1. Chapter 23: Nuclear Chemistry Nuclear Changes Nuclear Equations 238 92 U  234 90 Th + 4 2 He

2. Chapter 23: Nuclear Reactions & the Nucleus Nucleons: particles in the nucleus: –p + : proton –n 0 : neutron. –Mass number: the number of p + + n 0. Isotopes: have the same number of p + and different numbers of n 0. Nuclear Binding Energy –Mass Defect

3. Chapter 23: Energy Versus Mass Energy and Mass are interchangable. Law of Conservation of Mass and Law of Conservation of Energy are the same. –Sum of all energy and all mass is a constant. Nuclear Binding Energy –Mass Defect

4. Balancing Nuclear Equations In nuclear equations, the total number of nucleons is conserved: 238 92 U  ? + 4 2 He

5. Chapter 23: Radioactivity Isotopes: have the same number of p + and different numbers of n 0. Stable Radioisotopes –Natural –Manmade

6. Radiosotopes: Radioactivity Definition The three primary types of radioactivity are: –  -Radiation is the loss of 4 2 He from the nucleus, –  -Radiation is the loss of an electron from the nucleus, –  -Radiation is the loss of high-energy photon from the nucleus.

7. Radioactivity Nuclear Equations

8. Radioactivity: Types of radioactive Decay Alpha Particles/Alpha Emission: 4 2 He and 4 2  represent  -radiation. – 238 92 U  234 90 Th + 4 2 He.

9. Radioactivity: Types of radioactive Decay Beta Particles/Beta Emission: Nucleons undergoes decay: 1 0 n  1 1 p + + 0 -1 e - (  -emission) 14 6 C  0 -1 e - + ______

10. Radioactivity: Gamma Emission Electromagnetic Energy Spectrum Can be emitted with Particles

11. Additional Types of Nuclear Decay: Decay Equations Positron Emission 1 0 p +  1 0 n + 0 1 e + (positron or  + - emission) A positron is a particle with the same mass as an electron but a positive charge.

12. Additional Types of Nuclear Decay: Decay Equations ELECTRON CAPTURE 1 1 p + + 0 -1 e -  1 0 n (electron capture) 7 4 Be + 0 -1 e -  ______

13. Radioactive Series A nucleus usually undergoes more than one transition/decay on its path to stability. The series of nuclear reactions that accompany this path is the radioactive series. Nuclei resulting from radioactive decay are called daughter nuclei.

14. Patterns of Nuclear Stability Radioactive Series For 238 U, the first decay is to 234 Th (  -decay). The 234 Th undergoes  -emission to 234 Pa and 234 U. 234 U undergoes  -decay (several times) to 230 Th, 226 Ra, 222 Rn, 218 Po, and 214 Pb. 214 Pb undergoes  -emission (twice) via 214 Bi to 214 Po which undergoes  -decay to 210 Pb. The 210 Pb undergoes  -emission to 210 Bi and 210 Po which decays (  ) to the stable 206 Pb.

15. Chapter 23: Nuclear Reactions: Nuclear Transmutation Target + Particle  Products 14 N + 4   17 O + 1 p. The above reaction is written in short- hand notation: 14 N( ,p) 17 O.

16. Nuclear Transmutations Using Charged Particles

17. Chapter 22: Nuclear Reactions- Nuclear Fission Consider a neutron bombarding a 235 U nucleus:

18. Nuclear Fission During fission, the incoming neutron must move slowly because it is absorbed by the nucleus, The heavy 235 U nucleus can split into many different daughter nuclei, e.g. 1 0 n + 238 92 U  142 56 Ba + 91 36 Kr + 3 1 0 n releases 3.5  10 -11 J per 235 U nucleus. For every 235 U fission 2.4 neutrons are produced. Each neutron produced can cause the fission of another 235 U nucleus.

19. Nuclear Fission

20. Chapter 22: Nuclear Reactions-Nuclear Fusion Light nuclei can fuse to form heavier nuclei.. 2 1 H + 3 1 H  4 2 He + 1 0 n Fusion products are not usually radioactive, so fusion is a good energy source. High energies are achieved by high temperatures: the reactions are thermonuclear.

21. Chapter 23: Rates of Radioactive Decay: Half-Life (t 1/2 ) Definition Multiple Half-Lives

22. Chapter 23: Rates of Radioactive Decay: Half-Life (t 1/2 ) Definition Multiple Half-Lives

23. Chapter 23: Rates of Radioactive Decay: Half-Life (t 1/2 ) t 1/2Twenty Micrograms of a radioisotope exists at time zero. t 1/2 = 12 hours. How much of this radioisotope remains after 12 hours, 24 hours, 36 hours, 48 hours?

24. Rates of Radioactive Decay

26. Rates of Radioactive Decay: Half-Life (t 1/2 ) Each isotope has a characteristic half-life. Half-lives are not affected by temperature, pressure or chemical composition. Natural radioisotopes tend to have longer half- lives than synthetic radioisotopes.

27. Rates of Radioactive Decay: Half-Life (t 1/2 ) Half-lives can range from fractions of a second to millions of years. Naturally occurring radioisotopes can be used to determine how old a sample is. This process is carbon/radioactive dating.

28. Calculation of Half-Life Radioactive decay is a first order process: Rate = kN The rate of decay is called activity (disintegrations per unit time). If N 0 is the initial number of nuclei and N t is the number of nuclei at time t, then

29. Calculation of Half-Life Ln [A]= (-kt) + ln A º / A t

30. Half-Life Calculations First Order Rate Constant is Related to Half Life

31. Rates of Radioactive Decay Rates of Radioactive Decay Carbon Dating For us to detect 14 C the object must be less than 50,000 years old. The half-life of 14 C is 5,730 years. We assume the ratio of 12 C to 14 C has been constant over time. It undergoes decay to 14 N via  -emission: 14 6 C  14 7 N + 0 -1 e

32. Problem A wooden artifact from a Chinese temple has a 14 C activity of 24.9 counts per minute (CPM) as compared with an activity of 32.5 CPM for a standard of zero age. From the half-life of 14 C decay, 5715 years, determine the age of the artifact.

33. Problem 21.37 (9 th Edition) (t 1/2 )t= 3.3232 (t 1/2 ) log N º / N t (t 1/2 ) = 5715 years(t 1/2 ) = 5715 years N º = 32.5 cpm N t = 24.9 cpm time/age = 2.20E3 years

34. Example 23.1, Page 969 Balance the following nuclear equations and identify X 212 Po  208 Pb + X 137 Cs  137 Ba + X

35. Detection of Radioactivity Matter is ionized by radiation. Geiger counter determines the amount of ionization by detecting an electric current. A thin window is penetrated by the radiation and causes the ionization of Ar gas. The ionized gas carried a charge and so current is produced. The current pulse generated when the radiation enters is amplified and counted.

36. Detection of Radioactivity

37. Chapter 22: Mass-Energy Conversions in Nuclear Reactions Einstein showed that mass and energy are proportional: E = mc 2 If a system loses mass it loses energy (exothermic). If a system gains mass it gains energy (endothermic).

38. E = mc 2 Since c 2 is a large number (8.99  10 16 m 2 /s 2 ) small changes in mass cause large changes in energy. Mass and energy changed in nuclear reactions are much greater than in nonnuclear/chemical reactions.

39. Chapter 22: Energy Changes in Nuclear Reactions Nuclear Binding Energies The mass of a nucleus is less than the mass of their nucleons. Mass defect is the difference in mass between the nucleus and the masses of nucleons.

40. Calculation of Binding Energy Binding energy is the energy required to separate a nucleus into its nucleons. ∆E = ∆mc 2 c=2.99792458E8 meters/seconds

41. Calculation of Binding Energy MUST CONNECT MASS CHANGES TO ENERGY KINETIC ENERGY=1/2 (MASS VELOCITY) 2 Joule = kg (m 2 /second 2 )

42. Problem Calculate the binding energy per nucleon for the following nuclei. (a) 12 6 C (nuclear mass = 11.996708 amu) Proton = 1.0072765 amu Neutron = 1.0086649 amu Amu = 1.66053873E-24 grams

43. Problem Joule = kg (m 2 /second 2 ) Calculated mass = [(6) (1.0072765 amu) + (6) (1.0086649 amu)] = Mass defect = calculated mass – actual mass =

44. Energy Changes in Nuclear Reactions Nuclear Binding Energies

45. Binding Energy: Units The energy carried by radiation is usually given in electron volts (eV) 1 eV is the energy an electron receives when accelerated under the influence of 1 volt It is related to the joule as: 1 eV = 1.602E-19 Joules

46. Recognition of Radisotopes: Nuclear Stability 14 C, 12 C, 13 C Look for neutron/protron ratio Look for magic numbers Or, Refer to A Belt of Stability

47. Recognition of Radioistopes: Neutron/Proton Ratio If N/P ratio = 1, probably stable. If N/P ratio < 1, maybe radioactive –Positron Emission –Electron Capture If N/P ration > 1, maybe radioactive Probably Beta Emitter

48. Recognition of Radioisotopes: Neutron/Proton Ratio Elements with Atomic Number of 83 or higher are usually radioactive. –Alpha, and Gamma Emitters

49. Recognition of Radioisotopes: Magic Numbers Isotopes with specific numbers of protons or neutrons are more stable than the rest The magic numbers are: 2, 8, 20, 28, 50, 82, and 126 Magic numbers do not cancel the need for a favorable neutron:proton ratio 4 He 16 O 208 Pb Magic numbers supports the hypothesis that a nucleus has a shell structure with energy levels like those for electrons

50. Recognition of Radioisotopes: Band/Belt of Stability Band/Belt of Stability will be provided. Most Reliable method

51. Isotopes with atomic number greater than 83 tend to be alpha emitters Their nuclei have too many protons. The most efficient way to lose protons is by the loss of an alpha particle. Isotopes occurring above and to the left of the band of stability tend to be beta emitters The neutron:proton ration is apparently too high. A nucleus that undergoes beta decay loses a neutron and gains a proton which reduces the ratio. For example, beta decay by fluorine-20 decreases the neutron:proton ratio from 11/9 to 10/10.

52. The band of stability. A 1:1 ratio of neutrons to protons is indicated by the straight line. The band of stability curves away from this. More protons require more neutrons to provide a compensating nuclear strong force and to dilute the electrostatic proton- proton repulsions.

53. Isotopes lying below and to the right of the band are positron emitters These nuclei have too few neutrons to be stable. Positron emission increases the neutron:proton ratio. – For example, a fluorine-17 has a neutron:proton ratio of 8/9. Positron emission converts it into oxygen-17, an isotope in the band of stability with a neutron:proton ratio of 9/8.

54. Recognition of Radioisotopes Band/Belt of Stability Isotopes lying below and to the right of the band are positron emitters These nuclei have too few neutrons to be stable. Positron emission increases the neutron:proton ratio. – For example, a fluorine-17 has a neutron:proton ratio of 8/9. Positron emission converts it into oxygen-17, an isotope in the band of stability with a neutron:proton ratio of 9/8.

55. Biological Effect of Radioactivity The radiation can generate very reactive species with unpaired electrons called free radicals or simply radicals target + radioactivity  free radicals Once the radicals are generated, they can start a chain of undesirable chemical events

56. Terms Used to Measure Exposure to Radioactivity Must measure rate of radioactivity production. Must Also Measure Energy Deposited by Radioactivity. Must Also Measure Biological Damage.

57. The curie (Ci), named after Marie Curie, is an older unit equal to the activity of a 1.0 g sample of radium-226 SI Equivalent Unit is the Becquerel (Bq). Bq = one disintegration per second 1 Ci = 3.7E10 Bq

58. The energy equivalent to the quantity of radiation absorbed by some material is defined in units of absorbed dose or simply dose The SI unit of absorbed dose is the gray (Gy) 1 Gy = 1 J of energy absorbed per kg of material The Radiation Absorbed Dose or rad is an older unit 1 rad = 0.01 Gy

59. A different unit is used for measuring the biological effects of radiation because the effects depend on the type and energy of the radiation The sievert (Sv) is the SI unit of dose equivalent, H The dose equivalent (in Sv) is a product of the absorbed dose (D) from some radiation source times a factor (Q) that takes into account the biologically significant properties times any other factor (N) bearing on the net effect The rem is an older unit for dose equivalent: 1 rem = 0.01 Sv A typical X-ray involves about 0.007 rem

60. –A whole body dose of 25 rem (0.25 Sv) induces noticeable changes in blood –A set of symptoms called radiation sickness develops at about 100 rem, and becomes severe at about 200 rems Symptoms include: nausea, vomiting, a drop in white blood count, diarrhea, dehydration, prostration, hemorrhaging, and loss of hair –If everyone in a large group received 400 rem, half would die in 60 days –600 rem would kill everyone within a week

61. Natural and man made sources of radiation contribute to a combined background radiation total of about 360 mrem per person per year About 82% of the background radiation is from natural sources and the remaining 18% from medical sources (much in the form of X-rays) Radiation sources should be avoided because the intensity of radiation diminishes as 1/(distance from a source) 2

62. Problem 23.6-C: Page 1017 Complete the following nuclear equation and identify X. 59 27 Co + 1 0 n  56 25 Mn + X

63. Problem 23.14-A, B: Page 1017 For each pair of isotopes listed, predict which one is less stable. (A) 6 3 Li or 9 3 Li (B) 23 11 Na or 25 11Na

64. Problem 23.34-A: Page 1018 Write a balanced nuclear equation for the following reaction and identify X. d (deuterium) is 2 1 H (A) 80 34 Se(d, p)X

65. Problem 23.36: Page 1018 A long cherished dream of alchemists was to produce gold from cheaper and more abundant elements. This dream was finally realized when 198 80 Hg was converted into gold by nuclear bombardment. Write a balanced equation for this reaction.

66. Problem 23.86: Page 1020 A 0.0100 gram sample of a radioactive isotope with a half-life of 1.3E9 years decays at the rate of 2.9E4 dpm. Calculate the molar mass of the isotope.

67. Strategy: 23.86 Have number of grams. Need number of moles. Calculate value of k. Since the unit of this rate are DP minute (DPM), convert t1/2 to minutes Substitute in: Rate = k N –“N” can be number of Nuclei –Calculate “N” –Use Avogardo’s number to convert value of “N” to moles. –0.0100 grams/calculated number of moles

68. Problem 23.95-A: Page 1021 In 2006, an ex-KGB agent was murdered in London. Subsequent investigation showed that the cause of death was poisoning with the radioactive isotope 210Po, which was added to his drinks/food. (A) 210 Po is prepared by bombarding 209 Bi with neutrons. Write the equation for this reaction

69. Problem 23.95-D: Page 1021 In 2006, an ex-KGB agent was murdered in London. Subsequent investigation showed that the cause of death was poisoning with the radioactive isotope 210Po, which was added to his drinks/food. (D) t 1/2 = 138 d and decays by alpha emission. Calculate the energy of an emitted alpha particle. Assume both the parent and daughter nuclei to have zero kinetic energy. The atomic masses are: 210 Po (209.9825 amu); 206 Pb (205.97444 amu) ; 4 2 α (4.00150 amu)

70. Strategy: 23.86 D Need mass defect: ∆m = final – initial Δm = (mass 206Pb + mass α) − mass 210Po ∆E = (Δm) C 2 Calculated value of ∆E represents value of alpha particle since it is only on the reactant side ∆E = 1.04E-12 J

71. Problem 23.95-E: Page 1021 In 2006, an ex-KGB agent was murdered in London. Subsequent investigation showed that the cause of death was poisoning with the radioactive isotope 210 Po, which was added to his drinks/food. (E) Ingestion of 1 µg of 210 Po could prove fatal. What is the total energy released by this quantity of 210 Po?

72. Strategy: 23.86 E The energy calculated in part (d) is for the emission of one α particle. The total energy released in the decay of 1μg of 210 Po is: (1.0E-6 g 210 Po) (mole 210 Po /209.98285 g) (1 mole α /mole 210 Po) (6.022E23 α particles/1 mole α) (1.04E-12 J/1 α particle = 2.98E3 J