1 / 11

Binding energies of different elements

Binding energies of different elements. Quantum Tunnelling. Coulomb Potential: E c = Z A Z B e 2 / 4 πε 0 r Tunnelling probability: P tunnel  exp(-(E G /E) 0.5 ). Gamow Energy: E G = ( π Z A Z B ) 2 2m r c 2 m r = m A m B /(m A +m B )  = e 2 / (4 πε 0 ћ c)  1/137.

curran-weaver
Download Presentation

Binding energies of different elements

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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%

  7. 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.

  8. 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?

  9. 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

More Related