Download 13 – Synthesis of heavier elements

13 – Synthesis of heavier elements
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1
The triple α Reaction
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
4
4
8
He + He → Be
(13.1)
followed by
8
Be + 4 He → 12 C + γ
(13.2)
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.
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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
α + α ↔ 8 Be,
8
Be + α ↔ 12 C(7.7 MeV)
(13.3)
Due to the equilibrium conditions, the 12C*(7.7 MeV) abundance is given by
the Saha Equation. After some algebra one gets
3/2
! 2π ! $
Y12C(7.7 MeV) ##
&&
" m12C kT %
2
9/2
!
$
2
π
!
3
2 2
−Q/kT
= Y4He ρ N A ##
&& e
" m 4He kT %
2
(13.4)
with
Q / c 2 = m12C(7.7) − 3mα
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(13.5)
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The 3α à
12C
rate
m12C ~ 3m 4He
Neglecting the binding energy,
And one gets
(13.6)
3
! 2π ! $ −Q/kT
Y12C(7.7 MeV) = 3 Y ρ N ##
&& e
(13.7)
" m 4He kT %
3/2
3
4He
2
2
A
2
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.
The reaction rate for 3α is
λ3α = Y12C(7.7 MeV) ρ N A
1 3 3 3
r3α = Yα ρ N A < ααα >
6
Γγ
!
(13.8)
(13.9)
where the factor 1/3! = 1/6 arises because the α’s are 3 identical particles.
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The 3α à
12C
One then obtains
rate
3
"
%
Γγ −Q/kT
2
π
!
3/2
< ααα >= 6 ⋅ 3 $$
''
e
# m 4He kT & !
2
(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 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.
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The
12C(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 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.
Here is a high
lying resonance
Here is a sub
threshold
resonance
E2 DC
E1 E1 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.
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12C(α,γ)
rate
S(E) [keV b]
The
104
103
10
12C(!!")16O
2
101
100
E1
E2
1
2
3
4
5
Ecm [MeV]
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.
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The
12C(α,γ)
rate
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.
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8
12C
burning reactions
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
12
C + 12 C → 24 Mg → 23 Mg + n − 2.6 MeV
→ 20 Ne + α + 4.6 MeV
→ 23 Na + α + 2..2 MeV
(13.11)
Among these, the reaction
12
C + 12 C → 20 Ne + α + 4.6 MeV
(13.12)
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.
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Neon burning reactions
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,
16
20
O + α ↔ 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 20Ne to 16O and
20
16
24
2 Ne → O + Mg + 4.6 MeV
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24Mg
10(13.15)
10
Neon burning reactions
Even if one does not know the reaction cross section for
20
16
Ne + γ → 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
3/2
3/2
!
$
!
$
n i n j gig j mi m j
kT
−Q/kT
##
&& #
=
e
2&
nk
g k " m k % " 2π ! %
(13.18)
We denote by <σv> the i + j à k reaction rate, and by λk the k à i + j the decay
rate of k.
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Neon burning reactions
Since the abundances are in equilibrium, the following relation holds
dn k
= n i n j < συ > −λk n k = 0 leading to
dt
(13.19)
nin j
λk
=
< συ > n k
(13.20)
Combining this with the Saha equation yields
3/2
3/2
g i g j ! m i m j $ ! kT $ −Q/kT
λk
##
&& #
=
e
2&
< σ v > g k " m k % " 2π ! %
(13.21)
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
3/2
g i g j ! µ kT $ −Q/kT
λk
=
e
#
2&
< σ v > g k " 2π ! %
(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.
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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
g ≡ Z = ∑ g ie
−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 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.
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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
16
O + 16 O → 32 S → 31 P + p + 7.7 MeV
→ 28 Si + α + 9.6 MeV
(13.24)
→ 31S + n +1.5 MeV
→ 30 P + p − 2.4 MeV
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.
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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)
(13.25)
leads to the abundance of the nucleus (Z,N) given by
3/2 $
2 '
A
2π!
Y(Z, N) = YpZ YnN g(Z, N)(ρN A ) A−1 A &&
))
2 % m u kT (
3
(A−1)
2
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 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.
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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 Density
(109k)
(gcm-3)
Hydrogen
Helium
Carbon
Neon
Oxygen
Silicon
7x106years
5x105years
600years
1year
6months
1day
0.06
0.2
0.9
1.7
2.3
4.1
5
7x102
2x105
4x106
107
3x107
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¤.
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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.
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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.
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