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

2. Interior Zones of the Sun 6,000 K Convective Zone < 0.01 g/cm 3 500,000 K Radiative Zone < 0.01-10 g/cm 3 8,000,000 K Core < 10-160 g/cm 3 15,000,000 K

3. E I II III V where In simple 1-dimensional case an exponential decay connects the wave functions either side of the barrier

4. The Nuclear pp cycle 4 protons  4 He + 6  + 2 e + 2p 26.7 MeV

5. 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

6. vxvx vyvy vzvz And for each value of energy E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) Notice for any fixed E, m this defines a 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 = 4  v 2 dv

7. (1.7)(1.7) Maxwell Boltzmann distribution The probability distribution With a root mean square speed of

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

9. E I II III V probability of tunneling to here x = r 1 x = r 2 In simple 1-dimensional case an exponential decay connects the wave functions either side of the barrier

10. R E Where this time we’re tunneling in with an energy from r 2 where: r2r2 which we can just write as minimum energy to reach the barrier

11. hence then with the substitutions: with E=T  becomes

12. and for R « r 2 the term in the square brackets reduces to Performing the integral yields:

13. with E = ½ mv 2 can also write as

14. Cross sections for a number of fusion reactions which I will abbreviate as

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

16. which has a maximum where and at that maximum:

18. Since we can evaluate , for example, for the case of 2 protons: with kT in keV from which we can see:

19. If T ~ 10 6 K, kT ~ 0.1 keV for T ~ 10 8 K, kT ~ 10 keV with kT in keV This factor of 100 change in the temperature leads to a change in the fusion rate > 10 10 !!!

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

21. Q=5.49 MeV Q=0.42 MeV The sun 1 st makes deuterium through the weak (slow) process: Q=12.86 MeV then 2 passes through both of the above steps then can allow This last step won’t happen until the first two steps have built up sufficient quantities of tritium that the last step even becomes possible. I. The proton-proton cycle 2(Q 1 +Q 2 )+Q 3 =24.68 MeV plus two positrons whose annihilation brings an extra 4  m e c 2 = 4  0.511 MeV

22. Q=1.20 MeV Q=1.95 MeV Q=7.55 MeV II. The CNO cycle Q=7.34 MeV Q=1.68 MeV Q=4.96 MeV carbon, nitrogen and oxygen are only catalysts

23. Interior Zones of the Sun 6,000 K Convective Zone < 0.01 g/cm 3 500,000 K Radiative Zone < 0.01-10 g/cm 3 8,000,000 K Core < 10-160 g/cm 3 15,000,000 K

24. The 1 st 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 curves for solar densities 10 5 kg m -3 for protons and 10 3 kg m -3 for 12 C. Rate of energy production In later generations the relative importance of the two processes depends upon temperature.   T 4   T 17 CNO cycle p-p chain sun

25. 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.

26. At 10 8 K helium burning starts fusing: Q=190 keV Q=-91.9 keV At T= 10 8 K the fraction of helium nuclei meeting this threshold is 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. 8 Be exists as a resonance decaying with   10 -16 sec. Its formation requires 91.9 keV kinetic energy shared between the initial states.

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

28. As the helium supply in the core is exhausted further collapse leads to even higher temperatures. At ~5 x 10 8  10 9 K carbon and oxygen fusion can take place. Q=13.9 MeV Q=16.5 MeV and others, which yield protons, neutrons or helium nuclei

29. During the silicon burning phase (2  10 9 K) elements up to iron are finally produced. Even at these temperatures the Coulomb barrier remains too 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.

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

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