energy sources in stars (§10.3)

What makes a star shine?What makes a star shine?

Energy Sources in Stars (§10.3)Energy Sources in Stars (§10.3)Energy Sources in Stars (§10.3)Energy Sources in Stars (§10.3)

Energy Sources in Stars (§10.3)Energy Sources in Stars (§10.3)Energy Sources in Stars (§10.3)Energy Sources in Stars (§10.3)

Sun’s Energy OutputSun’s Energy OutputLLsunsun = 4 x 10 = 4 x 102626 W (J s W (J s-1-1))ttsunsun 4.5 x 10 4.5 x 1099 yr (1.4 x 10 yr (1.4 x 101717 s) - oldest rocks - s) - oldest rocks - radioactive datingradioactive dating

LLsunsun ~ constant over geological timescale (fossil evidence) ~ constant over geological timescale (fossil evidence)

Therefore total energy output ETherefore total energy output Etottot = L = Lsunsun x t x tsun sun = 6 x 10= 6 x 104343 J J

How might the Sun have generated this energy?How might the Sun have generated this energy?

What makes a star shine?What makes a star shine?

Energy Sources in StarsEnergy Sources in StarsEnergy Sources in StarsEnergy Sources in Stars

Energy Source #1:Gravitational contractionEnergy Source #1:Gravitational contraction We believe Sun collapsed from a large gas We believe Sun collapsed from a large gas

cloud (R ~ ∞) to its present size cloud (R ~ ∞) to its present size

No change in Mass, M

Rclou

d

R

Potential energy released Potential energy released is: is:

EEgravgrav = -(E = -(Enownow - E - Einitialinitial))P.E. is obtained from P.E. is obtained from

integrating grav. integrating grav. potential over all points potential over all points in the spherein the sphere

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P.E. = − G0

Rsun∫43

πρr3( ) 4πr2ρdr( )

r

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= − 1615( )π 2ρ 2GR5 for constant ρ

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With M = 43

πR3ρ, we get

P.E. = −3GMsun

2

5Rsun


However, from Virial Theorem (EHowever, from Virial Theorem (Ethth = -1/2 E = -1/2 Egravgrav), only 1/2 ), only 1/2 of the gravitational energy is available for energy of the gravitational energy is available for energy release, the other half heats the star.release, the other half heats the star.

So, So, EEgravgrav = -(E = -(Enownow -E -Einitialinitial)) = 3/10(GM= 3/10(GMsunsun

22)(1/R)(1/Rsunsun - 1/ - 1/) ) ~ 10~ 104141 J J ~ 1/600 E~ 1/600 Etottot needed over Sun’s lifetime needed over Sun’s lifetimeBy this process, Sun could only radiate for By this process, Sun could only radiate for

t = t = EEgravgrav/L/Lsun sun ~ 10~ 1077 years yearst = t = EEgravgrav/L/Lsun sun is Kelvin-Helmholtz timescale is Kelvin-Helmholtz timescale Conclusion: GravitationalConclusion: Gravitational Contraction not the main energy Contraction not the main energy

source in Sunsource in Sun

Energy Source #1:Gravitational contractionEnergy Source #1:Gravitational contraction

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P.E. = −3Gmsun

2

5Rsun


e.g. How much energy would be released if Sun was e.g. How much energy would be released if Sun was completely ionized and then all gas recombined completely ionized and then all gas recombined into neutral atoms? (binding energy of the H into neutral atoms? (binding energy of the H atom 13.6 ev: 1 ev = 1.6 x 10atom 13.6 ev: 1 ev = 1.6 x 10-19-19 J J 13.6 ev ~ 2 13.6 ev ~ 2 x 10x 10-18-18 J) J)

Assume Sun is 100% H (X = 1) Assume Sun is 100% H (X = 1) Total # H atoms NTotal # H atoms NHH = M = Msunsun/m/mHH ~ 10 ~ 105757

EEtottot ionization ~ 10 ionization ~ 105757 x (2 x 10 x (2 x 10-18-18) J ~ 2 x 10) J ~ 2 x 103939

JJThis is only ~ 1/30,000 EThis is only ~ 1/30,000 Etotaltotal of Sun’s energy output of Sun’s energy output

Sun’s energy output ~ burning 7000 kg coal each hour Sun’s energy output ~ burning 7000 kg coal each hour on every sq. metre Sun’s surface!on every sq. metre Sun’s surface!

Conclusion: Chemical Reactions cannot be source Conclusion: Chemical Reactions cannot be source Sun’s energySun’s energy

Energy Source #2:Chemical ReactionsEnergy Source #2:Chemical Reactions


Fission? Fission? Z = 0.02 (2% Sun consists of heavy elements). Z = 0.02 (2% Sun consists of heavy elements). Fissionable isotopes like Fissionable isotopes like 235235U are rare (isotopes U are rare (isotopes

are nucleii with the same # protons but different # are nucleii with the same # protons but different # neutrons)neutrons)

Fusion? Fusion? e.g. 4 protons get converted into 2p + 2n i.e. 4H e.g. 4 protons get converted into 2p + 2n i.e. 4H

1 1 44He He H is very abundant in the Sun (X = 0.73)H is very abundant in the Sun (X = 0.73)

4 H nuclei: mass = 4 x m4 H nuclei: mass = 4 x mpp = 6.693 x 10 = 6.693 x 10-27-27 kg kg1 1 44He nucleus: mass = 6.645 x 10He nucleus: mass = 6.645 x 10-27-27 kg (binding energy kg (binding energy included)included)

Difference is 0.048 x 10Difference is 0.048 x 10-27-27 kg = 0.7% of original mass kg = 0.7% of original mass

Where does this mass go?Where does this mass go?

Energy Source #3:Nuclear EnergyEnergy Source #3:Nuclear Energy


Example:Example:Suppose that Sun was originally 100% H and only 10% of that Suppose that Sun was originally 100% H and only 10% of that was available for fusion. Thus…was available for fusion. Thus…

MMHH available = 0.1 M available = 0.1 Msunsun

MMHH destroyed through fusion = 0.1 M destroyed through fusion = 0.1 Msunsun x 0.007 x 0.007

total nuclear energy available Etotal nuclear energy available Etottot = (7 x 10 = (7 x 10-4-4 M Msunsun)c)c2 2 ~ ~

1.3 x 101.3 x 104444 J J Sun’s total lifetime energy output = 6 x 10Sun’s total lifetime energy output = 6 x 104343 J J EEtottot ~ 2 x the total energy Sun has emitted!! ~ 2 x the total energy Sun has emitted!!

Sun could shine for at least another 5 x 10Sun could shine for at least another 5 x 1099 yr at its yr at its current luminositycurrent luminosity

Hence tHence tnuclearnuclear for Sun ~ 10 for Sun ~ 101010 yrs yrs

Energy Source #3:Nuclear EnergyEnergy Source #3:Nuclear EnergyThe “lost” mass is converted into energy according to Einstein’s equation for the rest energy of The “lost” mass is converted into energy according to Einstein’s equation for the rest energy of

matter: E = mcmatter: E = mc22 (E energy, m mass, c speed light) (E energy, m mass, c speed light)


Fission? Fission? Z = 0.02 (2% Sun consists of heavy elements). Z = 0.02 (2% Sun consists of heavy elements). Fissionable isotopes like Fissionable isotopes like 235235U are rare (isotopes U are rare (isotopes

are nucleii with the same # protons but different # are nucleii with the same # protons but different # neutrons)neutrons)

Fusion? Fusion? e.g. 4 protons get converted into 2p + 2n i.e. 4H e.g. 4 protons get converted into 2p + 2n i.e. 4H

1 1 44He He H is very abundant in the Sun (X = 0.73)H is very abundant in the Sun (X = 0.73)

4 H nuclei: mass = 4 x m4 H nuclei: mass = 4 x mpp = 6.693 x 10 = 6.693 x 10-27-27 kg kg1 1 44He nucleus: mass = 6.645 x 10He nucleus: mass = 6.645 x 10-27-27 kg (binding energy kg (binding energy included)included)

Difference is 0.048 x 10Difference is 0.048 x 10-27-27 kg = 0.7% of original mass kg = 0.7% of original mass

Where does this mass go?Where does this mass go?

Energy Source #3:Nuclear EnergyEnergy Source #3:Nuclear Energy

Can Fusion Occur in Sun’s Core?Can Fusion Occur in Sun’s Core? Can Fusion Occur in Sun’s Core?Can Fusion Occur in Sun’s Core?

Classical Approach:Classical Approach: EEkinkin proton > E proton > Ecoulombcoulomb

1/2 m1/2 mppvv2 2 > e> e22/r (cgs form)/r (cgs form) LHS is thermal gas LHS is thermal gas energy, thus, 3/2 kT energy, thus, 3/2 kT >e>e22/r/r

At TAt Tcentralcentral ~ 1.6 x 10 ~ 1.6 x 107 7 K K EEkinkin ~ 3.3 x 10 ~ 3.3 x 10-16 -16 JJ

For successful fusion, r For successful fusion, r ~ ~ 1010-15-15 m (1 fm) m (1 fm)

Protons must overcome mutual electrostatic Protons must overcome mutual electrostatic repulsion: Coulomb Barrierrepulsion: Coulomb Barrier

EEcoulombcoulomb ~ 2.3 x 10 ~ 2.3 x 10-13-13 J J EEkinkin ~ 10 ~ 10-3-3 E Ecoulombcoulomb Classically, would need T~10Classically, would need T~101010 to overcome to overcome Coulomb barrierCoulomb barrierSo, no Fusion??So, no Fusion??


Can we be helped by considering the distribution of Can we be helped by considering the distribution of energies? - not all protons will have just Eenergies? - not all protons will have just Ethth = 3/2 kT = 3/2 kT

Proton velocities are distributed according to the Proton velocities are distributed according to the Maxwellian equation: P(v) = 4Maxwellian equation: P(v) = 4(m/2(m/2kT)kT)3/23/2 v v22 e e-mv-mv22/2kT/2kT

At the high energy tail of the Maxwellian distribution, At the high energy tail of the Maxwellian distribution, the relative number of protons, with E > 10the relative number of protons, with E > 1033 x E x Ethth is: is:

N(EN(Ecoulombcoulomb ~ 10 ~ 1033 <E <Ethermalthermal>)/N(<E>)/N(<Ethermalthermal>) ~ e>) ~ e--E/ktE/kt ~ e ~ e-1000-1000 = 10 = 10-430-430!!!!

Number protons in Sun = MNumber protons in Sun = Msunsun/M/MHH ~ 10 ~ 1057 57 so 1 in so 1 in 1010430430 of these will have enough energy to of these will have enough energy to overcome Coulomb barrier.overcome Coulomb barrier.

So again,So again, no nuclear reactions??no nuclear reactions??


Called Quantum TunnellingCalled Quantum Tunnelling Particles have wavelength, (Particles have wavelength, ( = h/p ), = h/p ),

associated with them (like photons) called associated with them (like photons) called de Broglie wavelength de Broglie wavelength

Proved in many experiments - for example Proved in many experiments - for example diffraction electrons (wave phenomenon)diffraction electrons (wave phenomenon)

Eg. Eg. free electron (3 x 10 free electron (3 x 1066 m/s) = 0.242 nm m/s) = 0.242 nm (size atom) (size atom)

person (70 k gm) jogging at 3 m/s = 3 x person (70 k gm) jogging at 3 m/s = 3 x 1010-16-16 m (negligible - person won’t diffract!) m (negligible - person won’t diffract!)

Increased wavelength reflects loss momentum.Increased wavelength reflects loss momentum.

Quantum Approach:Quantum Approach: Heisenberg Uncertainty Principle - Heisenberg Uncertainty Principle - ppx > h/2x > h/2 If If p small, then p small, then x may be large enough that x may be large enough that protons have non-negligible probability of being protons have non-negligible probability of being located within 1 fm of another proton inside Coulomb located within 1 fm of another proton inside Coulomb barrierbarrier

Classically, the proton is reflected at RClassically, the proton is reflected at R00. .

Quantum mechanically, there exists a finite Quantum mechanically, there exists a finite probability for a proton to reach interaction probability for a proton to reach interaction radius, R.radius, R.

Can Fusion Occur in Sun’s Core?Can Fusion Occur in Sun’s Core? Can Fusion Occur in Sun’s Core?Can Fusion Occur in Sun’s Core?Even with tunnelling, are stars hot enough for nuclear reactions to Even with tunnelling, are stars hot enough for nuclear reactions to

proceed? proceed?

= h/p is the wavelength associated with a massive particle (p = = h/p is the wavelength associated with a massive particle (p = momentum)momentum)

In terms momentum, the kinetic energy of a proton is:In terms momentum, the kinetic energy of a proton is:1/2 m1/2 mppvv22 = p = p22/2m/2mpp

Assuming proton must be within 1 de Broglie Assuming proton must be within 1 de Broglie of its target to tunnel of its target to tunnel set distance, r, of closest approach to set distance, r, of closest approach to , (where barrier height = , (where barrier height =

original K.E. original K.E. incoming particle) gives:incoming particle) gives:

ee22/r = e/r = e22// = p = p22/2m/2mpp = (h/ = (h/))22/2m/2mpp

Solving for Solving for (=h (=h22/2m/2mppee22) and substituting r = ) and substituting r = into 3/2 kT = e into 3/2 kT = e22/r, we /r, we get the QM estimate of the temperature required for a nuclear get the QM estimate of the temperature required for a nuclear reaction to occur:reaction to occur:TTQMQM = 4/3(e = 4/3(e44mmpp/kh/kh22) - putting in numbers) - putting in numbers

TTQMQM = 10 = 1077 K which is comparable to T K which is comparable to Tcorecore. Therefore, fusion is . Therefore, fusion is feasible at centre of Sunfeasible at centre of Sun

Proton-Proton ChainProton-Proton Chain Proton-Proton ChainProton-Proton Chain

Conservation Laws in nuclear reactions:Conservation Laws in nuclear reactions:(1) mass-energy(1) mass-energy(2) charge(2) charge(3) difference between number particles and anti-(3) difference between number particles and anti-

particles conserved - ie particle cannot be created from particles conserved - ie particle cannot be created from anti-particle or vice-versa but a pair can be formed or anti-particle or vice-versa but a pair can be formed or destroyed without violating this ruledestroyed without violating this rule

ParticleParticle SymbolSymbol Rest mass Rest mass (kg)(kg)

Charge Charge (e-)(e-)

SpinSpin

photonphoton 0 0 0 0 11

neutrinoneutrino >0>0 0 0 1/21/2

anti-anti-neutrinoneutrino

-- >0>0 00 1/21/2

electronelectron e-e- 9 x 109 x 10-31-31 -1-1 1/21/2

positronpositron e+e+ 9 x 109 x 10-31-31 +1+1 1/21/2

proton proton ((11

11H)H)p+p+ 1.6 x 101.6 x 10--

2727

+1+1 1/21/2

neutronneutron nn 1.6 x 101.6 x 10--

2727

00 1/21/2

deuterondeuteron 2211HH 3.2 x 103.2 x 10--

2727

+1+1

Basic particles involved in Basic particles involved in nuclear reactionsnuclear reactions


Types of nuclear reactions Types of nuclear reactions

Beta Decay: n Beta Decay: n p p++ + e + e-- + + -- proceeds proceeds spontaneously - also for neutron inside spontaneously - also for neutron inside nucleusnucleuseg [Z-1, A] eg [Z-1, A] [Z, A] + e- + -- (Z = # p (Z = # p++, N = # n, , N = # n, A = Z+N)A = Z+N)

Inverse Beta Decay: pInverse Beta Decay: p++ + e + e-- n + eg [[1313N N 13C + e+ + ]

(p+, ) process: ) process: AAZ + pZ + p++ A+1(Z+1) + eg [12C + p+ 13N + ]

(, ) process: particle (4He) added to nucleus to make heavier particle [AZ + 4He A+4(Z+2) + ] eg. [8Be + 4He 12C + ]


Since the Sun’s mass consists mostly of H and He, Since the Sun’s mass consists mostly of H and He, we anticipate nuclear reactions involving we anticipate nuclear reactions involving these two elementsthese two elementseg eg pp++ + p + p++ 22He He pp++ + + 44He He 55LiLi 44He + He + 44He He 88BeBe

But there is a problem here - what is it?But there is a problem here - what is it?All these reactions produce unstable particlesAll these reactions produce unstable particles

eg peg p++ + p + p++ 22He He p p++ + p + p++

pp++ + + 44He He 55Li Li p p++ + + 44HeHe 44He + He + 44He He 88Be Be 44He + He + 44HeHe

So among light elements there are no two-particle So among light elements there are no two-particle exothermicexothermic reactions producing stable reactions producing stable particles. particles. We have toWe have to look to more peculiar look to more peculiar reactions or those involving rarer particlesreactions or those involving rarer particles

The proton-proton chainPPI Branch

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Repeat steps 1 & 2 to produce 2 23He nuclei

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STEP 3 : 23He + 2

3He → 24He + p+ + p+

NET RESULT

Total of 6

protons

1 He nucleus + 2 protons+ 2e+

+ neutrinos + energy

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STEP 1 : p+ + p+→12H + e+ + υ

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proton → neutron via "inverse Beta decay"

p+ → n + e+ + υ e

Deuterium(p+ + n)

positron neutrinoSloweststep

Weak nuclear force

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STEP 2 : p+ + 12H→2

3He + γLow-mass isotope

of He(2p+ + n)

Gamma ray photon

me+ ~ 5 x 10-4

mp+

Very low massme 10-5 mp+

Proton-Proton ChainProton-Proton ChainProton-Proton ChainProton-Proton Chain

Using Using considerations of: considerations of:

minimizing Coulomb minimizing Coulomb barriers, barriers,

crossections, crossections, and making sure and making sure conservation laws conservation laws are obeyed are obeyed

the nuclear the nuclear reaction chain at reaction chain at right produces the right produces the energy observed in energy observed in the Sun.the Sun.

4


Summary of proton-proton chain:Summary of proton-proton chain:• 6p6p++ 44He + 2pHe + 2p++ + 2e + 2e++ + 2 + 2 + 2 + 2 or or 4p4p++ 44He + 2e+ + 2He + 2e+ + 2 + + 22

• Mass difference between 4pMass difference between 4p++ and 1 and 1 44He = 26.7 Mev x 6.424 He = 26.7 Mev x 6.424 x 10x 101818 ev/J = 4.2 x 10 ev/J = 4.2 x 10-12-12 J J

• ~3% of this energy (0.8 Mev) is carried off by neutrinos ~3% of this energy (0.8 Mev) is carried off by neutrinos and does not contribute to the Sun’s luminosityand does not contribute to the Sun’s luminosity

• 2 e2 e++ immediately annihilate with 2 e immediately annihilate with 2 e-- and add 2 x 0.511 and add 2 x 0.511 Mev (= 1.02 Mev)Mev (= 1.02 Mev)

• So, the total energy available for the Sun’s luminosity So, the total energy available for the Sun’s luminosity per per 44He formed is: (26.7 - 0.8 +1.02) Mev = 26.9 Mev He formed is: (26.7 - 0.8 +1.02) Mev = 26.9 Mev = 4.2 x 10= 4.2 x 10-12-12 J J

• # # 44He formed/sec = LHe formed/sec = Lsunsun/4.2 x 10/4.2 x 10-12-12J = 3.9 x10J = 3.9 x1026 26 J/s J/s // 4.2x104.2x10-12-12 J = 9.3 x 10 J = 9.3 x 103737 44He/sHe/s

• Increase of Increase of 44He mass/time = dmHe mass/time = dmHeHe/dt = 9.3 x 10/dt = 9.3 x 103737 44He/s x He/s x 6.68 x 106.68 x 10-27-27 kg/ kg/44He = 6.2 x 10He = 6.2 x 101111 kg kg 44He /sHe /s

After 10After 101010 years (3 x 10 years (3 x 101717 sec), M( sec), M(44He) = 1.9 x 10He) = 1.9 x 102929 kg kg = 10% mass Sun= 10% mass Sun


The previous The previous nuclear reaction nuclear reaction chain (PPI) is not chain (PPI) is not the only way to the only way to convert H into He. convert H into He.

eg last step: eg last step: 33He + He + 33He He 44He + He + 2p2p++ can proceed can proceed differently if differently if there is there is appreciable appreciable 44He He presentpresent

A second branch (PPII)

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In solar centre, 69% of 23He combines with

another 23He, but 31% can fuse with 2

4He

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23He + 2

4He → 47Be + γ

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47Be + e− → 3

7Li + υ e

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37Li + p+ → 2

4He + 24He

PPI + PPII + PPIII operate simultaneously

A third branch (PPIII)

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In the Sun's core, a 47Be nucleus can capture

a p+ instead of an e− about 0.3% of the time.

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48Be → 2

4He + 24He

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47Be + p+ → 5

8B + γ

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58B → 4

8Be + e+ + υ eVery unstable

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p+ + p+ → 12H + e+ + υ e

12H + p+ → 2

3He + γ

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23He + 2

3He → 24He + 2p+

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23He + 2

4He → 47Be + γ

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47Be + e− → 3

7Li + υ e

37Li + p+ → 22

4He

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47Be + p+ → 5

8B + γ

58B → 4

8Be + e+ + υ e

48Be → 22

4He

(PPI)

(PPII) (PPIII

)

69% 31%

99.7% 0.3%

PPIIPPI PPIII1.0

0.5

00 10 20 30 40

Fraction of He4 produced

T (106)K)

Proton-Proton Chain: SummaryProton-Proton Chain: SummaryProton-Proton Chain: SummaryProton-Proton Chain: Summary

Alternate Fusion Reactions H Alternate Fusion Reactions H 44HeHeAlternate Fusion Reactions H Alternate Fusion Reactions H 44HeHe

• First step in PP chain has First step in PP chain has very low reaction rate (weak very low reaction rate (weak interaction).interaction).

• 1212C can act as a catalyst in C can act as a catalyst in fusion reaction. fusion reaction.

• Net result: Net result: 1212C + 4pC + 4p++ 1212C + C + 44He + 2eHe + 2e++ + + 22 + 3 + 3

• Note: Note: 1212C neither created nor C neither created nor destroyed. Also isotopes of destroyed. Also isotopes of N & O are temporarily N & O are temporarily produced.produced.

• CNO cycle dominates over PP CNO cycle dominates over PP chain if Tchain if Tcore core > 1.8 x 10> 1.8 x 1077

(slightly hotter than Sun)(slightly hotter than Sun)

106 yr

14 min

3 x 105 yr

3 x 108 yr

82 s

104 yr

X as frequently

Once for every 2500 instances of Cycle 1

THE CNO BI-CYCLE

(T < 108 K)

CNO Cycle (or bi-cycle)CNO Cycle (or bi-cycle)

Other Fusion ReactionsOther Fusion ReactionsOther Fusion ReactionsOther Fusion Reactions

p. 313p. 313~10~1099 K KMg, Si, S, Mg, Si, S,

etc.etc.1616OO

p. 313p. 3136 x 106 x 1088 K KO, Ne, Na, O, Ne, Na,

MgMg1212CC

p. p. 312312

~10~1088 K KBe, CBe, C44HeHe

triple triple process process

Ref.Ref.at Tat TcentralcentralintointoFusion ofFusion of

At higher TAt higher Tcentrecentre, heavier nuclei can fuse , heavier nuclei can fuse to produce energyto produce energy

We believe Universe began with only H, He, We believe Universe began with only H, He, (Li, Be)(Li, Be)

All other elements created in core of stars All other elements created in core of stars (stellar nucleosynthesis)(stellar nucleosynthesis)

We are all made of We are all made of “STARSTUFF”“STARSTUFF”

energy sources in stars (§10.3)

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