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Nuclear Chemistry

Nuclear Chemistry. Chapter 21. Review Chapter 3. Z = Atomic Number Atomic Number is the number of _______ . Mass Number Number of _______ + ________ Average Atomic mass Weighted average of mass numbers of isotopes What is an isotope? Why are electrons not included in the mass number?.

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Nuclear Chemistry

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  1. Nuclear Chemistry Chapter 21

  2. Review Chapter 3 • Z = Atomic Number • Atomic Number is the number of _______ . • Mass Number • Number of _______ + ________ • Average Atomic mass • Weighted average of mass numbers of isotopes • What is an isotope? • Why are electrons not included in the mass number?

  3. Hydrogen Isotopes • Protium (99.985%) • 1 proton, 0 neutons, 1 electron • Deuterium (0.015%) (Heavy Water) • __ proton, __ neutron, __ electron • Tritium (Rare) (Radioactive) • __ proton, __ neutron, __ electron

  4. Mass of an atom? • Mass of 1 atom is 2.657 x 10-23 • So use another method • Carbon 12 atom weighs 12 Atomic Mass Units • Atomic Mass Unit (AMU) • Easy to find the mass of an atom: • Find mass number or atomic mass + attach AMU as the units • Example: Oxygen = 16 amu OR 15.9994 amu

  5. First some vocab • Nucleons – particles in the nucleus • Nuclide – another name for an atom • Identified by the number of protons + neutrons • Nuclear Reaction – reaction that affects the nucleus of an atom • Transmutation – change in proton number • Change in the identity of a nucleus • Oxygen-16 transmutates via alpha emission to Carbon-12

  6. Mass Defect • When nucleons bind together into a nucleus, they LOSE mass • Mass Defect – (sum of the masses of the protons + neutrons + electrons) – (atomic mass) • Proton mass = 1.007 276 amu • Neutron mass = 1.008 665 amu • Electron mass = 0.000 5486 amu

  7. Find Mass Defect • Helium-4 atom (p. 681) • Helium atom = 2 protons, 2 neutrons, 2 electrons • 2 protons = 2(1.007 276 amu) = 2.014 552 • 2 neutrons = 2(1.008 665 amu) = 2.017 330 • 2 electrons = 2(0.000 5486 amu) = 0.001 097 • TOTAL: 4.032 979 amu • Periodic Table: 4.002 602 • Mass Defect = 4.032 979 amu – 4.002 602 amu • MASS DEFECT = 3.0377 x 10-2amu

  8. Nuclear Binding Energy (NBE) • Definition – The energy released when a nucleus is formed from its nucleons • Mass defect can be converted to NBE by Einstein’s famous equation: • E = mc2 • E = energy m = mass • c = speed of light = 3.00 x 108m/s • Now we will find nuclear binding energy in the previous problem.

  9. Finding Nuclear Binding Energy • Mass defect for Helium-4 = 3.0377 x 10-2amu • Step 1: Convert units: amu kg • Conversion Factor: 1 amu = 1.6605 x 10-27 kg • Calculation: (3.0377 x 10-2amu) (1.6605 x 10-27 kg/amu) • Mass = 5.0441 x 10-29 kg • E = mc2 & c = 3.00 x 108m/s • E = (5.0441 x 10-29 kg) (3.00 x 108 m/s)2 • E = 4.54 x 10-12 kg * m2/s2

  10. Nuclear Binding Energy • NBE is also the energy that must be input to break apart the nucleus into its constituent nucleons • Since energy is released when a nucleus forms, which is more stable the nucleus or the separated nucleons? • Nucleus, since energy is inversely proportional to stability • Lower energy = MORE stability

  11. Another Problem • Calculate the nuclear binding energy of a Sulfur-32 atom • Step 1: Calculate the mass defect • 16 protons (16*1.007276) + 16 neutrons (16*1.008665) + 16 electrons (16*0.0005486) = 16.116 416 + 16.138 64 + 0.0087776 = 32.263 833 6 = 32.263 83 Sig Figs !!! Mass Defect = 32.26383 – 32.065 = 0.1988336 = 0.199 amu Sig Figs !!!

  12. Another Problem (Page 2) • Step 2: Calculate the NBE • Mass in amu = 0.1988336 • 0.1988336 amu * (1.6605 x 10-27 kg/amu) • Mass in kg = 3.301 632 x 10-28 kg • E = mc2 • E = (3.301 632 x 10-28 kg)(3.00 x 108 m/s)2 • E = 2.971 468 x 10-11 kg * m2/s2 = 2.97 x 10-11kg * m2/s2

  13. Half-Life • Half-life – time required for ½ of a radioactive material to decay • Each radioactive nuclide has its own ½ life • Longer ½ life = more stable nuclide • After 1 Half-Life = 50% remain • 2 Half-Lives = 25% remain • 3 Half-Lives = 12.5% remain

  14. Potassium-40

  15. Half-Life = Math Problems • Phosphorous-32 has a ½ life of 14.3 days. How many milligrams (mg) remain after 57.2 days, if the sample began with 4.0 mg? • 57.2 / 14.3 = 4 Half-Lives • 4 Half-Lives = (1/2)(1/2)(1/2)(1/2) of original amount remains • 1/16 of the original amount remains • 4.0 * (1/16) = 0.25 mg remains

  16. Half-Life Problems (Page 2) • Complete problems from Packet “Practice Problems” which is next to the decay series page. • Complete PRACTICE Problems on pp. 689 in textbook

  17. Pp. 693 Bottom • Alpha particles cannot go through paper • Beta particles can go through paper but not aluminum • Gamma particles can go through both, but not lead or concrete

  18. Nuclear Fission • Nuclear Fission – heavy nucleus splits into more-stable nuclei of intermediate mass • Mass will be converted to energy, usually a lot of energy • Chain reaction – material that begins a reaction is also one of the products so it can begin another reaction • Critical Mass – minimum amount of nuclide that is required to sustain a chain reaction • Nuclear Power Generators use controlled-fission chain reaction to produce energy • Also produces unwanted radioactive nuclides • Makes fish (and humans) glow!!

  19. Nuclear Weapons • Fission weapons were actually used against Nagasaki and Hiroshima at the end of WW2

  20. Nuclear Fusion • Low mass nuclei combine to form a heavier, more stable nucleus • Immense energy production • Source of energy for the Sun and many stars • Thermonuclear or H-bombs • Fusion of Deuterium + Tritium • 100 times power of atomic bombs ¼ mile diameter & 320 feet deep  This blast contaminated more US residents than any other activity  Yucca Flats, NV

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