuclei to their most stable form 12 11 12 13 13 12 C(γ,n) C Sn( C)= 18721.66 keV >> Sn( C)= 4946.31 keV C(γ,n) C 12 11 12 13 13 12 C(γ,p) B Sp( C)= 17532.86 keV >> Sp( N)= 1943.49 keV N(γ,p) C More energy required Less energy required from γ to remove a n or p N=Z=6 from γ to remove a n or p Photodisintegration even-even Photodisintegration less probable more probable more stable 25 Oxygen burning Burning conditions: T~ 2 x 109 K ρ ~ 107 g/cm3 Major reaction sequences: (5%) (56%) (5%) (34%) plus recapture of n,p,d,α Main products: 28Si,32S (90%) and some 33,34S,35,37Cl,36.38Ar, 39,41K, 40,42Ca 26 Oxygen burning Iliadis 27 Silicon burning Burning conditions: T~ 3-4 GK ρ ~ 109 g/cm3 Reaction sequences: • fundamentally different to all other burning stages • direct burning 28Si+28Si would require too high temperatures (Coulomb barrier) • complex network of fast (γ,n), (γ,p), (γ,α), (n,γ), (p,γ), and (α,γ) reactions • The net effect : 2 28Si --> 56Ni 4 • photodisintegration plays a dominant role (Nγ � T ) need new concept to describe burning: Nuclear Statistical Equilibrium (NSE) Quasi Statistical Equilibrium (QSE) 28 The endpoint of the Silicon burning Rolfs&Rodney 29 Silicon burning Iliadis 30 Silicon burning Iliadis 31 Reaction chains in Si burning From Iliadis λ(α,γ) λ(γ,α) 32 Nuclear Statistical Equilibrium In NSE, each nucleus is in equilibrium with protons and neutrons Means: the reaction Z * p + N * n <−−−> (Z,N) is in equilibrium Or more precisely: Z ⋅ µ p + N ⋅ µn = µ(Z,N) for all nuclei (Z,N) NSE is established when both, photodisintegration rates of the type (Z,N) + γ −−> (Z-1,N) + p (Z,N) + γ −−> (Z,N-1) + n (Z,N) + γ −−> (N-2,N-2) + α and capture reactions of the types (Z,N) + p −−> (Z+1,N) (Z,N) + n −−> (Z,N+1) are fast (Z,N) + α −−> (Z+2,N+2) 33 NSE: timescale NSE is established on the timescale of the slowest reaction 18 10 16 10 102 g/cm3 107 g/cm3 approximation by Khokhlov MNRAS 239 (1989) 808 14 10 12 10 NSE if the timescale is 10 10 shorter than the timescale 8 10 for the temperature and 6 10 density being sufficiently 4 3 hours 10 high.
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