1. Chapter 21: Nuclear Chemistry Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop

2. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 2 How are atoms formed?  Big Bang—Intense heat ~10 9 K  Cooled quickly to 10 6 K—T of stars  e –, p, n formed and joined into nuclei—atoms  Mostly H and He (as in our sun)  Rest of elements formed by nuclear reactions  Fusion—two nuclei come together to form another heavier nucleus  Fission—one heavier nucleus splits into lighter nuclei  Various other types of reactions + +

3. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 3 Nuclear Shorthand  Nucleons  Subatomic particles found in the nucleus  Protons (p)  Neutrons (n)  Nuclide  Specific nucleus with given atomic number (Z )  Atomic Number (Z )  Number of protons in nucleus  Determines chemical properties of nuclide  Z = p  Mass Number (A)—mass of nuclide  A = n + p

4. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 4 Shorthand for Writing Nuclides  Where X = atomic symbol  e.g.  In the neutral atom: e – = p = Z  Isotopes  Nuclides with same Z (same number of p), but different A (different n) HydrogenDeuteriumTritium 1 p1 p + 1 n1 p + 2 n

5. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 5 Radioactivity  Radioactive isotopes  Isotopes with unstable atomic nuclei  Emit high energy streams of particles or electromagnetic radiation  Radionuclides  Another name for radioactive isotopes  Undergo nuclear reactions  Uses  Dating of rocks and ancient artifacts  Diagnosis and treatment of disease  Source of energy

6. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 6 Mass Not Always Constant  Mass of particle not constant under all circumstances  It depends on velocity of particle relative to observer  As approaches speed of light, mass increases  When v goes to zero  Particle has no velocity relative to observer  v/c  0  Denominator  1  and m = m o m = mass of particle v = velocity of particle m  = rest mass c = speed of light

7. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 7 Why don’t we observe mass change?  In lab and ordinary life, velocity of particle is small  Only see mass vary with speed as velocity approaches speed of light, c  As v  c,(v/c)  0 and m  ∞  In lab, m = m o within experimental error  Difference in mass too small to measure directly  Scientists began to see relationship between mass and total energy  Analogous to potential and kinetic energies

8. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 8 Law of Conservation of Mass and Energy  Mass and energy can neither be created nor destroyed, but can be converted from one to the other.  Sum of all energy in universe and all mass (expressed in energy equivalents) in universe is constant  Einstein Equation   E = (  m o )c 2  Where c = 2.9979 × 10 8 m/s

9. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 9 Mass Defect  Rest mass of nuclide is always less than sum of masses of all individual nucleons (neutrons and protons) in that same nuclide  Mass is lost upon binding of neutrons and protons into nucleus  When nucleons come together, loss of mass translates into release of enormous amount of energy by Einstein's relation Energy released = – Nuclear Binding Energy  Nuclear Binding Energy  Amount of energy must put in to break apart nucleus

10. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 10 What is Mass Loss? For given isotope of given Z and A or

11. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 11 Ex. 1 Binding Energy Calculation What is the binding energy of 7 Li 3+ nucleus? Step 1. Determine mass loss or mass defect A. Determine mass of nucleus mass of 7 Li 3+ = m ( 7 Li isotope) – 3m e = 7.016003 u – 3(0.0005485 u) = 7.0143573 u B. Determine mass of nucleons mass of nucleons = 3 m p + 4 m n = 3(1.007276470 u) + 4(1.008664904 u) = 7.056489026 u

12. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 12 Ex. 1 Binding Energy Calculation (cont.) C.  m = m nucleus – m nucleons = 7.0143573 u – 7.056489026 u = –0.0421317 u = mass lost by nucleons when they form nucleus Step 2. Determine energy liberated by this change in mass  E = (  m o )c 2  E = – 6.287817 × 10 –12 J/atom

13. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 13 Ex. 1 Binding Energy Calculation (cont.)  E = –6.287817 × 10  12 J/atom × 6.0221367 x 10 23 atoms/mole  E = –3.78655 × 10 12 J/mole = –3.78655 × 10 9 kJ/mole Compare this to:  10 4 – 10 5 J/mol (10 2 – 10 3 kJ/mol) for chemical reactions  Nuclear ~ 1 – 10 million times larger than chemical reactions!!

14. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 14 MeV (Energy Unit)  Nuclear scientists find it convenient to use a different Energy unit: MeV (per atom)  Electron volt (eV)  Energy required to move e  across energy potential of 1 V  1 eV = 1.602 × 10 –19 J  M(mega) = 1 × 10 6  So 1 MeV = 1 × 10 6 eV  1 MeV = 1.602 × 10 –13 J

15. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 15 Ex. 1 Binding Energy Calculation in MeV  For Ex. 1. Converting E to MeV gives  Often wish to express binding energy per nucleon so we can compare to other nuclei  For Li 3+ with 3 1 p and 4 n this would be

16. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 16 Ex. 2 Calculate E Released  The overall reaction in the sun responsible for the energy it radiates is  How much energy is released by this reaction in kJ/mole of He? m ( 1 H) = 1.00782 u m ( 4 He) = 4.00260 u m ( 0  + ) = 0.00054858 u

17. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 17 Ex. 2 Calculate E Released (cont.)   m = m products – m reactants   m = m ( 4 He) + 2m (e + ) – 4m ( 1 H)   m = 4.00260 u + 2(0.00054858 u) – 4(1.00782 u)   m = –0.02758 u  [We will convert u to kg, kg m 2 /s 2 to J, and atoms to moles in the following calculation]  E = –2.479 × 10 12 J/mol = –8.268 × 10 9 kJ/mol

18. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Determine the binding energy, in kJ/mol and MeV/atom, for an isotope that has a mass defect of –0.025861 u. A. –2.3243 × 10 9 kJ/mol; 24.092 MeV/atom B. –3.8595 × 10 –12 kJ/mol; 24.092 MeV/atom C. –7.7529 kJ/mol; 8.03620 × 10 –8 MeV/atom D. –2.3243 × 10 9 kJ/mol; 4.1508 × 10 –2 MeV/atom 18

19. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! - Solution 19

20. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 20  Divide binding energy E B by mass number, E B /A  Get binding energy per nucleon Binding Energies per Nucleon

21. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 21 Implications of Curve  Most E B /A in range of 6 – 9 MeV (per nucleon)  Large binding energy E B /A means stable nucleus  Maximum at A = 56  56 Fe largest known E B /A  Most thermodynamically stable  Nuclear mass number (A) and overall charge are conserved in nuclear reactions  Lighter elements undergo fusion to form more stable nuclei

22. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 22 Implications of Curve Fusion  Researchers are currently working to get fusion to occur in lab  Heavier elements undergo fission to form more stable elements Fission  Reactions currently used in bombs and power plants ( 238 U and 239 Pu)  As stars burn out, they form elements in center of periodic table around 56 Fe

23. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 23 Radioactivity  Spontaneous emission of high energy particles from unstable nuclei  Spontaneous emission of fundamental particle or light  Nuclei falls apart without any external stimuli  Discovered by Becquerel (1896)  Extensively studied by Marie Curie and her husband Pierre (1898  early 1920's)  Initially worked with Becquerel

24. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 24 Fun Facts  Marie and Pierre Curie discovered polonium and radium  Nobel Prize in Physics 1903  For discovery of Radioactivity  Becquerel, Marie and Pierre Curie—all three shared  Nobel Prize in Chemistry 1911  For discovery of Radium and its properties  Marie Curie only  Marie Curie - first person to receive two Nobel Prizes and in different fields

25. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 25 Discovery of Radioactivity  Initially able to observe three types of decay  Labeled them , ,  rays (after first three letters of Greek alphabet)  If they pass through an electric field, very different behavior

26. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 26 Discovery of Radioactivity   rays attracted to negative pole so its positively charged   rays attracted to positive pole so its negatively charged   rays not attracted to either so its not charged   

27. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 27 Nuclear Equations  Used to symbolize decay of nucleus e.g. 238 U  234 Th + parent daughter  Produce new nuclei so need separate rules to balance Balancing Nuclear Equations a.Sum of mass numbers (A, top) must be same on each side of arrow b.Sum of atomic numbers (Z, bottom) must be same on each side of arrow 9290

28. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 28 Types of Spontaneous Emission 1.Alpha (  ) Emission He nucleus 2 n + 2 p A = 4 and Z = 2  Daughter nuclei has: A decreases by 4  A = – 4 Z decreases by 2  Z = – 2  Very common mode of decay if Z > 83 (large radioactive nuclides)  Most massive particle  e.g.

29. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Balancing Nuclear Equations 29 1.The sum of the mass numbers (A; the superscripts) on each side of the arrow must be the same 2.The sum of the atomic numbers (Z; the subscripts; nuclear charge) on each side of the arrow must be the same  e.g. A: 234 = 230 + 4 Z: 92 = 90 + 2

30. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E  Emission of electrons  Mass number A = 0 and charge Z = –1  But How? No electrons in nucleus!  If nucleus neutron rich — nuclide is too heavy 2. Beta (  – or e – ) Emission 30

31. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 2. Beta (  – or e – ) Emission 31  Charge conserved, but not mass  m  E  Ejected e – has very high KE + emits  Antineutrino variable energy particle  Accounts for extra E generated e.g.

32. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 3. Gamma (  ) Emission  Emission of high energy photons  Often accompanies  or  emission  Occurs when daughter nucleus of some process is left in excited state  Use * to denote excited state  Nuclei have energy levels analogous to those of e – in atoms  Spacing of nuclear E levels much larger   light emitted as  -rays e.g. 32

33. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 4. Positron (  + or e + ) Emission  Emission of e +  Positive electron  Where does  + come from?  If nucleus is neutron poor)  Nuclide too light  Balanced for charge, but NOT for mass 33

34. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 4. Positron (  + or e + ) Emission  Product side has much greater mass!  Reaction costs energy  Emission of neutrino  Variable energy particle  Equivalent of antineutrino but in realm of antimatter  e + emission only occurs if daughter nucleus is MUCH more stable than parent 34

35. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 4. Positron (  + or e + ) Emission What happens to e + ?  Collides with electron to give matter anti-matter annihilation and two high energy  -ray photons  m  E  Annihilation radiation photons  Each with E  = 511 keV  What is antimatter?  Particle that has counterpart among ordinary matter, but of opposite charge  High energy light, massless  Detect by characteristic peak in  -ray spectrum 35

36. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 5. Electron Capture (EC)  e – in 1s orbital  Lowest Energy e –  Small probability that e – is near nucleus  e – actually passes through nucleus occasionally  If it does:  Net effect same as e + emission 36

37. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 37 Types of Spontaneous Emission 6. Neutron Emission = ( )  Fairly rare  Occurs in neutron rich nuclides  Does not lead to isotope of different element 7. Proton Emission = ( )  Very rare

38. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 38 Types of Spontaneous Emission 8. Spontaneous Fission  No stable nuclei with Z > 83  Several of largest nuclei simply fall apart into smaller fragments  Not just one outcome, usually several different—see distribution

39. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 39 Summary—Common Processes 1. Alpha (  ) Emission  Very common if Z > 83 2. Beta (   ) Emission e –  Common for neutron-rich nuclides—below belt of stability 3. Positron (  + ) Emission e +  Common for neutron-poor nuclides—above belt of stability 4. Electron Capture (EC)  Occurs in neutron-poor nuclides, especially if Z > 40 5. Gamma (  ) Emission  Occurs in metastable nuclei (in nuclear excited state)

40. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 40 Learning Check  Complete the following table which refers to possible nuclear reactions of a nuclide: Emission  Z =  p nn ee AA New Element?  –– ++ EC  –2–2–2–20–4–4yes +1 –1–10yes +1–1–10yes–1–1 0000no +1 –1–10yes

41. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 41 Learning Check Balance each of the following equations a. b. c. d. e.

42. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! What is the missing species,, in the following nuclear reaction? A. B. C. D. 42

43. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 43 What Holds Nucleus Together?  Consider nucleus  Neutrons and protons in close proximity  Strong proton-proton repulsions  Neutrons spread protons apart  Neutron to proton ratio increases as Z increases Strong Forces  Force of attraction between nucleons  Holds nuclei together  Overcomes electrostatic repulsions between protons  Binds protons and neutrons into nucleus

44. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 44 Table of Nuclides  Chart where Rows = different atomic number Columns = different number of neutrons  Symbol entered if element is known  Stable nuclei  Natural abundance entered below symbol  Shaded area  Trend of stable nuclei = Belt of Stability  Z ≈ number of neutrons (for elements 1 to 20)  Unstable nuclei  Give type(s) of radioactive decay (spontaneous)  Outer edges, most of atoms

45. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 45 Table of Nuclides Atomic number (Z = number of protons) Number of neutrons

46. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 46 Table of Nuclides  Shaded area = stable nuclei  Trend of stable nuclei = diagonal line = Belt of Stability  Z ≈ number of neutrons (for elements 1 to 20)  Note: only a small corner of table is shown. (The complete is in Handbook of Chemistry and Physics)

47. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 47 Belt of Stability  Each isotope is a dot  Up to Z = 20  Ratio n /Z = 1  As Z increases, n > Z and  By Z = 82, n/Z ~1.5  n = number of neutrons  Z = number of protons n Z = p 1n:1p Stable nuclide, natural  Unstable nuclide, natural  Unstable nuclide, synthetic Band of Stability 1n:1p 1.1n:1p 1.2n:1p 1.3n:1p 1.4n:1p 1.5n:1p e – emitters e + emitters

48. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 48 How To Predict if Nuclei are Stable 1) Atomic Mass = weighted average of masses of naturally occurring isotopes, i.e. most stable ones 2) Compare atomic mass of element to A (atomic mass number) of given isotope and see if it is more or less  Atomic Mass > Atoo light to be stable  Atomic Mass < Atoo heavy to be stable Ex.Atomic MassConclusion 180 Os 135 I Final note:  All nuclei with Z > 83 are radioactive 190.2 126.9 Too light, neutron poor Too heavy, neutron rich

49. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 49 More Patterns of Stability  If we look at stable and unstable nuclei, other patterns emerge  283 stable nuclides (out of several thousand known nuclides)  If we look at which have even and odd numbers of protons (Z) and neutrons (n); patterns emerge Zn# stable nuclides even 165 evenodd56 oddeven53 odd 5 2 H, 6 Li, 10 B, 14 N, 138 La

50. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 50 More Patterns of Stability  Clearly NOT random: even must imply greater stability  Not too surprising  Same is true of electrons in molecules  Most molecules have an even number of electrons, as electrons pair up in orbitals  Odd electron molecules, radicals, are very unstable, i.e. very reactive!!

51. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 51 Magic Numbers  Look at binding energies, see certain numbers of protons and neutrons result in special stability  Called Magic Numbers  1 n and 1 p in separate shells  Magic numbers (for both 1 n and 1 p) are 2, 8, 20, 28, 50, 82, 126…  For e – pattern of stability is: 2, 10, 18, 36, 54, 86…(Noble gases)

52. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 52 Magic Numbers  Special stability of noble gases due to closed shells of occupied orbitals  Structure of nucleus can also be understood in terms of shell structure  With filled shells of neutrons and protons having added stability  At some point adding more neutrons to higher energy neutron shells decreases stability of nuclei with too high a neutron to proton ratio

53. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Isotopes above the band of stability are more likely to: A. emit alpha particles B. emit gamma rays C. capture electrons D. emit beta particles 53

54. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 54 Radioactive Nuclei Found in Nature  Non-naturally occurring elements (man-made unstable) are denoted by having atomic mass in parentheses  All nuclei with Z > 83 are radioactive  Yet some elements with Z between 83 and 92 occur naturally  Atomic weight is NOT in parentheses How can this be?  There are three heavy nuclei, which have very long half-lives  Long enough to have survived for billions of years  Each parent of natural decay chain

55. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 55 Decay Chains 238 U half-life (  ½ ) = 4.5 billion years   emitter  Daughter 234 Th is also radioactive   – emitter  Half-life much shorter  Long sequence of emissions,  and  –  Recall that  emission changes A by 4, while  – emission  A = 0  Result: every member of chain has A = (4n + 2) where n = some simple integer

56. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 56 238 Uranium Decay Chain 238 U 234 Th 5  10 9 y  25 d ,  234 Pa 7 hr ,  234 U 5.7  10 5 y  230 Th ,  8  10 4 y 226 Ra ,  2  10 3 y 222 Rn  4 d 218 Po 3 m  214 Pb ,  27 m 214 Bi 20 m ,  214 Po 1.6  10 –4 s  210 Pb 22 y ,  210 Bi  5 d 210 Po  138 d 206 Pb 92 90 91 92 90 8886 84 82 83 84 82 84 83 A stable isotope

57. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 57

58. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 58 Decay Chains  Final stable member of sequence is 206 Pb  Some intermediate nuclides have reasonably short half-lives  Still found in nature because they are constantly being replenished by decay of nuclei further up chain  Uranium-containing minerals (pitchblende is most famous) contain many radioactive elements

59. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! When the reaction,, occurs, the particle emitted is: A. an alpha particle B. a beta particle C. an electron D. a gamma ray 59

60. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 60 Transmutation  Change of one isotope for another  Caused by 1.Radioactive decay 2.Bombardment of nuclei with high energy particles   particles from natural emitters  Neutrons from atomic reactors  Protons made by stripping electrons for hydrogen  Protons and  particles can be accelerated in electrical field to give higher E  Mass and energy of bombarding particle enter target nucleus to form compound nucleus

61. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 61 Non-Spontaneous Nuclear Processes  Fusion  Occurs in stars—right now  How elements formed  Induced Fission  Bombard heavy nuclei with neutron

62. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 62 Compound Nucleus  Designated with *  High energy due to velocity of incoming particle  Energy quickly redistributed among nucleons, but usually unstable  To get rid of excess energy, nucleus ejects something  Neutron▪ Proton  Electron▪ Gamma radiation  Decay leaves new nucleus different from original

63. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 63 Example: Transmutation Compound nucleus New nucleus Target nucleus Bombard- ing particle High energy particle

64. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 64 Transmutation  Can synthesize given nucleus in many ways:  Once formed, compound nucleus has no memory of how it was made  Only knows how much energy it has

65. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 65 Transmutation Decay pathway depends on how much energy

66. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 66 Transmutation  Used to synthesize new isotopes  > 900 total  Most not on band of stability  All elements above 93 (neptunium) are man- made  Includes actinides above 93 + 104 – 112 + 114  Heavier elements made by colliding two larger nuclei  Also known as fusion

67. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! What would be the element produced from the fusion of with ? The species would be in a high energy state and in time would undergo decay to other species. A. No B. Lr C. U D. Hs 67

68. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 68 Measuring Radioactive Decay  Atomic radiation = ionizing radiation  Creates ions by knocking off electrons  Geiger Counter  Consists of a tube with a mica window, low pressure argon fill gas and two high voltage electrodes  Detects  and  radiation with enough E to penetrate mica window  Inside tube, gas at low pressure is ionized when radiation enters  Ions allow current to flow between electrodes  Amount of current relates to amount of radiation

69. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 69 Measuring Radioactive Decay  Scintillation Counter  Surface covered with chemical  Emits tiny flash of light when hit by radiation  Emission magnified electronically and counted  Film Dosimeters  Piece of photographic film  Darkens when exposed to radiation  How dark depends on how much radiation exposure over time  Too much exposure, person using must be reassigned to other work

70. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 70 Activity  Number of disintegrations per second  Used to characterize radioactive material  A = kN  k = first order decay constant in terms of number of nuclei rate than concentration  N = number of radioactive nuclides  Law of radioactive decay  Radioactive decay is first order kinetics process

71. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 71 Units of Activity  SI unit  Bequerel (Bq)  1 disintegration per second (dps)  1 liter of air has ~ 0.04 Bq due to 14 C in CO 2  Older unit  Curie (Ci)  3.7 × 10 10 dps = 3.7 × 10 10 Bq  Activity in 1.0 g 226 Ra = 1 Ci

72. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 72 Half-Life  Time it takes for number of nuclides, N t, present at time, t, to fall to half of its value.  Half-lives are used to characterize nuclides  If you know half-life:  Can use to compute k  Can also calculate A of known mass of radioisotope

73. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 73 Ex. 3 Activity of Sr-90 What is the activity of 1.0 g of strontium-90? The half-life = 28.1 years Step 1. Convert t ½ to seconds Step 2. Convert t ½ to k

74. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 74 Ex. 3 Activity of Sr-90 (cont.) Step 3. Convert mass of 90 Sr to number of atoms (N) Step 4. Calculate Activity = kN A = 5.23  10 12 atoms Sr/s  1 disintegration/atom A = 5.23  10 12 dps or 5.23  10 12 Bq

75. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 75 Ex. 4 Mass of 3 H in Sample 3 H, tritium, is a   emitter with a half-life t ½ = 12.26 yrs. MW = 3.016 g/mol. How many grams of 3 H are in a 0.5 mCi sample? Step 1. Convert half-life to seconds as Ci is in disintegrations per second (dps) Step 2. Convert t ½ to k

76. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 76 Ex. 4 Mass of 3 H in Sample (cont.) Step 3. Convert Ci to dps Step 4. Calculate g 3 H to get this activity Step 5. Convert atoms to g = 5.2 × 10 –8 g

77. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 77 Exposure Units  Not all materials equally absorb radiation, thus activity doesn’t describe effect of exposure  1 gray (Gy) = 1 J absorbed energy/kg material  SI unit of absorbed radiation  1 rad = absorption of 10 −2 J/ kilogram of tissue  Older unit  1 Gy = 100 rad  These units don’t take into account type of radiation

78. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 78 Exposure Units Sieverts (Sv)  SI unit of dose equivalent, H  Depends on amount and type of radiation as well as type of tissue absorbing it  H=DQN  H = dose in Sv  D = dose in Gy  Q = radiation properties  N = other factors Rem = older unit  1 Rem = 10 –2 Sv  Still used in medicine

79. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 79 Exposure to Radiation  Typically X ray = 0.007 rem or 7 mrem  0.3 rem/week is maximum safe exposure set by US government  25 rem (0.25 Sv): Causes noticeable changes in human blood  100 rem (1 Sv):  Radiation sickness starts to develop  200 rem (2 Sv):  Severe radiation sickness  400 rem (4 Sv):  50% die in 60 days  Level of exposure or workers at Chernobyl when steam explosion tore apart reactor  600 rem (6 Sv): lethal dose to any human

80. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Workers cleaning up the Fukushima reactors were exposed to as much as 400 mSv units of radiation per hour. How many rems of exposure does this correspond to? A. 4000 rem B. 400 rem C. 40 rem D. 4 rem 80

81. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 81 Why is Radiation Harmful?  Not heat energy  Ability of ionizing radiation to form unstable ions or neutral species with odd (unpaired) electrons  Free radicals  Chemically very reactive  Can set off other reactions  Do great damage in cell

82. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 82 Which Types are Most Harmful?  High energy gamma (  ) radiation and X rays  Massless  High velocity  Penetrate everything but very dense materials, such as lead Which type is least harmful?  Alpha (  ) particles  Most massive  Quickly slow after leaving nucleus  Don’t penetrate skin

83. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 83 Background Radiation  Presence of natural radionuclides means we can’t escape exposure to some background radiation  Cosmic rays (from sun) hit earth  Turn 14 N  13 C  13 C emits  – particles  Incorporated into food chain from CO 2 via photosynthesis  Radiation from soil and building stone  From radionuclides native to Earth’s crust  Top 40 cm of soil hold 1 g radium (  emmiter) /sq kilometer  40 K emit  – particles  Total average exposure 360 mrem/year  82% natural radiation 18% man made

84. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 84 Radiation Intensity  Intensity of radiation varies with distance from the source  Farther from emitter, lower intensity of exposure  Relationship is governed by Inverse Square Law, where:  I is intensity and  d is distance from source

85. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 85 Ex. 5 Inverse Square Law If the activity of a sample is 10 units at 5 meters from the source, what is it at 10 m?  What distance is needed to reduce 1 unit at 1 yd to the 0.05 units?

86. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! How far away from a radioactive source producing 40 rem/hr at a distance of 10 m would you need to be to reduce your exposure to 0.4 rem/hr? A.32 m B.100 m C.200 m D.1000 m 86

87. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 87 Radioactive Decay—Kinetics  Spontaneous decay of any nuclide follows first order kinetics  May be complicated by decay of daughter nuclide  For now consider single step decay processes  Rate of reaction for first order process A  products  In nuclear reaction, consider rate based on number of nuclei N present

88. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 88 Radioactive Decay—Kinetics  The integrated form is: ln N – ln N o = – kt  N = number of nuclei present at time t  N o = number of nuclei present at t = 0  Plot ln N (y axis) versus t (x axis)  Yields straight line—indicative of first order kinetics  Plot of N vs. time gives an exponential decay.

89. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 89 Ex. 6 Activity Calculations 131 I is used as a metabolic tracer in hospitals. It has a half-life, t ½ = 8.07 days. How long before the activity falls to 1% of the initial value? t = 53.6 days

90. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! How many hours will it take a radioisotope with a half-life of 10.0 hours to drop to 12.5% of its original activity? A. 30.0 hrs B. 20.0 hrs C. 40.0 hrs D. 63.2 hrs 12.5% of original activity is 3 half-lives or 30.0 hrs. 90

91. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 91 Radioisotope Dating  How old is an object?  Fields — Geology, Archeology, and Anthropology  Nature provides us with natural clocks or stopwatches A) Radiocarbon Dating (Willard Libby— Nobel Prize in 1960)  Cosmic rays (from space) enter atmosphere  Some react with N in atmosphere forming radioisotope 14 C   – emitter with t ½ = 5730 yr

92. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 92 14 C Dating  14 C becomes incorporated into atmospheric CO 2 in very small quantities  14 C/ 12 C ratio in air is slightly greater than Earth’s crust because of ongoing enrichment  Living organisms breath, eat, etc…  14 C/ 12 C equilibrate with atmosphere  Radioactive 14 C is uniformly distributed around globe  Tested experimentally  Checked vs. counting tree rings, etc.  For precise work, use correction based on alternate methods

93. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 93  HOW? Freshly cut wood samples have ~15.3 cpm per gram of total carbon  cpm = counts per minute   A o = 15.3 cpm/g total C  Assumption: A o was always 15.3 cpm, i.e. cosmic radiation is constant  When organism dies  it stops eating, breathing, etc…  14 C starts to decrease 14 C Dating

94. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 94  Wooden implement in Egyptian tomb (~3000 BC)  Have about half activity of fresh sample  ~5000 years have elapsed  Method is applicable for objects  Few hundred to ~20,000 years  Beyond this  Activity of sample is very low  Experimental uncertainties too big  This method used for dating 1.Charcoal in cave paintings 2.Linen wraps on Dead Sea scrolls 14 C Dating

95. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 95 Ex. 7 C-14 Dating Geologists examine shells found in cliffs. Shells are CaCO 3 and are made by living organisms. The activity of the shells is found to be 6.24 cpm/g total C. How old is the cliff formation? A = 6.24 cpm/g total C A o = 15.3 cpm/g total C t ½ = 5730 yr  Can use N/N o and A/A o interchangeably as A = kN  Since ratio, k cancels

96. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 96 Ex. 7 C-14 Dating (cont.) Rearranging and solving for t t = 7414 yr

97. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 97 B) Other Isotopes Provide Natural Clocks  Minerals (moon rocks) dated using isotopes with much longer half-lives t ½ = 1.27 × 10 9 yr  Compare ratios in rock t ½ = 4.5 × 10 9 yr  Rock with no other source of Pb can be dated using ratios

98. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 98 Ex. 8 Dating with U A sample of lava is found to contain 0.232 g of 206 Pb and 1.605 g 238 U. Since lead is volatile at the temperature of molten lava, all the 206 Pb now present came from the decay of 238 U, calculate the time since the solidification of this rock. Step 1. Mass of 238 U that decayed = = 0.268 g 238 U decayed

99. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 99 Ex. 8 Dating with U (cont.)  Step 2. Mass of 238 U in lava initially (t = 0) N o = 1.605 g + 0.268 g = 1.873 g t = 1.0 × 10 9 yr

100. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! A wooden bowl fragment found at an old camp site thought to be approximately 11,000 years old was submitted for carbon-14 analysis. The sample was found to have 4.67 cpm/g total C. What is the actual age of the sample? A. 4260 yrs B. 3347 yrs C. 9810 yrs D. 2523 yrs t = (5730 yrs × ln(4.67/15.3))/(ln 2) = 9810 yrs 100

101. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 101 Fission  Induce by bombarding unstable nucleus with a slow neutron  Nuclear chain reaction  Neutrons generated keep going  With small mass of 235 U reaction continues, but easily controlled  Some neutrons are lost to environment

102. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 102 Fission  “Critical mass”  Too much 235 U in one place  Too many neutrons absorbed  Too few lost  Uncontrollable fission  Leads to explosion  Use control rods to absorb excess neutrons and keep reaction from going critical

103. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 103 Nuclear Reactor  No chance of nuclear explosion  Critical mass requires pure 235 U  Reactor rods 2 – 4% 235 U rest non-fissionable 238 U  Core meltdown possible  If heat of fission not carried away by cooling water  Or  Explosion possible  High heat of fission splits H 2 O into H and O, which recombine very exothermically and cause a chemical explosion  What happened at Chernobyl

104. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 104 Nuclear Reactors  Could it happen at U.S. reactors?  Extremely unlikely  Chernobyl only single containment system  U.S. has all double containment systems  U.S. extra backup systems - both computer and mechanical that would prevent

105. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 105 Nuclear Reactor  Use heat from nuclear reaction to heat steam turbine  Use to generate electricity

106. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Which of the following fission reactions is Balanced? A. B. C. D. 106

107. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Nuclear Fusion  Occurs when light nuclei join to form heavier nucleus  On a mass basis, fusion yields more than five times as much energy as fission  Source of the energy released in the explosion of a H-bomb  The energy needed to trigger the fusion is provided by the explosion of a fission bomb  Source of energy in stars 107

108. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Thermonuclear Fusion  Uses high temperatures to overcome electrostatic repulsions between nuclei  T required are >100 million °C  Atoms want to fuse stripped of electrons  High initial energy cost  Plasma  Electrically neutral, gaseous mixture of nuclei and electrons  Make plasma very dense (>200 g/cm –3 )  Brings nuclei within 2 fm = 2 × 10 –15 m  Pressures = several billion atm  Not there yet, major problem  Containment of high temperature and pressures  Magnetic field current approach 108