Nuclear Reactions. Nuclear Reaction Examples
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1 Nuclear Reactions Nomenclature Atomic Number (Z) 6C12 Atomic Weight (A) Chemical Symbol (C for Carbon) Atomic Number = Number of Protons(+) (Identifies the element: 6 Carbon ) Atomic Weight = Number of Protons (+) plus Neutrons (o) (Identifies the isotope: Carbon-12 ) Nuclear Fusion: Nuclear Fission: Nuclear Reaction Examples 92U235 + o n1 92 U23 neutron 92U Xe Sr94+ o n1+ o n1+ γ gamma ray The Simplest Fusion Reaction: 1H1+ 1 H1 1 H2 +e++ ν neutrino deuterium positron also Matter-Antimatter Annihilation: e- + e+ γ electron positron
2 Thermonuclear Fusion in Stars The Proton Proton Reactions 1H1+ 1 H1 1 H2 + e++ ν followed by 1H1+ 1 H2 2 He3 + γ The positron e+ just gets converted to a gamma ray γ by e++ e- γ The light isotope of helium 2 He3 finds another of its kind whereupon 2He3 + 2 He3 2 He4 + 1 H1+ 1 H1 and we get back two of the six hydrogen nuclei (protons) we started with. The net result is 1H1+ 1 H1 + 1 H1 + 1 H1 2 He4 + 4γ s + 2ν s So how much energy is released in this process?
3 Energy Production by Hydrogen Fusion The basic process is 1H1+ 1 H1 + 1 H1 + 1 H1 2 He4 + 4γ s + 2ν s as four hydrogen nuclei fuse to make one helium nucleus In the process, some mass disappears The mass of four hydrogen nuclei is x kg The mass of one helium nucleus is x kg The difference is only x kg... or of the mass of hydrogen we started with. SO WHERE DOES IT GO? This or 0.7% of the mass is converted to energy. This appears mostly in the form of gamma radiation, γ (some produced by the annihilation of positrons e+ ) A very small amount is effectively wasted in the neutrino ν But just how much energy is that?
4 Energy Yield in Hydrogen Fusion When 1 kg of hydrogen is converted to kg of helium kg of mass is converted to energy.!!!! E = mc2!!!! E = (0.007 kg) x (3 x 108 m/s)2 = 6.3 x 1014 joules Recollect that M Sun = 2 x 1030 kilograms L Sun = 3.8 x 1026 watts If the Sun was all hydrogen which was all converted to helium the energy released would be 0.007xM Sun c2 = 1.26 x 1045 joules The Sun could then shine at its present luminosity for 110 x 109 years But the Sun isn t all hydrogen and not all of the Sun s hydrogen can be fused to make helium...
5 The Lifetime of the Sun Why the Thermonuclear in Thermonuclear Fusion? High temperatures are required to overcome the electrostatic repulsion between (positively charged) nuclei. (An element s atomic number, Z, equals its nuclear charge.) Temperatures in excess of 10,000,000 K are usually needed. These temperatures are attained only in the deep interior. So not all of the Sun s fuel is the kitchen A more careful calculation reduces the Sun s lifetime from the above crude estimate by about a factor of ten to τ Sun = 11 x 109 years But there was, historically, another possibility considered...
6 The CNO Reactions The first reaction sequence proposed to power the Sun was the six-step catalytic Carbon-Nitrogen-Oxygen (CNO) cycle 1. 1H1+ 6 C12 7 N13 + γ 2. 7N13 6 C13 + e+ + ν 3. 6C13+ 1 H1 7 N14 + γ 4. 7N14+ 1 H1 8 O15 + γ 5. 8O15 7 N15 + e+ + ν 6. 7N H1 6 C He4 Again, the positrons get annihilated via e++ e- γ The Net Result is 1H1+ 1 H1 + 1 H1 + 1 H1 2 He4 + 5γ s + 2ν s which is almost identical to the Proton-Proton result: 1H1+ 1 H1 + 1 H1 + 1 H1 2 He4 + 4γ s + 2ν s The energy yield is the same!...but there was a problem with this catalytic process...
7 The CNO Problem Interior temperatures of the Sun don t get high enough to overcome the electrical repulsion between protons 1 H and the nuclei of 6 C and 7 N (in steps 1, 3, 4, and 6) Note: The electric force between charges Q and q separated by distance r is F electric = +Qq/r2 But The net result and energy yield is the same as the protonproton reactions: 4 x 1 H1 2 He4 + energy And This happens to be the process which powers more massive (>1.5 MSun) main sequence stars (which do get hot enough!)
8 Summary: Thermonuclear Fusion in Stars Main Sequence Stars are powered by thermonuclear fusion Hydrogen is fused to helium. Basically: 1H1+ 1 H1 + 1 H1 + 1 H1 2 He4 Approximately 0.7 % of the mass disappears kg of Hydrogen makes kg of Helium and yields 6.3 x 1014 joules of energy (E = mc2) There are two routes for this process The Proton-Proton Reaction dominates below 1.5 M Sun The CNO reaction dominates above 1.5 M Sun (They contribute equally to the energy at 1.5 M Sun ) This Hydrogen Burning continues until fuel exhaustion...when all core hydrogen has been converted to helium With hydrogen exhaustion the main sequence stage ends as the star begins to die...
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