13 – synthesis of heavier elements - faculty web sites...

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13 – Synthesis of heavier elements



When hydrogen fusion ends, the core of a star collapses and the temperature can reach 108 K (8.6 keV). Then the following reaction can follow followed by The first reaction forms 8Be which is unbound by 90 keV and decays back to 2α within 2.6 x 10-16 s (8Be is a resonance). But as soon as a small abundance of 8Be is produced, the second reaction, Eq. (13.2), can occur. The second reaction will create 12C 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 12C in stars if something else did not happen. In 1954 Sir Fred Hoyle had the idea that there should be a resonance in 12C at approximately 7.7 MeV so that this reaction can occur with an appreciable rate and explain the existence of 12C 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 7.654 MeV. This state is known as the “Hoyle state” or the “life state” (no carbon, no life!). à Nobel prize 1983 to William Fowler.

The triple α Reaction

4He+ 4He→ 8Be 8Be+ 4He→ 12C+γ(13.1) (13.2)


The 3α à 12C rate When He-burning sets in, the production (3a à 12C) and destruction (12C à 3α) reactions for the resonant state in 12C*(at 7.7 MeV) are very fast and the chain reaction is in equilibrium, i.e., the reactions below occur equally well

Due to the equilibrium conditions, the 12C*(7.7 MeV) abundance is given by the Saha Equation. After some algebra one gets

Y12C(7.7 MeV)

2π!2

m12CkT

!

"##

$

%&&

3/2

=Y4He3 ρ2NA

2 2π!2

m4HekT

!

"##

$

%&&

9/2

e−Q/kT

αmmcQ 3/ 12C(7.7)2 −=

with

α +α↔ 8Be, 8Be+α↔ 12C(7.7 MeV) (13.3)

(13.4)

(13.5)


Y12C(7.7 MeV) = 33/2 Y4He3 ρ2NA

2 2π!2

m4HekT

!

"##

$

%&&

3

e−Q/kT

Neglecting the binding energy, And one gets

m12C ~ 3m4He

Since the rate for production of 12C (7.7 MeV) is nearly equal to the rate for decay of 12C to 3α, the total 3α reaction rate (per second and cm3) is equal to the total gamma decay rate (per second and cm3) from the 7.7 MeV state.

The 12C à 3α rate λ3α is where Γγ is the width for the decay by gamma-emission.

λ3α =Y12C(7.7 MeV)ρNA

Γγ

!

The reaction rate for 3α is where the factor 1/3! = 1/6 arises because the α’s are 3 identical particles.

r3α =16Yα3ρ3NA

3 <ααα >

The 3α à 12C rate (13.6)

(13.7)

(13.8)

(13.9)


One then obtains

<ααα >= 6 ⋅33/2 2π!2

m4HekT

"

#$$

%

&''

3Γγ

!e−Q/kT

With the exception of masses, the only information needed to calculate this rate is the gamma width of the 7.7 MeV state in 12C. The branching ratio for decay by alpha emission over the decay by gamma emission is very small: Γα/Γγ > 103. Thus γ-decay of 12C (7.7MeV) to the ground state of 12C is very rare, but occurs. The triple-alpha and the 12C(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.

The 3α à 12C rate

(13.10)


The 12C(α,γ) rate

6

Here is a high lying resonance

Here is a sub threshold resonance

E1 E1 E2 DC

E2

-  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.

Helium burning lasts about 10% of the hydrogen burning phase. It occurs at temperatures of about 300 million K and densities of about 104 g/cm3. After 12C is formed another reaction, the 12C(α,γ) can occur.

The figure shows the states of 16O involved in the process. Some are in the continuum (i.e., α + 12C free states – resonances), and the others are bound states in 16O. When the α + 12C are captured to 16O 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.


The 12C(α,γ) rate

1 2 3 4 5

Ecm [MeV]

100

101

102

103

104

S(E

) [

ke

V b

]

12C(!!")16O

E1

E2

The figure shows the experimental data for E1 and E2 transitions in the 12C(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 SE1 + SE2 ~ 120 eV b.


The 12C(α,γ) rate

8

Stellar modelers believe that the uncertainty in the 12C(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 12C(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.


12C burning reactions

9

For stars with masses above 8M¤, temperatures can reach T ~ 7 x 108 K and the densities at the core can reach 106 g/cm3. Then the following can happen Among these, the reaction has the largest cross section and dominates carbon burning. Following this reaction, protons, neutrons and α’s can be recaptured by 23Mg, which follows by β-decay to 23Na. 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.

12C+ 12C→ 24Mg→ 23Mg+ n − 2.6 MeV→ 20Ne+α + 4.6 MeV→ 23Na +α + 2..2 MeV

12C+ 12C→ 20Ne+α + 4.6 MeV

(13.11)

(13.12)


Neon burning reactions

10

For stars with masses above 12Msun, temperatures can reach T ~ 1.5 x 109 K and the densities at the core can reach 106 g/cm3. It is interesting the 20Ne 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 20Ne. Then alpha capture on oxygen and photodisintegration of neon are in equilibrium. That is, Thus, α’s will be present in the medium and can induce the reaction The net effect of this phase will be the burning of two 20Ne to 16O and 24Mg

16O+α↔ 20Ne+γ

20Ne+α↔ 24Mg+γ (4.6 MeV)

220Ne→ 16O+ 24Mg+ 4.6 MeV

(13.13)

(13.14)

(13.15)


Neon burning reactions Even if one does not know the reaction cross section for one can obtain it from the inverse reaction 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 We denote by <σv> the i + j à k reaction rate, and by λk the k à i + j the decay rate of k.

nin jnk

=gig jgk

mimj

mk

!

"##

$

%&&

3/2kT2π!2!

"#

$

%&

3/2

e−Q/kT

20Ne+γ→ 16O+α

16O+α→ 20Ne+γ

(13.16)

(13.17)

(13.18)


Neon burning reactions Since the abundances are in equilibrium, the following relation holds leading to Combining this with the Saha equation yields If we use mk ~ mi + mj (i.e., neglecting the binding energies), and defining the reduced mass µ = mimj/mk we get the detailed balance equation 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.

dnkdt

= nin j <συ > −λknk = 0λk

<συ >=nin jnk

λk<σ v >

=gig jgk

mimj

mk

!

"##

$

%&&

3/2kT2π!2!

"#

$

%&

3/2

e−Q/kT

λk<σ v >

=gig jgk

µkT2π!2!

"#

$

%&

3/2

e−Q/kT

(13.19)(13.20)

(13.21)

(13.22)


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 Ei given by exp(-Ei/kT). Thus one introduces the partition function These are the factors g to be used in Eq. (13.22). Using them, the reaction cross section for 20Ne + γ à 16O + α can be calculated by using the experimental data on the reaction 16O + α à 20Ne + γ. Gamma induced reactions are much more difficult to measure due to the difficulty of producing γ-beams. But this trick solves the problem.

g ≡ Z = gie−Ei /kT

i∑ (13.23)


Oxygen and silicon burning reactions When temperatures reach 2 x 109 K and densities are of the order of 107 g cm-3, oxygen burning becomes viable through the reactions 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 109 K and densities are of the order of 109 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 28Si one gets one 56Ni nucleus.

14

16O+ 16O→ 32S→ 31P+ p+7.7 MeV→ 28Si+α +9.6 MeV→ 31S+ n +1.5MeV→ 30P+ p− 2.4 MeV

(13.24)


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

leads to the abundance of the nucleus (Z,N) given by 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 24Mg 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 56Fe and 56Ni which settle in the star core.

Y(Z,N) =YpZYn

Ng(Z,N)(ρNA)A−1 A3/2

2A2π!2

mukT

$

%&&

'

())

32(A−1)

eB(Z,N)/kT

Z× p+N× n↔ (Z,N) (13.25)

(13.26)


Summary of nuclear burning of (medium) heavy nuclei

Burning Dura*on Temperature(109k)

Density(gcm-3)

Hydrogen 7x106years 0.06 5Helium 5x105years 0.2 7x102

Carbon 600years 0.9 2x105

Neon 1year 1.7 4x106

Oxygen 6months 2.3 107

Silicon 1day 4.1 3x107

The table shows a summary of the conditions required for nuclear burning reactions, as well as the duration of the process under those conditions. 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¤.


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 4He 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


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

13 – synthesis of heavier elements - faculty web sites...

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