2. 7.2 Nuclear Stability and Nuclear Reactions 2

3. Nuclides above the band are too large - decay by . To the left  decay occurs. Nuclides below the band have too few n o, positron decay occurs. A p+ becomes a n o. 3

4. Nucleon Mass, u H-11.007825 H-21.00705 H-31.00535 He-41.000605 C-121.00000 Fe-580.99885 4

5. Protons and Neutrons weigh less when bound in the nucleus. p+ and n o fall toward each other, emit a ~1 MeV photon. Photon has mass E = mc 2. E lost from nucleus. 5

6. 6 More nucleons in nucleus emit higher E photons p+ and n o fall toward each other, emit a 7 MeV photon. Higher nucleon numbers, higher PE lost on impact! 6

7. Binding E (BE) The amount of E emitted per nucleon = amount of E needed to put in to tear nucleus apart. BE the amount of work (eV, J) to tear apart nucleus. 7

8. Nuclear Energy Well It’s as if nucleus is a well. BE must be added to remove p+ or n o from well. PE is lost as photons/mass when nucleons fall in. More nucleons = deeper well. More nucleons in nucleus have more strong F, more BE/nucl. 8

9. Binding E per Nucleon. Total E of Nucleus Num. Nucleons. Higher binding E per nucleon = more stable elements.

10. Can look at E like this too! 10

11. 11 The higher the BE between nucleons, the more E/work needed to split it up, the more stable the nucleus – the less likely to decay (fall apart) Which is the most stable?

12. What happens beyond Fe-56 BE/nucleon decreases. Nuclei get large F elc takes over. PE released in nucl creation decreased.

13. Film Clip 2:45 min BE https://www.youtube.com/watch?v=UkLki XiOCWUhttps://www.youtube.com/watch?v=UkLki XiOCWU 13

14. 1. If U-235 decays to Pb-208. a. How much BE/nucl is released? b. How much E is released for a whole U- 235 nucleus? c. How many kg of mass is represented? ~ 0.2 MeV 235 x 0.2 = 47 MeV 47 x 10 6 eV x 1.6 x 10 -19 J/eV 7.52 x 10 -12 J 7.52 x 10 -12 J = m (3x10 8 m/s) 2 8.4 x 10 -29 kg.

15. Nuclear Physics 9 9 minutes. https://www.youtube.com/watch?v=- YMgacsJyD0&playnext=1&list=PL019160 6751B22A12&feature=results_mainhttps://www.youtube.com/watch?v=- YMgacsJyD0&playnext=1&list=PL019160 6751B22A12&feature=results_main 15

16. Do Now: A. What of photon will be emitted for a photon released when a single 235 U nucleon when the nucleus decays to form 208 Pb assuming all the E is released as a EM radiation? B. How many AMUs is 25.32 g of anything? 2.6 x 10 -14 m. 16 1.524 x 10 25 u.

17. Energy from the Nucleus

18. Mass Defect (  m) Sum individual nucleons weigh more than the nucleus. Mass/ E difference = mass defect/deficit. The mass defect equals the BE. Find dif btw atomic mass and mass of constituents.

19. E and mass are equivalent. E = mc 2 (joules) 1 u = 931.5 MeV. (remember 1 u = 1/12 mass C-12) We can calculate the BE per nucleon for each element. Find the difference in mass between the element and the sum of the individual component nucleons. 19

20. Calculation of BE per nucleon 1. Calculate the BE per nucleon of 54 Fe which has an average mass of 53.9396 u. Mass p+ = 1.00782 u Mass n o = 1.00866 u Mass e- = 0.000549

21. Calculate mass defect & BE/nucleon 26 p + x 1.00782 u = 26.20332 28 n o x 1.00866 u = 28.24248 26 e- x 0.0005490.014274 Mass constituents54.4458 u mass Fe nucl53.9396 u subtract Mass defect0.5062 u energyx 931.5 MeV/u BE = 471.5 MeV ÷ 54 nucl = 8.7 MeV/nucl 21

22. 2. Find the binding energy of He-4 which has a known mass of 4.002602 u. (I’m including the e- this time). Neutral He-4 consists of 2e-, 2p+, and 2n o. Find the sum of the parts in u. Look up in table: e- rest mass p+ rest mass n o rest mass

23. 23 2 x (0.000549u)0.001098 +2 x (1.007277u)2.014554 +2 x (1.008665u)2.017300 constituents4.032982 u nucleus4.002602 u defect0.03038 u x 931.5 MeV/nucl 28.29897 MeV ÷ 4 nucl = 7.0747 MeV/nucl 23 Mass Defect

24. 3. The nucleus of a deuterium atom consists of a proton and a neutron. If the mass of deuterium is 2.014102 u, calculate the BE in MeV..  m = 0.002378 ~ 2.215107 MeV / 2 = ~ 1.10755 MeV /nuc. 24

25. 25 To calculate the BE/nucl: 1. Find mass defect - the difference between the mass of the separate nucleons (unbound) and the mass of the bound nucleus. 2.Calculate the mass in atomic mass units. 3. Each unit has an energy equal to 931.5 MeV. 4. Multiply the mass defect by 931.5 MeV to convert the mass to energy. 5. Divide by # nucleons.

26. What happens when nucleus in not stable? Strive for stability. 26 Nuclear Reactions and Energy

27. Transmutation Nuclear transmutation, the conversion of one chemical element or isotope into another through nuclear reaction. Spontaneous Artificial/Induced Bombard Nucleus with a particle.

28. Two types of Transmutation Fission – heavy elements split to smaller fragments. Fusion – light elements fuse to heavier. 28

29. Fission We looked at fission by spontaneous radioactive decay. Fusion nucleus splits to 2 or more parts. Parent mass # decreases. material is added and “fused” to the nucleus. 29

30. 4. Which type of reaction is this? Fission – parent has more mass than daughter. 30

31. 5. Which type of reaction is this? 4 He + 14 N 17 O + 1 H 278 1 Proton  31 Fusion - Mass of parent increases

32. Other types of particles can be used to bombard the nucleus. Neutrons, protons, and H-2 are common. Which rx is this? 32

33. 16 O + 1 n A X + 2 H 80z1 Identify X if a 2 H is emitted in the reaction. What type of reaction is it? 33

34. 16 O + 1 n A X + 2 H 80z1 16 O + 1 n 15 X + 2 H 8071 Element must have 7 p+. It will be 15 N. 7 Solve for the mass & proton number for X by balancing. 34 Fission.

35. Energy Considerations in Fission & Fusion 35

36. Nuclear decay seeks to stabilize nucleus. When nucleus goes from lower to higher binding energy ratio/nucleon, energy is released in process. 36

37. Release of E occurs when element goes to more stable state. 7. Use BE table to predict the total energy release. 235 U 2 117 Pd fragments 92 46 Find the energy released. Use the BE/nucleon table. 37

38. 235 U ~ 7.6 MeV per nucleon 117 Pd ~ 8.4 MeV per nucleon. Take difference. This E is released per nucleon. 8.4 – 7.6 = 0.8 MeV/nucleon. But 235 U has 235 nucleons so: 235 x 0.8 MeV = 188 MeV released! Compare this to ionization energies.

39. BE Released in Rx can be converted to KE and/or heat. 39

40. 8: Lithium can be bombarded with n o to induce the following rx: Mass Li-66.015126 Mass n o =1.008665 Mass H-33.016030 Mass He-44.002604 6 Li + 1 n 3 H + 4 He 3012 How much E is released?

41.  m = (7.023791 – 7.018634) u  m = 0.005175 u x 931.5 MeV/u = 4.804 MeV Available as KE to fragments. some may be released as heat and light. 41 Calculate mass for each side. (7.023791u) (7.018634) 6 Li + 1 n 3 H + 4 He 3012

42. 9. Energy From Reactions 10 min http://www.youtube.com/watch?v=- YMgacsJyD0&playnext=1&list=PL0191606751B22A12& feature=results_mainhttp://www.youtube.com/watch?v=- YMgacsJyD0&playnext=1&list=PL0191606751B22A12& feature=results_main 42 8.1 Natural transmutations and half lives 10 min http://www.youtube.com/watch?v=I7WTQD2xYtQ

43. Read Hamper 7.4 and 7.5 pg 165#20 and pg 166 #21,22