Binding energies of different elements

Quantum Tunnelling

Coulomb Potential:

Ec = ZAZBe2 / 4πε0r

Tunnelling probability:

Ptunnel exp(-(EG/E)0.5)

Gamow Energy:

EG = (πZAZB)2 2mrc2

mr = mAmB/(mA+mB)

= e2 / (4πε0ћc) 1/137

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 EG

1/3

Sun: EG = 493keV; T = 1.6x107 K kT = 1.4keV

Hence E0 6.2 keV ( 4.4 kT)

The PP chain: the main branch

9x

9x

The PP chain

~85% in Sun

~15% in Sun

~0.02% in Sun

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%

Recap

Equation of continuity:

dM(r)/dr = 4 π r2 ρ

Equation of hydrostatic equilibrium:

dP/dr = -G M(r) ρ / r2

Equation of energy generation:

dL/dr = 4 π r2 ρ ε

where ε = εpp + εCNO is energy generation rate per kg.

ε = ε0ρTα

where α~4 for pp chain and α~17-20 for CNO cycle.

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?

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