the interior of the stars: nuclear reactions

The interior of the stars: Nuclear Reactions

Ilídio Lopes

The interior of the stars: Nuclear Reactions

Omega Centauri globular (HST 2009). There are around 10,000 stars in this image. The differentcolours show stars in different stages of their life cycles. Yellow-white Sun-like stars becomeorange as they cool and grow larger, some becoming large red giant stars. Bright and hot stars areblue.

Last Lecture (summary)

4

These basic equations supplemented with the following ”material equations”:

• Equation of state (pressure of a gas as a function of its density and temperature)

• Opacity (how opaque the gas is to the radiation field)

• Core nuclear energy generation rate

For our stars – which are isolated, static, and spherically symmetric – there are four basic equations to describe structure. All physical quantities depend on the distance from the centre of the star alone

1) Equation of hydrostatic equilibrium: at each radius, forces due to pressure differences balance gravity

2) Conservation of mass3) Conservation of energy : at each radius, the change in the energy flux = local

rate of energy release4) Equation of energy transport : relation between the energy flux and the local

gradient of temperature

Equations of stellar structure(reminder)

5

Typical timescales

td is known as the dynamical time.

td =2r3

GM (r)!

"#

$

%&

12

€

tth ~ GMs2

Lsrs

tth is defined as the thermal timescale (or Kelvin-Helmholtz timescale)

Equations of the Stellar Structure

Stellar Energy Sources

Nuclear Reactions:binding energy of the

atomic nucleus

9

The binding energy of the atomic nucleus

The general description of a nuclear reaction is

I(Ai,Zi )+ J(Aj,Z j )↔K(Ak,Zk )+ L(Al,Zl )

Where Ai = the baryon number, or nucleon number (nuclear mass)And Zi = the nuclear chargeThe nucleus of any element is uniquely defined by the two integers Ai and Zi

Recall that in any nuclear reaction the following must be conserved:1. The baryon number – protons, neutrons and their anti-particles2. The lepton number – light particles, electrons, positrons, neutrinos,

and anti-neutrinos. 3. Charge

Note also that the anti-particles have the opposite baryon/lepton number totheir particles.

10

The total mass of a nucleus is known to be less than the mass of the constituent nucleons. Hence there is a decrease in mass if a companion nucleus is formed from nucleons, and from the Einstein mass-energy relation E=mc2 the mass deficit is released as energy.

This difference is known as the binding energy of the compound nucleus. Thus if a nucleus is composed of Z protons and N neutrons, it’s binding energy is

For our purposes, a more significant quantity is the total binding energy per nucleon. And we can then consider this number relative to the hydrogen nucleus

€

Q(Z,N) ≡ Zmp + Nmn −m(Z,N)[ ]c 2

€

Q(Z,N)A


11

The variation of binding energy per nucleon with baryon number A:

• General trend is an increase of Q with atomic mass up to A= 56 (Fe). Then slow monotonic decline

• There is steep rise from H through 2H, 3He, to 4He → fusion of H to He should release larger amount of energy per unit mass than say fusion of He to C.

• Energy may be gained by fusion of light elements to heavier, up to iron.

• Or from fission of heavy nuclei into lighter ones down to iron.

The binding energy of the atomic nucleusBinding energy

Binding Energy Per Nucleon vs. Atomic Number Energetically favorable Energetically unfavorable

Most bound nucleus isiron 56: up to A< 56 fusionreleases energy, but A> 56fusion requires energy input


13

The nuclear burning is able to produce only elements up to iron, since the creation of elements heavier than the “iron peak” is endothermic, and the electrostatic repulsion for charged particle reactions is increasing with nuclear charge. The various peaks reflects the stability of isotopes against further addition of neutrons and protons and are due to the structure of the nuclei, easily explained in the shell model of nuclear physics.


Abundances of nuclei

15

Abundances of nuclei


Nuclear Reactions(Basics)

In one fusion reaction, merging two forms of hydrogen — tritium and deuterium — yields helium plus a neutron and energy.

Energy Generation in Stars

In one fusion reaction, merging two forms of hydrogen — tritium and deuterium — yields helium plus a neutron and energy (example on Earth nuclear reactors).

Energy Production in Stars:Thermonuclear Reactions

Mass of nuclei with several protons and/or neutrons does notexactly equal mass of the constituents - slightly smaller becauseof the binding energy of the nucleus

The main process is hydrogen fusion into helium:

4 protons, total mass =4× 1.0081 = 4.0324 amu

helium nucleus, mass =4.0039 amu

Mass difference: 0.0285 amu = 4.7 x 10-26 g

ΔE = ΔMc2 = 4.3×10-50 erg = 27 MeVOr about 0.7% of the total rest mass

4 1H→ 4He + photons and neutrinos

• Energy release: photons (hν) + neutrinos– Only the photons contribute to the energy generation– Neutrinos are lost to the system– Note that the photon generation can be after the reaction:

specifically in ee+ annihilation• A reaction network is a specific set of nuclear

reactions leading to either energy production and/or nucleosynthesis.– pp hydrogen burning is one such network– CNO hydrogen burning is another– 3α (He burning) is another


Conservation Laws• Electric Charge• Baryon Number

– p = n = 1– e = ν = 0

• Spin

The Units in ε = ε0ρaTn• ε = ergs g-1 s-1 = g cm2 s-2 g-1 s-1 = cm2 s-3

• Assume a = 1 , ρa = ρ : g cm-3

• So cm2 s-3 = ε0 g cm-3 Kn

• ε0 = (cm2 s-3)/(g cm-3 Kn) = (cm2/g) cm3 s-3 K-n

• So how do we get ε0 (see the following references) ?– Bowers & Deming Section 7.3– Clayton


Nuclear Reactions(Fusion)

22

EnergypotentialE(r)

Coulomb barrier At r = r0, height of theCoulomb barrier is:

E = Z1Z2e2

ro~ Z1Z2 MeV

Overcoming the Electrostatic Barrier

Nuclear stronginteractions

Electrostaticrepulsion

+

r0

+ + +

r

Most particles have too lowthermal energies to overcomethe Coulomb barrier!

What makes the fusion possible is the quantum tunneling effect Probability of tunneling increases steeply with particle energy

Occurrence of fusion reactions

Nuclear Reactions (Fusion)

23

Occurrence of fusion reactionsNow will discuss the conditions under which fusion can occur –and whether such conditions exist in stellar interiors

• Nuclei interact through four forces of physics – only electromagnetic and strong nuclear forces are important here.

• Two positively charged nuclei must overcome coulomb barrier (long range force ~1/r2), to reach separation distances where strong force dominates (10-15 m, typical size of nucleus)

Schematic plots

• V (potential energy) vs nuclei separation distance

• Wave function representing penetration of a potential barrier by nucleus with kinetic energy of approach Ekin (below barrier height).

Nuclear Reactions (Fusion)

24€

z1z2e2

4πε0rz1, z2 = number of protons in each nuclei

e= charge on electron=1.6 10-19 C

𝜀0= permittivity of free space = 8.85 10-12 C2N-1 m-2

To fuse nuclei, must surmount the coulomb barrier. Height of barrier is estimated by:

Occurrence of fusion reactionsNuclear Reactions (Fusion)

Nuclear Reaction Rate(Basics)

Cross Section σ– Let X = Stationary nucleus– Let a = Moving nucleus

• Probability of a reaction per unit path length (MFP=Mean Free Path) is nXσ where nX is the density of particle X.

– nXσ = # cm-3 cm2 = # cm-1 (This is 1 / MFP)• MFP = (1 / nXσ) = vt where t = time between collisions and v =

speed of particles (the one(s) that are moving).• t = 1 /nXσv or Number of Reactions (per Stationary particle/s) =

nXσv

• Total number of reactions: r = na nX v σ(v)– r = cm-3 cm-1 cm s-1 = # / (cm3 s)

– The cross section σ must be a function of v.

Nuclear Reaction Rate

27

Integrate over Velocity• There actually exist a range of velocities f(v) so we

must integrate over all velocities:

• The total energy is then rQ/ρ.– r = reaction rate in # / (cm3 s)– Q = energy produced per reaction– ρ = density g cm-3

– r Q/ρ = cm-3 s-1 g-1 cm3 ergs = ergs s-1 g-1 = ε = ε0ρaTn

( ) ( )a X a Xr n n v v f v dv n n vs s= º < >ò


28

The Cross Section• We write the cross section thus:

• S(E) is the “area” of the reaction - set by the de Broglie wavelength of the particle

• S(E) varies slowly with E• Now change variable to E and rewrite <v>

1/ 2( )( )b

ES EE eE

s-

=( )

0

12 3/ 2

0

( ) ( ) ( )

( )

8 ( )E bkT E

v E v E f E dE

assume f e dE to be Maxwellian

v kT S E e dEm

s s

sp

¥

æ ö- -¥ ç ÷- è ø

< >=

æ ö< >= ç ÷è ø

ò

ò


29

Penetration Factor• The penetration factor is defined as:

• m is the reduced mass of the system

12 22

1 21/ 2

42

2

12 2

21 24

2

mc baZ ZE Ee e

eahc

mcb aZ Z

p

p

æ ö -- ç ÷ç ÷è ø º

=

é ùæ öê ú= - ç ÷ê úè øê úë û


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Analytically• S(E) varies slowly with E and its primary contribution is

at E0 ==> want S(E0)

• Next approximate the Peak by a Gaussian and you get:– See Clayton for the approximation formulae.

( )0

1312

0 203/ 2

8 ( ) 4( ) 23

EkTS Ev E kT e

m kTps

p-æ ö< >= ç ÷

è ø


Nuclear Reaction Rate(Gamow peak)

32

Quantum TunnellingAs derived in your Quantum Mechanics courses, there is a finite probability for a particle to penetrate the Coulomb barrier as if “tunnel” existed. Quantum effect discovered by George Gamow(1928) in connection with radioactivity.

Penetration probability (calculated by Gamow) is given as:

€

e−πZ1Z 2e

2

ε 0hv

Hence this increases with v (particle velocity), but we know v will be Maxwellian distribution for ideal gas. Hence fusion probability is product

€

prob( fusion)∝e−πZ1Z 2e

2

ε 0hv e−mv 2

2kT

Nuclear Reaction Rate (Gamow peak)

33

Gamow Peak• <σv> is the combination of

two pieces:– e-E/kT (Maxwellian-

green) decreases with increasing E and represents the decreasing number of particles with increasing E.

– e-b/Sqrt(E) (Penetration-blue) increases with E and represents the increasing probability of a reaction with E [speed].

E

E0

• E0 is the peak energy for <σV> and can be thought of as the most effective energy for that T

Nuclear Reaction Rate (Gamow peak)

34

Schematically this is plotted, and the fusion most likely occurs in the energy window defined as the Gamow Peak.

The Gamow peak is the product of the Maxwellian distribution and tunnelling probability. The area under the Gamow peak determines the reaction rate.

The higher the electric charges of the interacting nuclei, the greater the repulsive force, hence the higher the Ekin and T before reactions occur.Highly charged nuclei are obviously the more massive, so reactions between light elements occur at lower T than reactions between heavy elements.

Gamow PeakNuclear Reaction Rate (Gamow peak)

Nuclear energy:PP Chain

The PP Chain

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Reactions of the PP ChainsReactions of the PP Chains

Reaction Q (MeV)

Average ν Loss (MeV)

S0(KeV barns)

dS/dE(barns)

B τ (years)†

H(p,β+ν)D 1.442 0.263 3.78(10-22) 4.2(10-24) 33.81 7.9(109)

D(p,γ)He3 5.493 2.5(10-4) 7.2(10-6) 37.21 4.4(10-8)

He3(He3,2p)He4 12.859 5.0(103) 122.77 2.4(105)

He3(α,γ)Be7 1.586 4.7(10-1) -2.8(10-4) 122.28 9.7(105)

Be7(e-,ν)Li7 0.861 0.80 3.9(10-1)

Li7(p, α)He4 17.347 1.2(102) 84.73 1.8(10-5)

Be7(p,γ)B8 0.135 4.0(10-2) 102.65 6.6(101)

B8(β+ν)Be8*(α)He4 18.074 7.2 3(10-8)

† Computed for X = Y = 0.5, ρ = 100, T6 = 15

Nuclear Reaction Rates

38

Temperature dependence of Nuclear Reactions

Nuclear energy• If each reaction releases an energy L, the amount of energy released per unit

mass is justbare

re TXXr xiixix 0»÷÷

ø

öççè

æ L=

rep 24 rdrdLr =

• The sum over all reactions gives the nuclear reaction contribution to e in our fifth fundamental equation:

PP chainThe nuclear energy generation rate for the PP chain, including all three branches:

KTT 66 10/=

Near T~1.5x107 K (i.e. the central temperature of the Sun).

W/kg1007.1 46

25

7 TXpp re -´=

355 /10 mkgrr =

Temperature dependence of PP chain

Nuclear Burning Stages and Processes• Main Sequence Energy Generation• pp Hydrogen Burning– pp cycle: H(H,e+νe) D (H,γ) He3 then He3 (He3,2p) He4– Slowest Reaction is: H (H,e+νe) D• The cross section for the above has never been measured. It

also depends on the weak interaction which governs β decay: p → n + e+ + νe

– For pp T ~ 107 K so this is the energy source for the lower main sequence (mid F - 2 M⦿ and later)

– ε = ε0ρT4 for the pp cycle

Nuclear Reaction Rates

Nuclear energy:CNO cycle

42

The CNO CycleAt birth stars contain a small (2%) mix of heavy elements, some of the most abundant of which are carbon, oxygen and nitrogen (CNO). These nuclei may induce a chain of H-burning reactions in which they act as catalysts.

The process is known as the CNO Cycle.

There are alternative names that you may come across :• The CNO bi-cycle• The CNOF cycle• The CN and NO cycles• The CN and NO bi-cycles

In this course we will just refer to it all as the CNO cycle – and discuss the branches, but not specifically label them.

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1 12C + p ➞ 13N+ 𝛾2 13N ➞ 13C + e+ + 𝜈e3 13C + p ➞ 14N + 𝛾4 14N + p ➞ 15O + 𝛾5 15O ➞ 15N + e+ + 𝜈e6 15N + p ➞ 12C + 4He

In the steady state case, the abundances of isotopes must take values such that the isotopes which react more slowly have higher abundance. The slowest reaction is p capture by 14N. Hence most of 12C is converted to 14N.

CNO Cycle(main CNO-branch)

W/kg1024.8 9.1965

27 TXXCNOCNO re -´=

There is a second, independent cycle in which carbon, nitrogen and oxygen act as catalysts. The main branch (accounting for 99.6% of CNO reactions) is:

44

The two processes have very different temperature dependences. The rate of energy production in each (from D. Arnett – “Supernovae and nucleosynthesis”, Princeton University Press):

€

εPP = ε0ρXH2 TT0

$

% &

'

( )

4.6

€

εCNO = ε0ρXHXCNO fNT

25 ×106%

& '

(

) * 16.7

Equating this two gives the T at which they produce the same rate of energy production:

€

T ≈1.7 ×107 XH

50XCN

$

% &

'

( )

112.1

K

Below this temperature the PP chain is most important, and above it the CNO Cycle dominates. This occurs in stars slightly more massive than the sun e.g. 1.2-1.5M¤.

Temperature dependence (PP chain and CNO Cycle)

Nuclear energy:Triple alpha reaction

Helium collisions

• Recall that the temperature at which quantum tunneling becomes possible is:

( ) 2

422

21

20 34

41

kheZZT µ

pe=

• As hydrogen is converted into helium, the mean molecular weight increases. • To keep the star in approximate pressure equilibrium, the density and

temperature of the core must rise

HmkTP

µr

=

As H burning progresses, the temperature increases and eventually He burning becomes possible

K 109.1 22

21

7 ZZmH ÷

÷ø

öççè

æ´=

µ

Helium Burning: the triple-a reaction

47

Helium Burning: the triple-a reactionSimplest reaction in a helium gas should be the fusion of two helium nuclei. There is no stable configuration with A=8. For example the beryllium isotope 8Be has a lifetime of only 2.6 10-16 s4He + 4He ➞ 8Be

But a third helium nucleus can be added to 8Be before decay, forming 12C by the “triple-alpha” reaction4He + 4He ➞ 8Be8Be + 4He ➞ 12C +𝛾

48

Fred Hoyle (1952-54) suggested this small probability of 𝛼-capture by short lived 8Be would be greatly enhanced if the C nucleus had an energy level close to the combined energies of the reacting 8Be and 4He nuclei. The reaction would be a faster “resonant” reaction.

This resonant energy level of 12C was not experimentally known at the time. Hoyle’s prediction led to nuclear experiment at Caltech, and resonant level discovered.

Thus helium burning proceeds in a 2-stage reaction, and energy released is

€

Q3α = 3ΔM(4He) −ΔM(12C)[ ]c 2 = 7.275MeVIn terms of energy generated per unit mass ~ 5.8 1013 JKg-1 (I.e. 1/10 of energy generated by H-burning). But the T dependence is astounding:

€

ε3α ∝ρ2T 40


Triple-alpha process The burning of helium occurs via the triple alpha process:

g+®+

Û+

CHeBeBeHeHe

126

42

84

84

42

42

The intermediate product 8-Beryllium is very unstable, and will decayif not immediately struck by another Helium. Thus, this is almost abody interaction W/kg1085.3 0.41

832

58

3 TYre a-´=

Note the very strong temperature dependence. A 10% increase inincreases the energy generation by a factor 50.


As a side process, some carbon nuclei fuse with additional helium toproduce a stable isotope of oxygen and energy:

50

Temperature dependence (PP chain, CNO Cycle & Triple-Alpha)

ε ~ TηT

Nuclear energy:Fusion of heavy elements

52

Carbon and Oxygen burningCarbon burning (fusion of 2 C nuclei) requires temperatures above 5 ✕108 K, and oxygen burning in excess of 109 K.

Interactions of C and O nuclei are negligible – as at the intermediate temperatures required by the coulomb barrier the C nuclei are quickly destroyed by interacting with themselves

12C + 12C ➞ 24Mg + 𝛾➞ 23Mg + n ➞ 23Na + p➞ 20Ne + 𝛼➞ 16O + 2𝛼

16O + 16O ➞ 32S + 𝛾➞ 31S + n ➞ 31P + p➞ 28Si + 𝛼➞ 24Mg + 2𝛼

The branching ratios for these reactions are temperature dependent probabilities.12C + 12C ➞ ~13MeV (~5.2 1013 JKg-1)16O + 16O ➞ ~16MeV (~4.8 1013 JKg-1)These reactions produce p, n, 𝛼, which are immediately captured by heavy nuclei, thus many isotopes created by secondary reactions.

53

Silicon burning: nuclear statistical equilibrium

But now very high Coulomb barrier, at T above O burning, but below that required for Si burning, photodisintegration takes place

Si disintegration occurs around (2.7–3.5 109K) ~ 3109 K, and the light particles (proton and alpha) emitted are recaptured by other Si nuclei.

Two Si nuclei could fuse to create 56Fe – the end of the fusion chain.

16O + 𝛾⟷ 20Ne + 𝛾

Although the reactions tend to a state of equilibrium, a leakage occurs towards the stable iron group nuclei (Fe, Ni, . . .), which resist photodisintegration up to 7 109 K.

photodisintegration rearrangement,creates new elements by addingone of these freed 4He. per capturestep in the following sequence(photo-ejection of alphas notshown).

This produces Ne at T~109K but reverses above 1.5 109 K.

54

Major nuclear burning processesCommon feature is release of energy by consumption of nuclear fuel. Rates of energy release vary enormously. Nuclear processes can also absorb energy from radiation field, we shall see consequences can by catastrophic.

Nuclear Fuel

Process Tthreshold106K

Products Energy per nucleon (Mev)

H PP ~4 He 6.55

H CNO 15 He 6.25

He 3a 100 C,O 0.61

C C+C 600 O,Ne,Ma,Mg 0.54

O O+O 1000 Mg,S,P,Si ~0.3

Si Nuc eq. 3000 Co,Fe,Ni <0.18

Neutron-Capture Nucleosynthesis

(Extra)

56

The s-process and r-processThe necessary condition for the s- and r-process is the presence of neutrons.Interaction between nuclei and free neutrons (neutron capture) – the neutrons are produced during C, O and Si burning. Neutrons capture by heavy nuclei is not limited by the Coulomb barrier – so could proceed at relatively low temperatures. The obstacle is the scarcity of free neutrons. If enough neutrons available, chain of reactions possible:

I(A, Z) + n ➞ I1(A+1, Z) I1(A+1, Z) + n ➞ I2(A+2, Z) I2(A+2, Z) + n ➞ I3(A+3, Z) …etc

If a radioactive isotope is formed it will undergo 𝛽-decay, creating new element.

IN(A+N, Z) ➞ J(A+N, Z+1) + e- + anti-𝝼If new element stable, it will resume neutron capture, otherwise a undergo series of 𝛽-decaysJ(A+N, Z+1) ➞ K(A+N, Z+2) + e- +anti-𝝼K(A+N, Z+2) ➞ L(A+N, Z+3) + e- +anti-𝝼

–

The binding energy of the atomic nucleusThe rapid neutron capture process or r-process responsible for the creation (nucleosynthesis) of approximately half the atomic nuclei heavier than iron. The captures must be rapid in the sense that the nucleus does not have time to undergo radioactive decay before another neutron arrives to be captured.The slow neutron capture process or s-process is a series of reactions in nuclear astrophysics which occur in stars, particularly AGB stars. The s-process is responsible for the creation (nucleosynthesis) of approximately half the atomic nuclei heavier than iron.

The s-process is secondary,meaning that it requires pre-existing heavy isotopes as seednuclei to be converted into otherheavy nuclei.

Taken together, the r- and s-processes account for themajority of abundanceevolution of elements heavierthan iron.

58

In the process two types of reactions and two types of nuclei are involved:Neutron captures and 𝛽-decays ; stable and unstable nucleiStable nuclei may undergo only neutron captures, unstable ones my undergo both, with the outcome depending on the timescales for the two processes.What can we say about the timescales of these processes ?

Hence neutron capture reactions may proceed more slowly or more rapidlythan the competing 𝛽-decays.

The different chains of reactions and products are called the slow/rapid neutron-capture process (s-process/r-process).

The s-process and r-process

59

Some basic examples

The s-process and r-process

The astrophysical site for the r-process is not clearly identified, but is probably to be found in supernova explosions or similar energetic events.

The s-process is certainly taking place in stars of intermediate mass (2-5 M sun stars) in an advanced phase ofevolution

Common nuclear processes in astrophysics energy of the atomic nucleus

Recommended References:

• “Section 18.3 Thermonuclear Reaction Rates”, page:182 – 187, from ”Stellar Structure and Evolution” R. Kippenhahn et al. (extra)

• “Section 18.5 The Major Nuclear Burning Stages”, page:192 – 202, from ”Stellar Structure and Evolution” R. Kippenhahn et al. (extra)

• “Principles of Stellar Evolution and Nucleosynthesis”, Donald D. Clayton, McGraw-Hill, 1968 (and Chicago: University of Chicago Press, 1983), 612 pages, ISBN-10: 0070112959, ISBN-13: 978-0070112957.

• ”Supernovae and Nucleosynthesis (Princeton Series in Astrophysics)”, David Arnett, 1996, Published by Princeton University Press, ISBN 10: 0691011478 ISBN 13: 9780691011479

Summary

63

Summary and conclusion

from "Thermonuclear Processes in Stars and Stellar Neutrinos" Georg Wolschin

Recalling that we wanted to determine the physics governing 𝜀 (energy) will be defined by the nuclear energy source in the interiors:

• We have covered the basic principles of energy production by fusion• The PP chain and CNO cycle have been described, • He burning by the triple-alpha reaction was introduced• Later burning stages of the heavier elements (C,O, Si) discussed• The r- and s-processes – origin of the elements heavier than Fe• Further reading provided from the more advanced, like, D. Arnett – “Supernovae and nucleosynthesis”, Princeton University Press

64

Summary and conclusionNuclear Reactions and Nucleosynthesis

65

Summary and conclusionNuclear Reactions and Nucleosynthesis

Periodic table showing the cosmogenic origin of each element. Elements from carbon up tosulfur may be made in small stars by the alpha process. Elements beyond iron are made in largestars with slow neutron capture (s-process), followed by expulsion to space in gas ejections(see planetary nebulae). Elements heavier than iron may be made in neutron star mergers orsupernovae after the r-process, involving a dense burst of neutrons and rapid capture by theelement.

https://en.wikipedia.org/wiki/Alpha_process

https://en.wikipedia.org/wiki/S-process

https://en.wikipedia.org/wiki/Planetary_nebulae

https://en.wikipedia.org/wiki/R-process

Week Problems

Problem . . .

Ilídio Lopes 66

Thank you for your attention!

Any Questions?

the interior of the stars: nuclear reactions

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