What temperature would provide a mean kinetic energy of 0.5 MeV?By comparison, the temperature of the surface of the sun 6000 K.

Interior Zones of the Sun 6,000 KConvective Zone < 0.01 g/cm3500,000 KRadiative Zone < 0.01-10 g/cm38,000,000 KCore < 10-160 g/cm315,000,000 K

EIIIIIIVwhereIn simple 1-dimensional case an exponential decay connects the wave functions either side of the barrier

The Nuclear pp cycle4 protons 4He + 6g+ 2e + 2p 26.7 MeV

The effective kinetic energy spectrum of nuclei in a gas (at thermal equilibrium) is given by the high energy part of the Maxwell-Boltzmann distribution

vxvyvzAnd for each value of energyE = mv2 = m(vx2 + vy2 + vz2 )Notice for any fixed E, m this definesa sphere of velocity points all which give the same kinetic energy.The number of states accessible by that energy are within the infinitesimal volume (a shell a thickness dv on that sphere).dV = 4v2dv

Maxwell Boltzmann distribution

The probability distributionWith a root mean square speed of

(1.7)

While the cross-section will have an energy dependence dominated by the barrier penetration probability where the details follow

EIIIIIIVprobability of tunneling to herex = r1x = r2In simple 1-dimensional case an exponential decay connects the wave functions either side of the barrier

REWhere this time were tunneling in with an energy from r2 where:

r2which we can just write as minimum energy to reach the barrier

hencethen with the substitutions:with E=Ta becomes

and for R r2 the term in the square brackets reduces to Performing the integral yields:

with E = mv2 can also write as

Cross sections for a number of fusion reactionswhich I willabbreviate as

The probability of a fusion event at kinetic energy E is proportional to the product of these two functions. kinetic energy EEmN(E)P (E) (E)

which has a maximum whereand at that maximum:

Sincewe can evaluate , for example, for the case of 2 protons: with kT in keVfrom which we can see:

If T ~ 106 K, kT ~ 0.1 keV

for T ~ 108 K, kT ~ 10 keV with kT in keVThis factor of 100 change in the temperature leads to a change in the fusion rate > 1010 !!!

There are actually two different sequences of nuclear reactions which lead to the conversion of protons into helium nuclei.

Q=5.49 MeVQ=0.42 MeVThe sun 1st makes deuterium through the weak (slow) process:Q=12.86 MeVthen2 passes through both of the above steps then can allowThis last step wont happen until the first two steps have built up sufficient quantities of tritium that the last step even becomes possible.I. The proton-proton cycle2(Q1+Q2)+Q3=24.68 MeVplus two positrons whose annihilation brings an extra4mec 2 = 40.511 MeV

Q=1.20 MeVQ=1.95 MeVQ=7.55 MeVII. The CNO cycleQ=7.34 MeVQ=1.68 MeVQ=4.96 MeV carbon, nitrogen and oxygen are only catalysts

Interior Zones of the Sun 6,000 KConvective Zone < 0.01 g/cm3500,000 KRadiative Zone < 0.01-10 g/cm38,000,000 KCore < 10-160 g/cm315,000,000 K

The 1st generation of stars (following the big bang) have no C or N.The only route for hydrogen burning was through the p-p chain. Shown are curvesfor solar densities 105 kg m-3 for protons and 103 kg m-3 for 12C. Rate of energy productionIn later generationsthe relative importance of the two processes depends upon temperature. T4 T17CNO cycle p-p chainsun

The heat generated by these fusion reactions raises the temperature of the core of the star.The pressure of this "black body" radiation is sufficient to counteract gravitational collapse.

However once the hydrogen in the central region is exhausted gravitational collapse resumes. The temperature will rise as gravitational potential energy converts to kinetic energy of the nuclei.

At 108 K helium burning starts fusing: Q=190 keVQ=-91.9 keVAt T= 108 K the fraction of helium nuclei meeting this thresholdis given by the Boltzmann factor e-91.9/kT ~ 2.2 10-5 (with kT= 8.6 keV, the mean thermal energy). Note: these reactions are reversible. 8Be exists as a resonance decaying with 10-16sec. Its formation requires 91.9 keV kinetic energy shared between the initial states.

the small branching ratio for the -decaymakes it only 4 x 10-4 as likely as a return to the initial state: Once stable 12C has been produced, further absorption can occur through Q=4.73 MeVQ=7.16 MeVQ=9.31 MeV

As the helium supply in the core is exhausted further collapse leads to even higher temperatures. At ~5 x 108 - 109 K carbon and oxygen fusion can take place. Q=13.9 MeVQ=16.5 MeVand others, which yield protons, neutrons or helium nuclei

During the silicon burning phase (2109 K)elements up to iron are finally produced.Even at these temperaturesthe Coulomb barrier remainstoo high to allow direct formation:Instead it is done in an equilibrium process of successive alpha particle absorptions balanced against photo-disintegration: At these temperatures the thermal photons have an average energy of 170 keV and their absorption can easily lead to the break up of nuclei.

Then absorption of the 4He by other 28Si nuclei eventually leads to the build up of 56Ni etc.

Beyond iron, nickel and cobalt there are no more exothermic fusion reactions possible. Heavier elements cannot be built by this process.