13 Synthesis of heavier elements. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 1

Transcription

1 13 Synthesis of heavier elements introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 1

2 The triple α Reaction When hydrogen fusion ends, the core of a star collapses and the temperature can reach 10 8 K (8.6 kev). Then the following reaction can follow 4 He + 4 He 8 Be (13.1) followed by 8 Be + 4 He 12 C +γ (13.2) The first reaction forms 8 Be which is unbound by 90 kev and decays back to 2α within 2.6 x s ( 8 Be is a resonance). But as soon as a small abundance of 8 Be is produced, the second reaction, Eq. (13.2), can occur. The second reaction will create 12 C with an excitation energy of approximately 7.7 MeV. But the rate estimate for this reaction is very small and there would be no way to appreciably form 12 C in stars if something else did not happen. In 1954 Sir Fred Hoyle had the idea that there should be a resonance in 12 C at approximately 7.7 MeV so that this reaction can occur with an appreciable rate and explain the existence of 12 C in the universe (as we have discussed before, resonance reaction cross sections are larger). In 1957, Fowler and collaborators in the Kellogg Radiation laboratory at Caltech discovered this resonance at the correct energy of MeV. This state is known as the Hoyle state or the life state (no carbon, no life!). à Nobel prize 1983 to William Fowler. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 2

3 The 3α à 12 C rate When He-burning sets in, the production (3a à 12 C) and destruction ( 12 C à 3α) reactions for the resonant state in 12 C * (at 7.7 MeV) are very fast and the chain reaction is in equilibrium, i.e., the reactions below occur equally well α +α 8 Be, 8 Be +α 12 C(7.7 MeV) (13.3) Due to the equilibrium conditions, the 12 C * (7.7 MeV) abundance is given by the Saha Equation. After some algebra one gets! 2π! 2 Y 12C(7.7 MeV) # " m 12C kt $ & % 3/2! 3 = Y 4He ρ 2 2 2π! 2 N A # " m 4He kt $ & % 9/2 e Q/kT with (13.4) 2 Q c = m 3 / 12C(7.7) m α (13.5) introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 3

4 The 3α à 12 C rate Neglecting the binding energy, m 12C ~ 3m 4He (13.6) And one gets! Y 12C(7.7 MeV) = 3 3/2 3 Y 4He ρ 2 2 2π! 2 N A # " m 4He kt $ & % 3 e Q/kT (13.7) Since the rate for production of 12 C (7.7 MeV) is nearly equal to the rate for decay of 12 C to 3α, the total 3α reaction rate (per second and cm 3 ) is equal to the total gamma decay rate (per second and cm 3 ) from the 7.7 MeV state. The 12 C à 3α rate λ 3α is where Γ γ is the width for the decay by gamma-emission. λ 3α = Y 12C(7.7 MeV) ρn A Γ γ! (13.8) The reaction rate for 3α is r 3α = 1 6 Y 3 ρ 3 N 3 α A < ααα > (13.9) where the factor 1/3! = 1/6 arises because the α s are 3 identical particles. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 4

5 The 3α à 12 C rate One then obtains " < ααα >= 6 3 3/2 2π! 2 $ # m 4He kt % ' & 3 Γ γ! e Q/kT (13.10) With the exception of masses, the only information needed to calculate this rate is the gamma width of the 7.7 MeV state in 12 C. The branching ratio for decay by alpha emission over the decay by gamma emission is very small: Γ α /Γ γ > Thus γ-decay of 12 C (7.7MeV) to the ground state of 12 C is very rare, but occurs. The triple-alpha and the 12 C(α,γ) (to be discussed next) produced within stars are the largest source of carbon and oxygen in the universe. The figure (from NASA) questions what would be of life, if the 3α reaction did not exist. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 5

6 The 12 C(α,γ) rate Helium burning lasts about 10% of the hydrogen burning phase. It occurs at temperatures of about 300 million K and densities of about 10 4 g/cm 3. After 12 C is formed another reaction, the 12 C(α,γ) can occur. The figure shows the states of 16 O involved in the process. Some are in the continuum (i.e., α + 12 C free states resonances), and the others are bound states in 16 O. When the α + 12 C are captured to 16 O a photon is emitted. A photon carries away several angular momenta (l = 1, 2,..) and several components of the electric field. They are labeled E1 (electric, l = 1), E2, etc. The complications to calculate or measure this cross section are numerous. E2 DC E1 E1 Here is a high lying resonance Here is a sub threshold resonance - Due to the Coulomb barrier, the cross section at the needed Gamow energy of 300 kev is very small. Direct measurements are impossible. - Also, the subthreshold resonances cannot be measured at resonance energy. - Quantum interference between the E1 and the E2 photon components are also an additional complication. 6 introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 6 E2

7 The 12 C(α,γ) rate 10 4 S(E) [kev b] C(!!") 16 O E2 E E cm [MeV] The figure shows the experimental data for E1 and E2 transitions in the 12 C(α,γ) reaction. The curves are theory guided fits. Notice that the S-factor scale is logarithmic. Thus, the extrapolation of the fits to the Gamow energy (the zero in this scale, i.e. 300 kev) can easily change by a factor of 2, or more. The presently accepted values at 300 kev are S(E1) ~ 50 kev b and S(E2) ~ 80 kev b, giving a total of S E1 + S E2 ~ 120 ev b. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 7

8 The 12 C(α,γ) rate Stellar modelers believe that the uncertainty in the 12 C(α,γ) rate is the most important nuclear physics uncertainty in astrophysics. Calculations show that the C/O ratio determines the subsequent stellar evolution. - The star will evolve further by carbon burning or by oxygen burning? - The remaining iron core sizes after the supernova explosion are very important to determine if the left over from a SN explosion becomes a neutron star or a black-hole. - Evidently, this reaction also plays strong role on nucleosynthesis in general. The figure, from Weaver and Woosley, Phys. Rep. 227 (1993) 65 shows the sensitivity on the production of several elements in a massive star as a function of the 12 C(α,γ) reaction cross section multiplied by a factor to accommodate for experimental uncertainties. We clearly see that the abundance of the various elements vary by huge factors depending on the value of this reaction cross section. 8 introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 8

9 12 C burning reactions For stars with masses above 8M, temperatures can reach T ~ 7 x 10 8 K and the densities at the core can reach 10 6 g/cm 3. Then the following can happen 12 C + 12 C 24 Mg 23 Mg + n 2.6 MeV Among these, the reaction 20 Ne +α MeV 23 Na +α MeV (13.11) 12 C + 12 C 20 Ne +α MeV (13.12) has the largest cross section and dominates carbon burning. Following this reaction, protons, neutrons and α s can be recaptured by 23 Mg, which follows by β-decay to 23 Na. At the end of this chain, Ne, Mg, Na and Al isotopes are formed. Oxygen is also present in the plasma, but does not start burning until something else occurs. 9 introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 9

10 Neon burning reactions For stars with masses above 12M sun, temperatures can reach T ~ 1.5 x 10 9 K and the densities at the core can reach 10 6 g/cm 3. It is interesting the 20 Ne burns before oxygen, because neon has a larger charge (Z = 10) than oxygen (Z = 8). The Coulomb barrier for oxygen burning is therefore smaller than for neon burning. But the temperatures are sufficiently high to initiate the photodisintegration of 20 Ne. Then alpha capture on oxygen and photodisintegration of neon are in equilibrium. That is, 16 O +α 20 Ne +γ (13.13) Thus, α s will be present in the medium and can induce the reaction 20 Ne +α 24 Mg +γ (4.6 MeV) (13.14) The net effect of this phase will be the burning of two 20 Ne to 16 O and 24 Mg 2 20 Ne 16 O + 24 Mg MeV 10 (13.15) introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 10

11 Neon burning reactions Even if one does not know the reaction cross section for 20 Ne +γ 16 O +α (13.16) one can obtain it from the inverse reaction 16 O +α 20 Ne +γ (13.17) as long as they are in thermal equilibrium. This is based on the detailed balance theorem and the Saha equation. If the two reactions i + j ßà k are in equilibrium and have a Q-value then the abundance ratios are given by the Saha equation n i n j n k = g i g j g k! # " m i m j m k $ & % 3/2! kt # " 2π! 2 $ & % 3/2 e Q/kT (13.18) We denote by <σv> the i + j à k reaction rate, and by λ k the k à i + j the decay rate of k. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 11

12 Neon burning reactions Since the abundances are in equilibrium, the following relation holds dn k dt = n leading to i n j < συ > λ k n k = 0 λ k (13.19) Combining this with the Saha equation yields < συ > = n i n j n k (13.20) λ k < σ v > = g i g j g k! # " m i m j m k $ & % 3/2! kt # " 2π! 2 $ & % 3/2 e Q/kT (13.21) If we use m k ~ m i + m j (i.e., neglecting the binding energies), and defining the reduced mass µ = m i m j /m k we get the detailed balance equation λ k < σ v > = g i g j g k! µkt # " 2π! 2 $ & % 3/2 e Q/kT (13.22) This means that if we know the degrees of freedom (or number of possible angular momentum and spin states, g), we can calculate the reaction rate for γ + k à i+j from the inverse reaction rate i +j à k + γ. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 12

13 Neon burning reactions Usually, the number of degrees of freedom for a state with angular momentum J is just g = 2J + 1. But in a plasma particles are excited and can occupy several energy states, increasing the number of possible rearrangements (# of degrees of freedom) of the system. In statistical physics, the way to count the number of degrees of freedom is based on the weight carried by the amount of excitation of the nucleus (also applies to atoms and other complex systems). This is due to the Boltzmann probability to have the energy state E i given by exp(-e i /kt). Thus one introduces the partition function g Z = g i e E i /kt (13.23) i These are the factors g to be used in Eq. (13.22). Using them, the reaction cross section for 20 Ne + γ à 16 O + α can be calculated by using the experimental data on the reaction 16 O + α à 20 Ne + γ. Gamma induced reactions are much more difficult to measure due to the difficulty of producing γ-beams. But this trick solves the problem. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 13

14 Oxygen and silicon burning reactions When temperatures reach 2 x 10 9 K and densities are of the order of 10 7 g cm -3, oxygen burning becomes viable through the reactions 16 O + 16 O 32 S 31 P + p MeV 28 Si +α MeV 31 S+ n +1.5 MeV 30 P + p 2.4 MeV (13.24) The first two reactions dominate the process. The main product of these reactions end being silicon and sulfur mostly (at the 90% level). But some small amounts of chlorine, argon, potassium and calcium are also produced. Silicon burning When temperatures reach 4 x 10 9 K and densities are of the order of 10 9 g cm -3, silicon burning starts. But, unlike the previous cases, this occurs through a sequence of (γ,n), (γ,p), (γ,α), (n, γ), (p, γ), and (α,γ) reactions. In the end for every 2 28 Si one gets one 56 Ni nucleus. 14 introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 14

15 Silicon burning reactions Silicon burning occurs in an environment with protons and neutrons in equilibrium with the other nuclei. The high density in the environment favors the production of heavy nuclei. But the high temperature favors (because of their lower mass) the energy sharing to free nucleons and lighter nuclei, with higher binding energy. A long but straightforward calculation based on equilibrium conditions between free protons and neutrons forming a nucleus (Z,N) and its disintegration back to protons and neutrons Z p + N n (Z,N) leads to the abundance of the nucleus (Z,N) given by Y(Z,N) = Y Z p Y N n g(z,n)(ρn A ) A 1 A 3/2 $ 2π! 2 ' 2 A & % m u kt ) ( (13.25) 3 2 (A 1) e B(Z,N)/kT (13.26) Where g(z,n) are the partition functions for the (Z,N) nucleus and B(Z,N) its binding energy. During silicon burning, nuclei heavier than 24 Mg are in statistical equilibrium, given by Eq. (13.26). This is because the high density favors heavy nuclei over free nucleons. The main product of silicon burning is 56 Fe and 56 Ni which settle in the star core. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 15

16 Summary of nuclear burning of (medium) heavy nuclei The table shows a summary of the conditions required for nuclear burning reactions, as well as the duration of the process under those conditions. Burning Dura*on Temperature (10 9 k) Hydrogen 7 x 10 6 years Density (g cm -3 ) Helium 5 x 10 5 years x 10 2 Carbon 600 years x 10 5 Neon 1 year x 10 6 Oxygen 6 months Silicon 1 day x 10 7 Hydrogen and helium burning occur for stars with mass 0.8 M 8 M Carbon to silicon burning occur during the later stages of stars with masses larger than 8 M. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 16

17 What happens to low mass stars? Low mass stars (< 2.3 M ) can leave the main sequence (MS) after the hydrogen burning stops. - The hydrogen shell increases and the star leaves the MS in the HR diagram (going up). - Its core shrinks, and the outer layers expand and cool. Its luminosity increases. - Later, its core starts 4 He burning called helium flash. - The core expands due to the huge temperature in the core. - The luminosity decreases. - The slower rate of energy production makes the outer layers contract and thus they also heat up, leading to an increase of the surface temperature T. Low-mass stars end up their productive life by ejecting their outer layers and creating planetary nebulae. The name planetary nebulae was given because they look as a small planet in a small telescope. A low mass star with one solar mass ejects as much as 40% of its material to the nebula. It is believed that the sun will become a red giant in about 5 billion years. It will grow and reach the orbit of Mars. wikipedia introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 17

18 What happens to very massive stars? For stars with mass larger than 8 M the temperatures can get very large and they can also fuse elements up to iron. Then they use up their nuclear fuel very quickly. This happens within a few million years as compared to the sun s burning timescale of 10 billion years. But iron fusion does not occur because no energy is released. The supergiant star starts to collapse and its core heats up. The iron core continues to collapse until it is stopped by compressed into neutrons. The infaling material bounces off the core, leading to a supernova explosion. Before the explosion, the star looks like an onion structure of layers where the burning of increasingly heavier elements occur from the surface inward to the core. The nuclear burning happening during the explosion is worth another lecture. wikipedia introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 18