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Nuclear Chemistry

Nuclear Chemistry. Chapter 23. Radioactivity . Emission of subatomic particles or high-energy electromagnetic radiation by nuclei Such atoms/isotopes said to be radioactive. Its discovery. Discovered in 1896 by Becquerel Called strange, new emission uranic rays

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Nuclear Chemistry

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  1. Nuclear Chemistry Chapter 23

  2. Radioactivity • Emission of subatomic particles or high-energy electromagnetic radiation by nuclei • Such atoms/isotopes said to be radioactive

  3. Its discovery • Discovered in 1896 by Becquerel • Called strange, new emission uranic rays • Because emitted from uranium • Marie Curie discovered two new elements both of which emitted uranic rays • Po & Ra • Uranic rays became radioactivity

  4. Types of radioactivity • Rutherford and Curie found that emissions produced by nuclei • Different types: • Alpha decay • Beta decay • Gamma ray emission • Positron emission • Electron capture

  5. Isotopic symbolism • Remember from 139/141? • Let’s briefly go over it • Nuclide = isotope of an element • Proton = 11p • Neutron = 10n • Electron = 0-1e

  6. Types of decay: alpha decay • Alpha () decay • Alpha () particle: helium-4 bereft of 2e- • = 42He (don’t write He+2) • Parent nuclide  daughter nuclide + He-4 • 23892U  23490Th + 42He • Daughter nuclide = parent nuclide atomic # minus 2 • Sum of atomic #’s & mass #’s must be = on both sides of nuclear equation!

  7. Alpha decay • Has largest ionizing power • Ability to ionize molecules & atoms due to largeness of -particle • But has lowest penetrating power • Ability to penetrate matter • Skin, even air, protect against -particle radiation

  8. Beta decay • Beta () decay • Beta () particle = e- • How does nucleus emit an e-? •  neutron changes into proton & emits e- •  10n  11p + 0-1e • Daughter nuclide = parent nuclide atomic number plus 1

  9. Beta decay • Lower ionizing power than alpha particle • But higher penetration power • Requires sheet of metal or thick piece of wood to arrest penetration •  more damage outside of body, but less in (alpha particle is opposite)

  10. Gamma ray emission • Gamma () ray emission • Electromagnetic radiation • High-energy photons • 00 • No charge, no mass • Usually emitted in conjunction with other radiation types • Lowest ionizing power, highest penetrating power • Requires several inches lead shielding

  11. Positron emission • Positron = antiparticle of e- •  same mass, opposite charge • (Collision with e- causes -ray emission) • Proton converted into neutron, emitting positron • 0+1e • 11p 10n + 0+1e • 3015P  3014Si + 0+1e • Atomic # of parent nuclide decreases by 1 • Positrons have same ionizing/penetrating power as e-

  12. Electron capture • Particle absorbed by, instead of ejected from, an unstable nucleus • Nucleus assimilates e- from an inner orbital of its e- cloud • Net result = conversion of proton into neutron • 11p+ 0-1e  10n • 9244Ru + 0-1e 9243Tc • Atomic # of parent nuclide decreases by 1

  13. Problems • Write a nuclear equation for each of the following: • 1. beta decay in Bk-249 • 2. electron capture in I-111 • 3. positron emission in K-40 • 4. alpha decay of Ra-224

  14. The valley of stability • Predicting radioactivity type • Tough to answer why one radioactive type as opposed to another • However, we can get a basic idea • Neutrons occupy energy levels • Too many lead to instability

  15. Valley of stability • In determining nuclear stability, ratio of neutrons to protons (N/Z) important • Notice lower part of valley (N/Z = 1) • Bi last stable (non-radioactive) isotopes • N/Z too high • Above valley: too many n, convert n to p • Beta-decay • N/Z too low • Below valley, too many p, convert p to n • Positron emission/e—capture and, to lesser extent, alpha-decay

  16. Predict type of radioactive decay • 1. Mg-28 • 2. Mg-22 • 3. Mo-102

  17. Magic numbers • Actual # of n & p affects nuclear stability • Even #’s of both n & p give stability • Similar to noble gas electron configurations • 2, 10, 18, 36, etc. • Since nucleons (= n+p) occupy energy levels within nucleus • Magic numbers • N or Z = 2, 8, 20, 28, 50, 82, and N = 126

  18. Radioactive decay series

  19. Detecting radioactivity • Particles detected through interactions w/atoms or molecules • Simplest  film-badge dosimeter • Photographic film in small case, pinned to clothing • Monitors exposure • Greater exposure of film  greater exposure to radioactivity

  20. Geiger counter • Emitted particles pass through Ar-filled chamber • Create trail of ionized Ar atoms • Induced electric signal detected on meter and then clicks • Each click = particle passing through gas chamber

  21. Scintillation counter • Particles pass through material (NaI or CsI) that emits UV or visible light due to excitation • Atoms excited to higher E state • E releases as light, measured on meter

  22. Radioactive decay kinetics • All radioactive nuclei decay via 1st-order kinetics •  rate of decay  to # of nuclei present • Rate = kN • Half-life = time taken for ½ of parent nuclides to decay to daughter nuclides

  23. Decay of Rn-220

  24. Problem • Pu-236 is an -emitter w/half-life = 2.86 years. If sample initially contains 1.35 mg, what mass remains after 5.00 years? • How long would it take for 1.35 mg sample of Pu-236 above to decay to 0.100 mg? • Assume 1.35 mg/1 L air

  25. Solution

  26. Devised in 1949 by Libby at U of Chicago Age of artifacts, etc., revealed by presence of C-14 C-14 formed in upper atmosphere via: 147N + 10n  146C + 11H C-14 then decays back to N by -emission: 146C  147N + 0-1e; t1/2 = 5730 years There is an approximately constant supply of C-14 Taken up by plants via 14CO2 & later incorporated in animals Living organisms have same ratio of C-14:C-12 Once dead, no longer incorporating C-14  ratio decreases 5% deviation due to variance of atmospheric C-14 Bristlecone pine used to calibrate data Carbon-dating good for 50,000 years Radiometric dating: radiocarbon dating

  27. Problem • Artifact is found to have C-14 decay rate of 4.50 disintegration/min  g of carbon. • If living organisms have a decay rate of 15.3, how old is the artifact? • Given decay rate is  to amount of C-14 present.

  28. Solution

  29. Radiometric dating: uranium/lead dating • Relies on ratio of U-238:Pb-206 w/in igneous rocks (rocks of volcanic origin) • Measures time that has passed since rock solidified • t1/2 = 4.5 x 109 years

  30. Example • A meteor contains 0.556 g Pb-206 (to every 1.00 g U-238). Determine its age.

  31. Solution

  32. Problem • A rock from Australia was found to contain 0.438 g of Pb-206 to every 1.00g of U-238. Assuming that the rock did not contain any Pb-206 at the time of its formation, how old is the rock?

  33. Solution

  34. Fission • Meitner, Strassmann, and Hahn discovered fission • Splitting of uranium-235 • Instead of making heavier elements, created a Ba and Kr isotope plus 3 neutrons and a lot of energy • Sample rich in U-235 could create a chain rxn • To make a bomb, however, need critical mass = enough mass of U-235 to produce a self-sustaining rxn

  35. Nuclear power • In America, about 20% electricity generated by nuclear fission • Imagine: • Nuclear-powered car • Fuel = pencil-sized U-cylinder • Energy = 1000 20-gallon tanks of gasoline • Refuel every 1000 weeks (about 20 years) • !

  36. Nuclear power plant • Controlled fission through U fuel rods (3.5% U-235) • Rods absorb neutrons • Retractable • Heat boils water, making steam, turning turbine on generator to make electricity

  37. Comparing • Typical nuclear power plant makes enough energy for city of 1,000,000 people and uses about 50 kg of fuel/day • No air pollution/greenhouses gases • But, nuclear meltdown (overheating of nuclear core) is a potential threat • No problem! • Also, waste disposal • Location, containment problems

  38. Comparing • Coal-burning power plant uses about 2,000,000 kg of fuel to make same amount of energy • But, releases huge amounts of SO2, NO2, CO2

  39. Mass to energy • E = mc2 • Explains relationship between energy formation and matter loss • Amount of energy released in U-235 fission per atom of U-235 is 2.8 x 10-11 J • BUT, amount of energy released in U-235 fission per 1 mole of U-235 is 1.7 x 1013 J! • More than a million times more energy per mole than a chemical rxn!

  40. Mass defect • Mass products < mass reactants • Difference in mass due to conversion of mass into energy • Called mass defect • Nuclear binding energy is energy corresponding to mass defect • Amount of energy required to break apart nucleus into nucleons (n + p)

  41. Some more stuff • Nuclear physicists use eV or MeV (mega eV) instead of joules • 1 MeV = 1.602 x 10-13 J • 1 amu = 931.5 MeV • Energy per nucleus and not per mole • To compare energy of 1 nucleus to another, calculate binding energy per nucleon • = nuclear binding energy of nuclide per #of nucleons (n + p) in nuclide • As binding energy per nucleon increases so does stability of species

  42. Example • Calculate the mass defect and nuclear binding energy per nucleon (in MeV and in J) for: • 42He • Made from: 211H + 210n • 11H = 2 x 1.00783 amu • 210n = 2 x 1.00866 amu • Net mass = 4.03298 amu

  43. Solution

  44. Problem • Calculate the mass defect and nuclear binding energy per nucleon (in MeV) for C-16 • Consider C-16 being made from 6 11H & 10 10n • C-16 mass = 16.014701 amu

  45. Solution

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