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Constraints on Nuclear Fusion Reactions

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Presentation on theme: "Constraints on Nuclear Fusion Reactions"— Presentation transcript:

1 Constraints on Nuclear Fusion Reactions
Binary reactions dominate Energy Conservation (Energy release) Charge conservation (d=2H nucleus) pp  d+e++Energy +?? Angular momentum conservation Proton, electron spin = (1/2)[h/(2π)] Deuteron spin = h/(2π) Neutrino predicted by E.Fermi pp  d+e++ +Energy Detected in 1950s by Reines and Cowen next to a nuclear reactors Weak Interactions (lepton number conservation) n  p + e+ anti-. Half Life 12,33 year Weak interactions are Weak  Reaction Rates are slow

2 Light Nuclei, http://www.nndc.bnl.gov/chart/
No stable A=5 nuclei

3 pp chains in Sun 31% 69% 99.7% (pp I) 0.3% (pp II) (pp III)

4 pp chains H+H+H+H4He+2e+2e++2
First Step pp  D + e+ +  + MeV Weak interaction process D=(pn) = nucleus of 2H. Neutron half life = min Suppressed by pp Coulomb repulsion Quantum Tunneling required Protons are waves If Kinetic energy = E-U > 0 = h/p If Kinetic energy = E-U < 0 Wave is exponentially damped exp(-r)  = [U-E]1/2.

5 Billiard Ball model of pp collisions
Proton density , Temperature T Velocity distribution (Maxwell-Boltzman) P(v;T) = exp(-mv2/(2KT)) / [2πkT]3/2. <mv2/2>= <mvx2/2> +<mvy2/2> +<mvz2/2> =3kT/2 vrms ≈ [kT/m]1/2. Collision rate per unit volume: R. ≈ (10-15m)2. Rate R ≈ 2v|≈ 2 [kT/m]1/2. Volume = vt

6 Ballistic Burn Time of Sun
Number density of sun = /MH. Average = 8 1029 /m3. Central = 1032 /m3. Temperature, rms Velocity vrms ≈ [kT/mc2]1/2 Surface kT = 0.5 eV. vrms ≈ 7000 m/s Central kT = 1KeV. vrms ≈ 3 105 m/s Proton Mass = 0.9 GeV Time for all protons to collide / R = /[|v|] Surface 1/[|v|] ≈ 10-4 s Center ≈ 30 ns Crucial role of Coulomb Repulsion and Weak interactions to regulate nuclear fusion rate in sun

7 pp fusion slowed by coulomb repulsion
Kinetic energy required to bring two protons to within 1 fm e2/(4π0 r) = 1.4 MeV Center of sun kT ≈ 1 KeV Maxwell Boltzmann distribution P(v) = exp{-[mv2/(2kT)]} P[-(1.4MeV)/(1 KeV)] ≈ exp(-1000) Classical pp Collision rate in sun ≈ 0 No pp fusion, even worse for heavier elements Quantum barrier penetration


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