1. Sun Nuclear Reactions If the mass in the center of the solar nebula is large enough, gravity will collapse more and more material, producing higher and higher pressures, producing higher and higher temperatures as a result of higher and higher energies of the particles in the gas. The high energies break apart any elements, resulting in the region of the protostar being composed mostly of hydrogen gas.

2. Sun Nuclear Physics

3. The work of Einstein – A new view of mass: Mass: A highly concentrated form of energy. E = m c 2 Sun Nuclear Physics

4. The work of Einstein – A new view of mass: E = m c 2 Therefore: All previous arguments about energy, including conversions and conservation must hold for mass-energy as well as kinetic energy and potential energy. Implication – mass can be converted to energy and energy can be converted to mass. Sun Nuclear Physics

5. The work of Einstein – A new view of mass: Mass: A highly concentrated form of energy. E = m c 2 Note: c is a VERY large number, and c 2 is HUGE. Therefore, a small amount of mass can be converted to a LARGE amount of energy. MASS conversion is the source of the energy produced in the core of the sun. Sun Nuclear Physics

6. Sun Nuclear Physics How much mass would be required to operate a 500 Megawatt power plant for one year? Note: The total amount of energy required to operate the facility for a year is 1.6 x 10 16 Joules.

7. Sun Nuclear Physics How much mass would be required to operate a 500 Megawatt power plant for one year? Note: The total amount of energy required to operate the facility for a year is 1.6 x 10 16 Joules. Solution: m c 2 = 1.6 x 10 16 Joules m = 1.6 x 10 16 (3 x 10 8 ) 2 1.6 x 10 16 9 x 10 16 =.177 kg = 177 g = Note: the mass of a paperclip is about 1 g

8. The Protostar Gravitational Collapse

9. The Protostar Gravitational Collapse Gravitational collapse causes an increase in pressure at the center of the protostar

10. The Protostar Gravitational Collapse High pressures strip electrons from nuclei, and break nuclei apart so that the interior is composed primarily of hydrogen

11. The Strong Force Puzzle: Nuclei should not be able to exist. Reason: The electrical force of repulsion between the protons should cause the nucleus to fly apart.

12. The Strong Force Puzzle: Nuclei should not be able to exist. Reason: The electrical force of repulsion between the protons should cause the nucleus to fly apart. Proposal: There is another force, called the Strong Force, which negates electrical repulsion over VERY short distance (the size of a nucleus).

13. The Strong Force Proposal: There is another force, called the Strong Force, which negates electrical repulsion over VERY short distance (the size of a nucleus). If protons can physically get close enough (not easy to do because of electrical repulsion) this strong force will dominate and allow the protons to “bind” and form a nucleus.

14. The Nucleus Constituents of the nucleus: * Protons * Neutrons

15. The Strong Force Description of elements: A Z E A – Number of Nucleons (Atomic Mass Number) Z – Number of Protons (Atomic Number) A nucleon is a proton or neutron

16. The Nucleus Constituents of the nucleus: * Protons – hold electrons in orbit * Neutrons - ?

17. Neutrons * Mass is a bit bigger than that of a proton * Electrically neutral

18. Neutrons * Mass is a bit bigger than that of a proton * Electrically neutral So what is the neutron? It doesn’t seem to do anything and it doesn’t seem to have any reason to exist???

19. Neutrons * Mass is a bit bigger than that of a proton * Electrically neutral So what is the neutron? It doesn’t seem to do anything and it doesn’t seem to have any reason to exist??? The answer comes from the study of radioactive decay and nuclear reactions

20. The Strong Force Reactions: A Z X → A’ Z’ Y + Particles + Stuff

21. The Strong Force Reactions: Beta Decay: neutron → proton + electron + antineutrino Inverse Beta Decay: positron + electron + proton → neutron + positron + neutrino

22. The Strong Force What is a neutrino? In examining beta decay, it was determined that he energies of the emitted electrons seemingly violated conservation of energy. However, charges and masses were pretty well accounted for. If energy was to be conserved, there needed to be a “hidden” particle, which carried no (or very little) mass, and also carried no charge. Since this particle was similar to a neutron, except for the mass issue, Enrico Fermi proposed that they be called “neutrinos” (little neutrons). Current thought – recent experiments indicate the neutrino has a very small mass.

23. The Strong Force What is an antiparticle? Observed particles have an associated particle with the same mass and spin but opposite electric charge and magnetic moment (not a part of this class). This associated particle, called an antiparticle, is an “opposite” to the particle in the sense that if the particle and antiparticle collide, their mass is converted into energy (ie, they annihilate each other). Under the right conditions, energy can be converted into a particle- antiparticle pair.

24. The Strong Force Inverse Beta Decay: positron + electron + proton → neutron + positron + neutrino Positron Proton Neutron Electron neutrino

25. The Protostar Gravitational Collapse Once high enough, the high pressures can cause protons to overcome electrical repulsion, and get close enough to “bind” that is, the high pressure can trigger nuclear reactions

26. Sun Nuclear Physics – The Sun’s Core The solar nuclear reactions: The intermediate steps - 1 H + 1 H ―› 2 H + positron + neutrino (2 reactions required) 2 H + 1 H ―› 3 He + energy (2 reactions required 3 He + 3 He ―› 4 He + 1 H + 1 H + energy 2 H is called deuterium

27. Sun Nuclear Physics – The Sun’s Core The solar nuclear reactions: The net result - 4 ( 1 H ) ―› 4 He + energy + 2 neutrinos

28. Sun Nuclear Physics – The Sun’s Core The solar nuclear reactions: The net result - 4 ( 1 H ) ―› 4 He + energy + 2 neutrinos Δ m = 4.77 x 10 -29 kg

29. Sun Nuclear Physics – The Sun’s Core The solar nuclear reactions: The net result - 4 ( 1 H ) ―› 4 He + energy + 2 neutrinos Δ m = 4.77 x 10 -29 kg Energy = Δ m c 2 = 4.3 x 10 -12 Joules per reaction

30. Sun Nuclear Physics – The Sun’s Core The solar nuclear reactions: The net result - 4 ( 1 H ) ―› 4 He + energy + 2 neutrinos Mass of 1 H = 1.6736 x 10 -27 Kg Mass of four 1 H = 6.6943 x 10 -27 Kg Mass of 4 He = 6.6466 x 10 -27 Kg Δ m = 4.77 x 10 -29 kg Energy = Δ m c 2 = 4.3 x 10 -12 Joules per reaction

31. Sun Nuclear Physics – The Sun’s Core Solar data: Luminosity (amount of energy per second in watts) = 4 x 10 26 W Therefore, in one second the sun produces 4 x 10 26 J of energy (this is a big number).

32. Sun Nuclear Physics – The Sun’s Core Therefore, in one second the sun produces 4 x 10 26 J of energy (this is a big number). For the sun to produce this much energy requires (energy / second) = (energy / reaction) * (number of reactions / second) Therefore, by rearranging this expression (number of reactions / second) = (Energy / second) / (Energy / reaction) = (4 x 10 26 J/sec) / 4.3 x 10 -12 Joules/reaction = 9.3 x 10 37 reactions per second

33. Sun Nuclear Physics – The Sun’s Core Therefore, in one second the sun produces 4 x 10 26 J of energy (this is a big number). Finally, (mass converted / second) = (number of reactions / second) * (Mass converted / reaction) = ( 9.3 x 10 37 reactions/second) * ( 4.77 x 10 -29 kg/reaction) = 4.4 x 10 9 kg/sec !!!

34. Sun Nuclear Physics – The Sun’s Core Therefore, in one second the sun produces 4 x 10 26 J of energy (this is a big number). Finally, (mass converted / second) = (number of reactions / second) * (Mass converted / reaction) = ( 9.3 x 10 37 reactions/second) * ( 4.77 x 10 -29 kg/reaction) = 4.4 x 10 9 kg/sec !!! On Earth, this mass conversion is equivalent to converting over 4 millions tons of material PER SECOND.

35. Sun Nuclear Physics – The Sun’s Core

36. Sun Nuclear Physics – The Sun’s Core 9.3 x 10 37 reactions per second

37. Sun Nuclear Physics – The Sun’s Core 9.3 x 10 37 reactions per second 4.4 x 10 9 kg/sec !!!

38. Sun Nuclear Physics – The Sun’s Core 9.3 x 10 37 reactions per second 4.4 x 10 9 kg/sec !!! On Earth, this mass conversion is equivalent to converting over 4 millions tons of material PER SECOND.

39. Sun Nuclear Physics – The Sun’s Core Some numbers (help for the homework problems) There are 4 H that participate in each reaction. There are 9.3 x 10 37 reactions/sec Therefore, the number of hydrogen nuclei converted into Helium every second is: ( 9.3 x 10 37 reactions/second ) * ( 4 Hydrogen/reaction ) = 3.7 x 10 38 Hydrogen/second

40. Sun Nuclear Physics – The Sun’s Core Some numbers (help for the homework problems) Since the mass of hydrogen is 1.67 x 10 -27 kg, The mass of hydrogen being converted into Helium every second is ( 3.7 x 10 38 Hydrogen/second ) (1.67 x 10 -27 Kg/Hydrogen) = 6.18 x 10 11 Kg/second That is, about 6 x 10 11 Kg of hydrogen is being converted to Helium every second in the core of the sun.

41. Sun Nuclear Physics – The Sun’s Core Some numbers (help for the homework problems) That is, about 6 x 10 11 Kg of hydrogen is being converted to Helium every second in the core of the sun. On the earth, this would be equivalent to Over 600,000 million tons of Hydrogen into Helium every second

42. Sun Nuclear Physics – The Sun’s Core Do not misinterpret the previous numbers ! The sun is converting 6 x 10 11 kg/sec of hydrogen through nuclear fusion. This does NOT mean that the sun is losing 6 x 10 11 kg/sec of its mass Most of the mass of the hydrogen remains as helium.

43. What’s New June 2011 Scientists at the Cern laboratory in Geneva have found a way to 'trap' elusive antimatter atoms for more than 15 minutes. Photograph: Cern

44. What’s New August 2011 Antimatter belt around Earth discovered by Pamela spacecraft may be enough to implement a scheme using antimatter to fuel future spacecraft Published in Astrophysical Journal Letters