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1) Big Bang ( nucleosintesi cosmologica )

1) Big Bang ( nucleosintesi cosmologica ). ????. Quali elementi. 700 000 anni T> 3000K (l’universo diventa trasparente) Tutta la materia come una miscela di particelle nucleari e di fotoni in equilibrio termico. Temperatura. Tempo. Origin of atoms in the solar system.

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1) Big Bang ( nucleosintesi cosmologica )

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  1. 1) Big Bang (nucleosintesi cosmologica) ???? Quali elementi

  2. 700 000 anni T> 3000K (l’universo diventa trasparente) Tutta la materia come una miscela di particelle nucleari e di fotoni in equilibrio termico.

  3. Temperatura Tempo

  4. Origin of atoms in the solar system • Two sources of nuclei: nucleosynthesis in the Big Bang and in Stars • The Big Bang made only H and He

  5. Stellar Nucleosynthesis • Until stars form, there is nothing except H and He • Gravitational instabilities develop which lead to formation of galaxies and collapse of molecular clouds to form stars

  6. Instabilità gravitazionale This cloud of atoms continued to expand for the next 700-500 x106 years, during which time inhomogeneities started to form in the distribution of atoms. Gravitational attraction between these more closely grouped atoms led to their growth into dense patches. These denser areas developed larger gravitational fields that further fueled their growth.

  7. Until stars form, there is nothing except H and He • Gravitational instabilities develop which lead to formation of galaxies and collapse of molecular clouds to form stars • At sufficient temperature and density (~107 K), nuclear fusion begins in star cores • Due to Coulomb repulsion between positively charged nuclei, non-resonant nuclear reaction rates obey a law of the form: r1r2 F∞ d2

  8. Two nuclei with Z1 and Z2 protons separated by distance r have electrostatic potential = (Energy barrier) where is the (magnitude of the) charge on the electron (and proton) and the permittivity of free space, is the standard constant that enters the SI version of Coulomb's Law. r1r2 F∞ d2 Setting or Z1 = Z2 = 1 we can estimate the barrier faced by two protons. This works out to be or . This is the Coulomb barrier.

  9. Most stars rely on this process to produce energy at some point. It is the main process occurring inside our Sun and other lower-mass stars (mass < 5 M ). Helium nuclei are built by adding Protons to a Hydrogen nucleus.

  10. Stellar Nucleosynthesis : Hydrogen Burning Coulomb Barrier for Fusion

  11. 1st Generation Stars - H fusion and production of 4He PP cycles (ppI) 1H + 1H 2H + b+ + venergy MeV 2H + 1H 3He + genergy MeV 3He + 3He 4He + 1H + 1Henergy MeV (stable)

  12. PP cycles (ppII) 3He + 4He7Be + genergy MeV 7Be + -e 7Li+ venergyMeV 7Li + p 4He + 4He energy MeV (stable) PP cycles (ppIII) 7Be + p 8B + genergy MeV 8B 8Be + v + +e energy MeV 8Be 4He + 4He energy MeV

  13. Stellar Nucleosynthesis : Hydrogen Burning • None of the two-particle reactions between the major species in juvenile H+He matter produce a stable product: • 1H + 1H = 2He (unstable) = 1H + 1H • 1H + 4He = 5Li (unstable) = 1H + 4He • 4He + 4He = 8Be (unstable) = 4He + 4He (0.1 femtoseconds) • However, Hans Bethe (1939) showed how hydrogen burning can begin with the exothermic formation of deuterium: • 1H + 1H = 2H + b+ + n + 1.442 MeV • This reaction initiates the PPI chain: 2 (1H + 1H = D + b+ + n) 2 ( 1H + D = 3He + g) 3He + 3He = 4He + 2 1H Net: 4 1H = 4He + 2 n + g • 2D/1H quickly approaches equilibrium value, but this is 1013 times smaller than the terrestrial value…terrestrial 2D is made elsewhere!

  14. The energy realised in these reactions …. Differences in the masses of each particle .. And the masses of the nuclei can be obtained from the atomic masses of their isotopes. atomic mass unit: (amu) m (12C )= 12 amu 1amu= 1.660540 x 10-27 Kg m (e- ) = 9.109 x 10-31 Kg atomic mass unit: (amu) m (1H )= 1.007825 amu--------- 1.673534 x 10-27 Kg (including 1 e-) -- 1.672623 x 10-27 Kg m (4He )= 4.00260 amu ---------- 6.646478 x 10-27 Kg (including 2 e-) -- 6.644656 x 10-27 Kg pp1------Δm ≈ 0.0066x 10-27 Kg ΔQ ≈ (Δm) c2 ≈ 4.2833 x 10-12 J = 26.74 MeV ΔQ ≈ (Δm) c2 ≈ 26.22 MeV = 2.26 K 1 degree kelvin= 8.621738 X10-5  eV                             = 0.0862          MeV

  15. 1.4-1.5 x106K >2.3 x106K 1.4-2.3 x106K

  16. Limite di massa perché un oggetto diventi stella, o resti un ammasso inerte (browndwarfs)- -core di una stella deve essere approx 10 volte più calda della T interna. -core density 100 volte più grande della densità media della stella Dati computazionali accurati hanno modellizzato che una stella possa “accendersi”. E quindi non rimanere una inerte browndwarf M= 0.08 M ☉ PP chain dominates in low mass main sequence stars. (M~M ☉) T?

  17. Protostar to…star …. The protostar, at first, only has about 1% of its final mass. But the envelope of the star continues to grow as infalling material is accreted. After a few million years, thermonuclear fusion begins in its core, then a strong stellar wind is produced which stops the infall of new mass. The protostar is now considered a young star since its mass is fixed, and its future evolution is now set

  18. http://imagine.gsfc.nasa.gov/docs/science/know_l2/stars.html Diagramme Hertzprung-Russel

  19. A star’s color is related to its surface temperature. - The coolest, smallest, dimmest stars are red dwarfs. Hotter, medium-sized stars are yellow stars, and the hottest, largest, most luminous main-sequence stars are blue stars.

  20. By the Net… “The Sun converts 600 million tons of Hydrogen into Helium every second – this means that 4 million tons of matter turns into energy every second. The Sun loses 4 million tons of mass every second – don’t worry, this is an infinitesimally small fraction of the mass of the Sun – even after 10 billion years it will hardly notice it! With this quantity of mass converted into energy each second, the Sun produces 4 x 1026 watts of energy – far more energy is released by the Sun in one second than humanity has used in its history! We will look at two processes which occur inside of stars to convert Hydrogen into Helium. The method of converting Hydrogen into Helium does not affect the energy output. In every case 0.71% of the mass of the Hydrogen is converted into Helium.”

  21. (T= 107 K) In many more-massive main sequence stars there are two closed cycled of reactions which not only convert H in He, but also convert 12 C into 13 C and various isotopes of nitrogen and oxygen. triple alpha processis the process by which three 4He nuclei are transformed into 12C 4He + 4He 8Be 8Be + 4He 12C + g + energy (7.4 MeV)

  22. In hotter stars, the CNO cycle plays a more important role in converting Hydrogen into Helium (i.e. It does it quicker than the Proton-Proton chain). CNO bi-cycle (1.5 x107 Kelvin) CN 1) NO → (stable)

  23. Meno frequente, reazioni +”lente” (probabilità 1/2500) 15N + 1H ------ 16O + γ 16O + 1H ------ 17F(min) + γ 17F ----------- 17O + e+ + υ 17O + 1H ----- 14N + 4He NO 2) 2 x107 Kelvin) Note that the net sum of these reactions is 12C + 4 1H ------12C + 4He + 2β+ + 2 υ In other words, four protons are converted into two positrons, two neutrinos, and one 4He nucleus; 12C serves only as a catalyst. Moreover, since the cycle can begin with any of the intermediate produces, it will occur if any of them exists.

  24. Energy production of the CN-CNO L’energia liberata per gr di ognuna delle sei reazioni del ciclo CNO ρ densità del gas Ni numero di nuclei Qi energia termica (calore) tì vita media dei nuclei (1.5 x107 Kelvin)

  25. The p-p and CNO cycles are the characteristic processes taking place during the main sequence lifetime of a star. In addition to providing the energy to support the star, they produce the fuel (Helium-4 nuclei) for all further stages of stellar nucleosynthesis. As the Hydrogen is converted into Helium, the density of Helium-4 increases and the core burning of heavier nuclei can start. This process occurs in stars with a mass greater than about 0.5 solar-masses and begins the next stage of stellar evolution and the next stage of stellar nucleosynthesis.

  26. CNO cycle dominates in higher –mass stars (M> a few M ☉) T? Mass balance??

  27. SUN The temperature in the coreof the sun; that is the inner quarter of its radius, must be at least 8 x 106 K with densities ranging from 1.6 x 105 kg m-3 in the centre to about 2 x 104 kg m-3 at the core perimeter. Sun produces (these days) 98-99% of its energy by means of PP reaction and 1-2% from the CNO cycle.

  28. Stellar Nucleosynthesis – Stars 0.8 – 1.5 M☉ • (Sun evolution) • · Two phases – hydrogen to helium, and then helium to carbon and oxygen. • · If the star does not have enough mass (>0.8 M☉, the second phase isn’t done. • · End result – White Dwarf star • First phase: “Proton-Proton chain” ~ (6 x106) °K • Second phase: Fusing Helium • · Minimum core temperature ~ (6 x106) K • · “CNO Cycle” • “Triple α” (3-4 x107) °K

  29. T Kelvin

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