General Physics (PHY 2140)

1. General Physics (PHY 2140) Lecture 39 • Modern Physics • Nuclear and Particle Physics • Nuclear Energy • Elementary particles http://www.physics.wayne.edu/~apetrov/PHY2140/ Chapter 30

2. Lightning Review • Last lecture: • Nuclear physics • Nuclear reactions Review Problem: A beam of particles passes undeflected through crossed electric and magnetic fields. When the electric field is switched off, the beam splits up in several beams. This splitting is due to the particles in the beam having different A. masses. B. velocities. C. charges. D. some combination of the above E. none of the above

3. Processes of Nuclear Energy • Fission • A nucleus of large mass number splits into two smaller nuclei • Fusion • Two light nuclei fuse to form a heavier nucleus • Large amounts of energy are released in either case

4. Nuclear Fission • A heavy nucleus splits into two smaller nuclei • The total mass of the products is less than the original mass of the heavy nucleus • First observed in 1939 by Otto Hahn and Fritz Strassman following basic studies by Fermi • Lisa Meitner and Otto Frisch soon explained what had happened • Fission of 235U by a slow (low energy) neutron • 236U* is an intermediate, short-lived state • X and Y are called fission fragments • Many combinations of X and Y satisfy the requirements of conservation of energy and charge

5. Sequence of Events in Fission • The 235U nucleus captures a thermal (slow-moving) neutron • This capture results in the formation of 236U*, and the excess energy of this nucleus causes it to undergo violent oscillations • The 236U* nucleus becomes highly elongated, and the force of repulsion between the protons tends to increase the distortion • The nucleus splits into two fragments, emitting several neutrons in the process

6. Energy in a Fission Process • Binding energy for heavy nuclei is about 7.2 MeV per nucleon • Binding energy for intermediate nuclei is about 8.2 MeV per nucleon • Therefore, the fission fragments have less mass than the nucleons in the original nuclei • This decrease in mass per nucleon appears as released energy in the fission event • An estimate of the energy released • Assume a total of 240 nucleons • Releases about 1 MeV per nucleon • 8.2 MeV – 7.2 MeV • Total energy released is about 240 Mev • This is very large compared to the amount of energy released in chemical processes

7. QUICK Problem In the first atomic bomb, the energy released was equivalent to about 30 kilotons of TNT, where a ton of TNT releases an energy of 4.0 × 109 J. The amount of mass converted into energy in this event is nearest to: (a) 1 g, (b) 1 mg, (c) 1 g, (d) 1 kg, (e) 20 kilotons (c). The total energy released was E = (30 ×103 ton)(4.0 × 109 J/ton) = 1.2 × 1014 J. The mass equivalent of this quantity of energy is:

8. Chain Reaction • Neutrons are emitted when 235U undergoes fission • These neutrons are then available to trigger fission in other nuclei • This process is called a chain reaction • If uncontrolled, a violent explosion can occur • The principle behind the nuclear bomb, where 1 g of U can release energy equal to about 20000 tons of TNT

9. Nuclear Reactor • A nuclear reactor is a system designed to maintain a self-sustained chain reaction • The reproduction constant, K, is defined as the average number of neutrons from each fission event that will cause another fission event • The maximum value of K from uranium fission is 2.5 • In practice, K is less than this • A self-sustained reaction has K = 1

10. Basic Reactor Design • Fuel elements consist of enriched uranium • The moderator material helps to slow down the neutrons • The control rods absorb neutrons • When K = 1, the reactor is said to be critical • The chain reaction is self-sustaining • When K < 1, the reactor is said to be subcritical • The reaction dies out • When K > 1, the reactor is said to be supercritical • A run-away chain reaction occurs

11. Elementary Particles

12. 1. The Big Question of Particle Physics… How did we get from here… … to here? And what does it have to do with heavy quarks?

13. Time Seems like…

14. Just after the Big Bang: • symmetric Universe • equal number of particles and antiparticles Now: • asymmetric Universe • planets, stars, galaxies, Wayne State, … Note: macroscopic laws of Nature do not distinguish matter and antimatter

15. A 10,000,000.00 Swedish Kronor question:Where did all the antimatter go? • The “Onion paradigm”: • identify degrees of freedom • see if the problem has a solution • if not, dig deeper…

16. What are the right degrees of freedom? • Fire • Water • Earth • Air … that is, according to the Greeks!

17. What would be the modern picture? Imagine that we have a very powerful microscope…

18. Modern understanding: the ``onion’’ picture Atom Let’s see what’s inside!

19. Modern understanding: the ``onion’’ picture Nucleus Let’s see what’s inside!

20. Modern understanding: the ``onion’’ picture Protons and neutrons Let’s see what’s inside!

21. Modern understanding: the ``onion’’ picture Collective name for particles containing 3 quarks Mesons and baryons Collective name for particles containing quark and antiquark Let’s see what’s inside!

22. Modern understanding: the ``onion’’ picture Collective name for particles containing 3 quarks (such as proton and neutron) Mesons and baryons Collective name for particles containing quark and antiquark Let’s see what’s inside! Note: apparent excess of matter over antimatter can be traced to excess of the number of baryons over antibaryons. Thus our Big Problem is called Problem of Baryon Asymmetry of the Universe.

23. Modern understanding: the ``onion’’ picture Quarks and gluons Let’s see what’s inside!

24. Modern understanding: the ``onion’’ picture ? … so the answer depends on the energy scale!

25. … same thing about the interactions

26. Unification of forces

27. The Standard Model of particle physics

28. The Standard Model of Elementary Particle Physics • ``Periodic table’’ of matter • Interactions: electromagnetic, weak, strong, (gravity)… • Contains 26 parameters: needs experimental input + Higgs particle

29. Conditions for baryon asymmetry Matter-antimatter imbalance in the Universe A.D. Sakharov • Baryon (and lepton) number - violating processes to generate asymmetry • Universe that evolves out of thermal equilibrium to keep asymmetry from being washed out • Matter interactions differ from antimatter interactions (“Microscopic CP-violation”) to keep asymmetry from being compensated in the “anti-world”

30. Can Standard Model explain baryon asymmetry? does it have “the right stuff”? what are the conditions for the baryon asymmetry? does it have enough of “the right stuff”?

31. Experimental methods video

32. Experimental methods

33. Experimental Facilities I Cornell University SLAC

34. Experimental Facilities II KEK (Japan) Fermilab (Batavia, IL)

35. What do physics PhD’s do? • Science route • Research in physics (national lab, research university) • Teaching and research (college) • Industry route • Computing/engineering jobs in companies • Finance industry (problem solving) • Scientific Publishing route

36. A couple of review problems and notes to remember…

37. Remember: • Electricity: • Electric field and electric potential are different things • Moreover, field is a vector while the potential is a scalar • Remember the difference between parallel and series connections • Remember that formulas for capacitors and resistors are “reversed” • Magnetism: • Use right hand rule properly • Special relativity • If the problem involves speeds close to the speed of light, use relativistic formulas for momentum, energy, addition of velocities • In particular, KE=mv2/2 is a NONRELATIVISTIC expression for KE • Atomic and nuclear physics • In a way of handling, nuclear reactions are very similar to chemical reactions

38. Example : Proton moving in uniform magnetic field A proton is moving in a circular orbit of radius 14 cm in a uniform magnetic field of magnitude 0.35 T, directed perpendicular to the velocity of the proton. Find the orbital speed of the proton. Given: r = 0.14 m B = 0.35 T m = 1.67x10-27 kg q = 1.6 x 10-19 C Recall that the proton’s radius would be Thus Find: v = ?