A. Thermodynamic Conditions (Massive Stars) Review: The helium core evolves like a separate star, mass Mα 6 tHe ~ 10 years (+- factor of three); inside RSG, BSG, or as WR star n= 1.5 to 3; assume 3 Lecture 10 1− β 4 M = 18 µ = (for pure helium) solve for β α µ2β 4 3 2/3 ⎛ M ⎞ Nucleosynthesis During Helium T = 4.6 ×106 µβ α ρ 1/3 K C ⎜ M ⎟ c Burning and the s-Process ⎝ ⊙ ⎠ E.g., Mα =6 (a 20 M⊙ main sequence star) need temperatures > 108 β =0.87 to provide significant 1/3 ⎛ ρ ⎞ energy generation by 3a T = 1.8 ×108 c K so this really just sets the c ⎝⎜ 1000g cm-3 ⎠⎟ density. So, typical temperatures are 1.5 - 2 x 108 K (higher in shell burning later). As the core evolves, the temperature and density go up significantly. Note non-degenerate. 25 MO Woosley (2019) helium stars evolved at constant mass Mass Density Temperature/108 3.5 1430 1.75 He 4.0 1260 1.79 4.5 1140 1.82 He 6.0 910 1.89 7.0 814 1.92 Most mass loss occurs during helium core burning for single stars The helium core itself can grow (single stars) or lose mass Z = 0.02. Helium burning at central helium mass (close binary plus winds) fraction of 50%. Helium burns in the star’s center and later in a shell. Helium burning in the shell is incomplete and makes more carbon relative to oxygen. At 1.9 x 108 K, the temperature sensitivity of the 3a rate is approximately T20. During helium burning the star is a red or blue supergiant depending on semiconvection, or a WR star if it loses its envelope. See previous lecture. B. Major Nucleosynthesis – C and O He shell He core In massive stars after helium burning in the stars center Heavy (calculations included semiconvection). Woosley et al (2007) CO-core Elements Neutron stars 12 16 20 M/M⊙ X( C) X( O) X( Ne) In the sun, 12 0.229 0.752 1.48(-3) 12C/16O = 0.32 15 0.214 0.765 2.05(-3) 19 0.202 0.776 3.15(-3) 25 0.184 0.792 5.62(-3) 35 0. 169 0.801 1.03(-2) Buchman 12C(α,γ)16O multiplied by 1.2. If the star contains appreciable metals there is, as we shall see also 22Ne and 18O.. Sukhbold and Woosley (2014) C. TRACE ELEMENT NUCLEOSYNTHESIS Complications: AND NEUTRONIZATION Prior to the ignition of the 3α reaction the star goes through a • If the helium core grows just a little bit towards the brief stage of "nitrogen burning". All the initial CNO in the core 14 end of helium burning, the extra helium convected in has been converted to N which captures an α-particle and greatly decreases the 12C synthesis. experiences a weak decay The number of CNO nuclei is 14 18 + 18 • Mass loss from very massive WR stars can greatly N(α,γ ) F(e ν) O preserved 12 16 X(14 N) 14. Y(14 N) 14. Y (14 N) Y (12C) Y (16O) increase the synthesis of both C and O for stars over = i = i( i + i + i ) 35 solar masses 0.0095 Z / Z = ( ⊙ ) 12 16 • The uncertain rate for C(α,γ) O This reaction is very important both as a source of18O in nature and because it creates a neutron excess • 12C/16O ratio may be affected by post-helium burning evolution and by black hole formation above some =1- 2Y 1 2 ZY N Z Y 2 ZY η e = − ∑ i i = ∑( i + i ) i − ∑ i i critical main sequence mass. 16O is made in the more N Z Y 1 1 η = ∑( i − i ) i − < η < massive (massive) stars. 4 14 Pr ior to this reation Ye = 0.50 because both He and N have Massive stars are definitely the site of oxygen nucleosynthesis equal numbers of neutrons and protons and η=0. After this reaction and contribute appreciably to carbon. η > 0 with a value that depends on the initial metallicity (i.e., initial mass fractions of C, N, and O) of the star. 14 18 18 During helium shell burning, which does not go to completion, Before N(α,γ ) F(e+ν) O the composition was 18O remains undestroyed and this is the source of 18O in nature. almost entirely 4He and 14N, hence η ≈ 0 (actually Convection helps to preserve it a small posive value exists because of 56Fe and the like). -3 Lifetimes in years Xα = 0.5, ρ = 1000 g cm After this reaction T8 = 1.8 T8 = 2.0 Z 14N(α,γ) 15 1.1 destroyed η = 0.00136 18O(α,γ) 2700 98 destroyed in core Z⊙ 22Ne(α,γ) + (α,n) 4.1(6) 2.2(5) partly burned 18 During helium core burning, O is later mostly 16O(α,γ) 3.5(10) 2.3(9) survives destroyed by 18O(α,γ )22Ne, but the neutron excess is unchanged. η can only be changed by weak interactions. Note that the reaction rate for 22Ne(α,n)25Mg is This neutron excess is very important to the subsequent uncertain and this is an important source of error in s-process nucleosynthesis. Need all the 22Ne nucleosynthesis and one of the main reasons nucleosynthesis to burn in low Z stars is different! So one expects that, depending on mass and rates, some but not all D. The s-Process in Massive Stars of the 22Ne will burn when the temperature goes up towards the end of helium burning. Late during helium burning, when the temperature rises to about 3.0 x 108 K, 22Ne is burned chieFly by the reaction The following table gives the temperature at the center of the 22 25 22 26 given model and the mass fractions of 22Ne, 25Mg, and 26Mg Ne(a,n) Mg (with some competition From Ne(a,g) Mg). each multiplied by 1000, when the helium mass fraction is 1% and zero Where do the neutrons go? 22 25 26 56 M TC Ne Mg Mg Some go on Fe but that Fraction is small: The remainder of the Y 22 σ 56 56 12 2.42 13.4 0.51 0.61 Ne will burn early f = Woosley, during carbon burning, σ Y 12.3 1.17 1.05 ∑ i i and Heger but then there will be 17 20 0.5 −2 −3 But, O(a,n) Ne (2007) 15 2.54 12.7 0.98 0.91 more abundant neutron 16 Y ≈ =3.1× 10 Y σ ≈1.2×10 17 O 16 16 16 16 destroys the O and 11.1 1.99 1.80 poisons, Na and Mg. restores the neutron 19 2.64 11.5 1.73 1.54 0.005 −4 −5 22 Y ≈ = 2.3 × 10 Y σ ≈ 1.3×10 9.4 2.90 2.87 These numbers are quite Ne 22 22 22 22 25 2.75 9.8 2.67 2.59 sensitive to the uncertain 0.005 −4 −3 For σ in mb 6.96 4.05 4.54 reaction rate for 25 Y25 ≈ = 2× 10 Y25 σ 25 ≈1.3×10 22 25 Mg 25 35 2.86 7.37 3.87 4.22 Ne(a,n) Mg and may be lower limits to the 22Ne 0.0013 4.41 5.18 6.39 Y ≈ = 2.3 × 10−5 Y σ ≈ 2.7×10−4 consumption. 56Fe 56 56 56 56 30 keV neutron capture cross sections Y σ (mostly from Bao et al, ADNDT, 2000) So a fraction 56 56 ~10 − 20% capture on iron. How many Y25 σ 25 + ... neutrons is this? If all the neutrons captured on iron one would have Nucleus s Nucleus s Yn / YFe ~ Y22 / Y56 ≈ 30 neutrons per iron (mb) (mb) where we have assumed a mass fraction of 0.015 for 22Ne 12 54 C 0.0154 Fe 27.6 and 0.0012 for 56Fe and that all 22Ne burns by (α,n). *16O 0.038 56Fe 11.7 20Ne 0.119 57Fe 40.0 22Ne 0.059 58Fe 12.1 But only 10 - 20% of these actually add to iron. 24Mg 3.3 58Ni 41.0 This is about 3 - 6 neutrons per iron and obviously not 25Mg 6.4 64Zn 59 26Mg 0.126 65Zn 162 nearly enough to change e.g., Fe into Pb, but the 28Si 2.9 66Zn 35 neutron capture cross sections of the isotopes generally 88Sr 6.2 (closed shell) increase above the iron group and the solar abundances The large cross section of 25Mg is particularly significant since it is made by 22Ne(a,n) 25Mg.. decrease. A significant s-process occurs that produces * Igashira et al, ApJL, 441, L89, (1995); factor of 200 upwards revision significant quantities of the isotopes with A < 90. At the end .... Composition of a 25 solar mass star at the end of helium burning compared with solar abundances (Rauscher et al 2001). 25 solar mass supernova model post-explosion 63Cu 65Cu Results depend most sensitively upon the reaction rate for 60Ni 61 62 64 22 25 Ni Ni Ni Ne(α,n) Mg. 59 If 22Ne does not burn Co until later (i.e., carbon burning there are much more abundant neutron 56 poisons Fe 57Fe 58Fe ruled out by subsequent experiment Here "s" stands for "slow" neutron capture 1 τ 22 22 22 1/2 dY τ β τ nγ τ β = = i Yα Y ( Ne)λαn ( Ne) Yα λαn ( Ne) = −Yi Yn ρ λnγ (i) + ..