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The Physics of Energy sources Stellar fusion B. Maffei [email protected] Stellar Fusion 1 Introduction ! Many sources of renewable energy rely on the Sun ! From radiation directly • Solar cells • Solar water heating ! From radiation indirectly • Wind created from radiative heating • Photosynthesis • Waves ! We will then see where the Sun gets its energy from. ! Fusion: link between real fusion application renewable energies ! But actually the real source of energy is gravity • Fusion is a consequence Stellar Fusion 2 Quick summary of stellar formation ! Due to gravity a cloud of interstellar gas contracts ! The interstellar matter contains mainly hydrogen ! About 90% is hydrogen ! About 9% helium ! 1% is the rest ! The density increases and so is the temperature ! Gravitational energy Kinetic energy ! The star emits radiation ~ like a blackbody ! The temperature reaches ~ 107K ! The temperature is high enough to start thermonuclear reaction • First reaction to take place is fusion of hydrogen into helium ! Equilibrium reached ! After a while the energy released by fusion counterbalance gravity ! The star stops collapsing and reach a stable diameter (at least for a while…) Stellar Fusion 3 Proton-Proton chain ! Fusion of hydrogen into helium results from a sequence of reactions ! This sequence is known as the proton-proton chain The first reaction is the combination of 2 protons to form the only stable two-nucleon system: the deuteron. Note: simple fusion of 2 protons is impossible because diproton is unstable. So we have β+ decay (proton neutron + positron) (1) p + p → d + e + +ν or 1 H +1H →2H + e + +ν Q=0.93MeV The probability of β decay is low making the cross-section of this reaction very small σ~ 10-33b at keV energies and ~ 10-23b at MeV energies Even with a high proton density at the centre of the Sun (~7.5x1025 protons.cm-3) the reaction rate is still fairly low. What makes it work is the enormous number of protons in the Sun (~1056) Stellar Fusion 4 Proton-Proton chain (2) ! The previous reaction is then quickly followed by: (2) p + d →3He + γ or 1 H + 2H →3He + γ Q=5.49MeV This is the most likely reaction to happen after formation of deuteron. D-D reaction is very unlikely due to the small number of deuterons compared to protons The previous reaction is very slow 1 deuteron formed /sec for every 1018 protons. Once formed the lifetime of deuteron is 2sec. Much more probable to collide with proton than another deuteron Deuterons are transformed into 3He almost as soon as they are formed Stellar Fusion 5 Proton-Proton chain (3) ! Reaction of 3He with proton is not possible ! The isotope 4Li is unbound and breaks up as soon as it is formed 3 He+1H →4Li →3He+1H ! It is unlikely for 3He to react with 2H ! Density of 2H is very low and transformed into 3He very rapidly 3He (3) 3 Stellar Fusion He+ 3He " # + 2 p ! will have to react with another 3He or 3 He+ 3He"4 He +21H Q=12.86MeV 6 Summary of p-p chain p + p "d + e + + # 3 1 H+1H "2 H + e + + # Q=0.93 MeV H+ 2H "3 He + $ Q=5.49 MeV p + d" He + $ 1 3 3 He+ 3He " % + 2 p Figure from wikipedia He+ 3He"4 He + 2 1H x2 Q=12.86 MeV The complete process is the conversion of 4 protons to helium 4 p → α + 2e + +!2ν + 2γ 41H →4He + 2e + + 2ν + 2γ Q=25.7 MeV Note: reactions with nuclear particles In order to find the total Q we need to take into account the electrons: ! We have a plasma, mix of protons and electrons, neutral overall ! We add 4 electrons on each side of the equation ! 4 neutral H atoms give a neutral He atom and the 2 remaining e- annihilate the 2 e+ Calculation of Q gives 26.7MeV Stellar Fusion 7 Nucleosynthesis continues ! Once a star has exhausted its hydrogen, helium fusion starts ! Hydrogen in stellar core becomes depleted not enough energy produced to counter-balance gravity ! Gravitational contraction resume until T reaches ~ 108K ! Overcome the Coulomb barrier between α particles helium burning ! Several reactions are possible using 4He Triple α process 4 He+ 4 He"8 Be 4 He+ 8Be"12 C + # 4Be is short lived but can react with α particle to give carbon α process ! Stellar Fusion 4 He+12C"16 O + # 4 He+16O"20 Ne + # 4 He+ 20Ne"24 Mg + # 8 What next? ! Low mass stars (< 0.5 solar masses) will not go beyond H fusion ! After burning all their hydrogen they will finish as red dwarfs ! Many mid-sized stars will end when He burning is complete ! We get a white dwarf star or a giant red ! For much more massive stars (several times solar masses) ! Several more burning stages can occur ! Each preceded by gravitational contraction increase in the core temperature T > 5x108K carbon fusion T > 2x109K oxygen fusion 12 C +12C →20Ne+ 4He 12 C +12C →23Na +1H 16 O+16O→28Si + 4He 16 O+16O→32S + γ 12 C +12C →23Mg + n 16 O+16O→31P +1H ...... T > 3x109K silicon fusion for very massive star (~10 solar masses) 28 Si + 28Si →56Ni But the Coulomb barrier is very high so most probably α process occurs at the same time 16 O→20Ne→24Mg →28Si 28 Stellar Fusion Si →32S →36Ar →40Ca →44Ti →48Cr →52Fe→56Ni 9 Nucleosynthesis in massive stars Sequence of stellar burning stops when core is mainly composed of elements with A~56 Fe, Ni Above A=60, no energy to be gained by fusion Last elements to be formed are Fe and Ni The various fusion reaction stages will not be exactly sequential: ! ! ! ! ressure and T will vary across the star P The highest T will be in the core More advanced fusion will start in the core Creation of layers of different elements Stellar Fusion 10 CNO cycle ! ! ! ! Most of the interstellar material is hydrogen but stars are synthesising heavier atoms These heavier elements will then be released when stars explode Second generation stars will already contain heavier elements This can lead to other reactions such as CNO cycle at the same time than H burning 12 C +1H →13N + γ 1.95 MeV 13 N →13C + e + + ν 2.22 MeV C +1H →14N + γ 7.54 MeV 13 14 N +1H →15O + γ 7.35 MeV O →15N + e + + ν 2.75 MeV 15 15 N +1H →12C + 4He 4.96 MeV Heavy elements stay the same but: Fusion of hydrogen Energy released Creation of 4He Stellar Fusion 11 Energy emitted 6000K While the star is consuming its hydrogen, its luminosity increases with time Main sequence represented by Hertzsprung-Russell diagram I(λ,T) 5500K 5000K Wavelength (m) The star surface being like a plasma, it will radiate like a blackbody at temperature T (lower than core temperature) Blackbody emission = Planck function 2hν 3 I (ν , T ) = ⎛ hν ⎞ ⎜ ⎟ ⎝ kT ⎠ in W .m − 2 .Hz −1.sr −1 ⎛ ⎞ ⎜ c e − 1⎟ ⎜ ⎟ ⎝ ⎠ 2hc 2 I (λ , T ) = in W .m − 2 .m −1.sr −1 ⎛ hc ⎞ ⎛ ⎜ ⎟ ⎞ 5 ⎜ ⎝ λkT ⎠ λ e − 1⎟ ⎜ ⎟ ⎝ ⎠ 2 Stellar Fusion h : Planck constant k : Boltzmann constant ν : frequency λ : wavelength T : temperature 12 Energy from the Sun ! The Sun has a surface temperature of ~ 5500K- 5700K ! The integration of the Planck function at T~5600K over the full spectral domain, over the area of the Sun and over the whole solid angle gives the solar luminosity (ref to lecture 1) L = *** 2hc 2 # #% hc &( & "5 %%e$ "kT ' )1(( $ ' d+dAd" = 3.83 , 10 26 W Knowing the luminosity and the mass of the Sun (~2x1030kg), taking into account hydrogen fusion only, we can then estimate how long the Sun will burn (see example on website) How much!energy do we get from the Sun? Average Sun-Earth distance re = 1.5x1011m Power per surface unit (irradiance) delivered on the Earth by the Sun : re L 3.83 # 10 26 $2 F = = = 1.35kW.m 4 " re2 4 " (1.5 # 1011 ) 2 If we suppose that 70% (average atmospheric transmission) of the radiation reaches the Earth s surface, we get ~ 950W per square meter Stellar Fusion ! 13 References ! Most of the material of this lecture is coming from ! Lilley, J. Nuclear Physics Principles and Applications (Wiley 2006) ! Kenneth S. Krane, Introductory Nuclear Physics (Wiley 1988) Stellar Fusion 14 Summary ! Could you explain what is the source of energy of a Star? ! What is the proton-proton chain? ! Why can it work? ! What is the slowest process in this chain? ! What is the CNO cycle? ! What are the last elements to be formed though fusion in massive stars? ! How would you derive the radiative power emitted by the Sun and received on Earth? Stellar Fusion 15