1 / 70

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

Download Presentation

Nuclear Chemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related