Binding energies of different elements

1. Binding energies of different elements

2. Quantum Tunnelling Coulomb Potential: Ec = ZAZBe2 / 4πε0r Tunnelling probability: Ptunnel exp(-(EG/E)0.5) Gamow Energy: EG = (πZAZB)2 2mrc2 mr = mAmB/(mA+mB)  = e2 / (4πε0ћc)  1/137

3. Energy-dependent fusion rates Dashed line: Boltzmann factor, PBoltz  exp(-E/kT) Dot-dash line: Tunnelling factor, Ptun exp(-(EG/E)0.5 Solid line: Product gives the reaction rate, which peaks at about E0 = (kT/2)2/3 EG1/3 Sun: EG = 493keV; T = 1.6x107 K  kT = 1.4keV Hence E0  6.2 keV ( 4.4 kT)

4. The PP chain: the main branch 9x 9x

5. The PP chain ~85% in Sun ~15% in Sun ~0.02% in Sun

6. The CNO Cycle Slowest reaction in this case: 14N + p  15O + γ. 14N lives for ~ 5x108 yr. Abundances: 12C ~ 4%, 13C ~ 1%, 14N ~ 95%, 15N ~ 0.004%

8. Recap Equation of continuity: dM(r)/dr = 4 π r2ρ Equation of hydrostatic equilibrium: dP/dr = -G M(r) ρ / r2 Equation of energy generation: dL/dr = 4 π r2ρε where ε = εpp + εCNOis energy generation rate per kg. ε = ε0ρTα where α~4 for pp chain and α~17-20 for CNO cycle.

9. Discussion: random walk Consider tossing a coin N times (where N is large). It comes as “heads” NH times and “tails” NT times. Let D = NH – NT. • What do you expect the distribution of possible values of D to look like (centre, shape)? • How do you expect the distribution of D to change with N? • Would you ever expect D to get larger than 100? If so, for what sort of value of N?

10. Random Walk Photon scattered N times: • = ř1 + ř2 + ř3 + ř4 +……+ řN Mean position after N scatterings: <> = <ř1> + <ř2> + <ř3>+ ……+ <řN> = 0 But, mean average distance travelled || comes from . = ||2, hence || = (.)0.5 . = (ř1 + ř2 + ř3 + ř4 +……+ řN).(ř1 + ř2 + ř3 + ř4 +……+ řN) = (ř1.ř1 + ř2.ř2 + ř3.ř3 +……+ řN.řN) + (ř1.ř2 + ř1.ř3 + ř1.ř4 +……+ ř2.ř1 +ř2.ř3 + ř2.ř4 + …… ř3.ř1 + ř3.ř2 + ř3.ř4 + ……) =  (ři.ři) = Nl2  || = (√N) l }=0