Energy Sources in Stars (§10.3)

Energy Sources in Stars (§10.3)What makes a star shine?

Energy Sources in Stars (§10.3)What makes a star shine?Sun’s s Energy OutputL sun = 4 x 10 26 W (J s -1 )sun 4.5 x 10 9 yr (1.4 x 10 17 s) - oldest rocks - radioactive datingt sunL sun ~ constant over geological timescale (fossil evidence)Therefore total energy output E tot = L sun x t sun = 6 x 10 43How might the Sun have generated this energy?43 J

Energy Sources in StarsEnergy Source #3:Nuclear EnergyFission?Z = 0.02 (2% Sun consists of heavy elements).Fissionable isotopes like 235 U are rare (isotopes are nucleiiwith the same # protons but different # neutrons)Fusion?e.g. 4 protons get converted into 2p + 2n i.e. i4H → 1 4 HeH is very abundant in the Sun (X = 0.73)4 H nuclei: mass = 4 x m p = 6.693 x 10 -27 kg1 4 He nucleus: mass = 6.645 x 10 -27 kg (binding energy included)Difference is 0.048 x 10 -27 kg = 0.7% of original massWhere does this mass go?

Energy Sources in StarsEnergy Source #3:Nuclear EnergyThe “lost” mass is converted into energy according to Einstein’sequation for the rest energy of matter: E = mc 2 (E energy, mmass, c speed light)Example:Suppose that Sun was originally 100% H and only 10% of that wasavailable for fusion. Thus…M H available = 0.1 M sunΔM H destroyed through fusion = 0.1 M sun x 0.007total nuclear energy available E tot = (7 x 10 -4 M sun )c 2 ~ 1.3 x 10 44Sun’s s total lifetime energy output = 6 x 10 43 JE tot ~ 2 x the total energy Sun has emitted!!→ Sun could shine for at least another 5 x 10 9 yr at its current luminosityHence t nuclear for Sun ~ 10 10 yrs44 J

Energy Sources in StarsEnergy Source #3:Nuclear EnergyFission?Z = 0.02 (2% Sun consists of heavy elements).Fissionable isotopes like 235 U are rare (isotopes are nucleiiwith the same # protons but different # neutrons)Fusion?e.g. 4 protons get converted into 2p + 2n i.e. i4H → 1 4 HeH is very abundant in the Sun (X = 0.73)4 H nuclei: mass = 4 x m p = 6.693 x 10 -27 kg1 4 He nucleus: mass = 6.645 x 10 -27 kg (binding energy included)Difference is 0.048 x 10 -27 kg = 0.7% of original massWhere does this mass go?

Can Fusion Occur in Sun’s s Core?Protons must overcome mutual electrostatic repulsion: CoulombBarrierClassical Approach: E kin proton > E coulomb 1/2 m p v 2 > e 2 /r (cgs(form) LHS is thermal gas energy,thus, 3/2 kT >e 2 /r At T central ~ 1.6 x 10 7 K → E kin~ 3.3 x 10 -16 J For successful fusion, r ~10 -15 m (1 fm) →E coulomb ~ 2.3 x 10 -13 J E kin ~ 10 -3 E coulombClassically, would need T~10 10 to overcome Coulomb barrierSo, no Fusion??

Can Fusion Occur in Sun’s s Core?Can we be helped by considering the distribution of energies? - notall protons will have just E th = 3/2 kTProton velocities are distributed according to the Maxwellianequation: P(v) = 4π(m/24(m/2πkT)3/2 v 2 e -mv2 /2kTAt the high energy tail of the Maxwellian distribution, the relativenumber of protons, with E > 10 3 x E th is:N(E coulomb ~ 10 3 )/N() ~ e -ΔE/kt~ e -1000 = 10 -430 !!Number protons in Sun = M sun /M H ~ 10 57 so 1 in 10 430 of thesewill have enough energy to overcome Coulomb barrier.So again, no nuclear reactions??

Can Fusion Occur in Sun’s s Core?Even with tunnelling, , are stars hot enough for nuclear reactions to proceed?λ = h/p is the wavelength associated with a massive particle (p = momentum)In terms momentum, the kinetic energy of a proton is:1/2 m p v 2 = p 2 /2m pAssuming proton must be within 1 de Broglie λ of its target to tunnelset distance, r, of closest approach to λ, (where barrier height = original K.E.incoming particle) gives:e 2 /r = e 2 /λ = p 2 /2m p = (h/λ) 2 /2m pSolving for λ (=h 2 /2m p e 2 ) and substituting r = λ into 3/2 kT = e 2 /r, we get theQM estimate of the temperature required for a nuclear reaction to occur:QM = 4/3(e 4 m p /kh 2 ) - putting in numbersT QMT QMQM = 10 7 K which is comparable to T core . Therefore, fusion is feasible atcentre of Sun

Proton-Proton ChainBasic particles involved in nuclear reactionsParticlephotonneutrinoanti-neutrinoelectronpositronproton ( 1 1H)neutronSymbolγνν-e-e+p+nRest mass(kg)0>0>09 x 10 -319 x 10 -311.6 x 10 -271.6 x 10 -27Charge (e-)000-1+1+10Spin11/21/21/21/21/21/2deuteron21H3.2 x 10 -27+1Conservation Laws in nuclear reactions:(1) mass-energy(2) charge(3) difference between number particles and anti-particles conserved -ie particle cannot be created from anti-particle or vice-versa but a pair can beformed or destroyed without violating this rule

Proton-Proton ChainTypes of nuclear reactionsBeta Decay: n → p + + e - + ν - proceeds spontaneously - also forneutron inside nucleuseg [Z-1, A] → [Z, A] + e - + ν - (Z = # p + , N = # n, A = Z+N)Inverse Beta Decay: p + + e - → n + νeg [ 13 N → 13 C + e + + ν](p + , γ) ) process: A Z + p + → A+1 (Z+1) + γeg [ 12 C + p + → 13 N + γ](α, γ) process: α particle ( 4 He) added to nucleus to make heavierparticle [ A Z + 4 He → A+4 (Z+2) + γ]eg. [ 8 Be + 4 He → 12 C + γ]

Proton-Proton ChainSince the Sun’s s mass consists mostly of H and He, we anticipatenuclear reactions involving these two elementsegp + + p + → 2 Hep + + 4 He → 5 Li4 He + 4 He → 8 BeBut there is a problem here - what is it?All these reactions produce unstable particlesegp + + p + → 2 He → p + + p +p + + 4 He → 5 Li → p + + 4 He4 He + 4 He → 8 Be → 4 He + 4 HeSo among light elements there are no two-particle exothermicreactions producing stable particles. We have to look tomore peculiar reactions or those involving rarer particles

Proton-Proton ChainSummary of proton-proton chain:• 6p + → 4 He + 2p + + 2e + + 2ν 2 + 2γ 2 or 4p + → 4 He + 2e+ + 2ν 2 + 2γ2• Mass difference between 4p + and 1 4 He = 26.7 Mev x 6.424 x 10 18 ev/J= 4.2 x 10 -12 J• ~3% of this energy (0.8 Mev) ) is carried off by neutrinos and does notcontribute to the Sun’s s luminosity• 2 e + immediately annihilate with 2 e - and add 2 x 0.511 Mev (= 1.02Mev)• So, the total energy available for the Sun’s s luminosity per 4 He formedis: (26.7 - 0.8 +1.02) Mev = 26.9 Mev = 4.2 x 10 -12 J• # 4 He formed/sec = L sun /4.2 x 10 -12 J = 3.9 x10 26 J/s / 4.2x10 -12 J= 9.3 x 10 37 4 He/s• Increase of 4 He mass/time = dm He /dt = 9.3 x 10 37 4 He/s x 6.68 x 10 -27kg/ 4 He = 6.2 x 10 11 kg 4 He /sAfter 10 10 years (3 x 10 17 sec), M( 4 He) = 1.9 x 10 29 kg = 10% massSun

Proton-Proton ChainThe previous nuclearreaction chain (PPI) is notthe only way to convert Hinto He.A second branch (PPII)In solar centre, 69% of 2 3 He combines withanother 2 3 He, but 31% can fuse with 2 4 He32He + 4 2He " 7 4Be + #eg last step:3 He + 3 He → 4 !He + 2p +can proceed differently ifthere is appreciable 4 Hepresent!!!!!7 4Be + e " # 7 3Li + $ e73Li + p + " 4 2He + 4 2HeA third branch (PPIII)In the Sun's core, a 4 7 Be nucleus can capturea p + instead of an e " about 0.3% of the time.4Very unstable7 Be + p + " 5 8 B + #85B " 8 4Be + e + + # e8 Be " 4 2He + 4 2He4PPI + PPII + PPIII operatesimultaneously!!

Proton-Proton Chain: Summaryp + + p + " 1 2 H + e + + # e21H + p + " 3 2He + $69% 31%32He + 3 2He " 4 2He + 2p + 3 2He + 4 2He " 7 4Be + #!(PPI) 99.7% 0.3%!!7 4Be + e " # 7 3Li + $ e!7 Li + p + # 2 4 2He3Fraction of He 4 produced(PPII)1.00.50PPI!PPII7 4Be + p + " 8 5B + #85B " 8 4Be + e + + $ e84Be " 2 4 2He(PPIII)PPIII0 10 20 30 40T (10 6 )K)

Alternate Fusion Reactions H → 4 HeCNO Cycle (or bi-cycle)• First step in PP chain has very lowreaction rate (weak interaction).• 12 C can act as a catalyst in fusionreaction.• Net result:12 C + 4p + → 12 C + 4 He + 2e + + 2ν2+ 3γ3• Note: 12 C neither created nordestroyed. Also isotopes of N & Oare temporarily produced.• CNO cycle dominates over PPchain if T core > 1.8 x 10 7 (slightlyhotter than Sun)THE CNO BI-CYCLE(T < 10 8 K)10 6 yr}14 min3 x 10 5 yr3 x 10 8 yr82 s10 4 yrX as frequentlyOnce forevery 2500instances ofCycle 1

Other Fusion ReactionsAt higher T centre , heavier nuclei can fuse to produce energyFusion ofintoat T centralRef.4Hetriple α processBe, C~10 8 Kp. 31212 CO, Ne, Na, Mg6 x 10 8 Kp. 31316 OMg, Si, , S, etc.~10 9 Kp. 313We believe Universe began with only H, He, (Li, Be)All other elements created in core of stars (stellar nucleosynthesis)We are all made of “STARSTUFF”