1 / 40

Isotopes

Isotopes. Variants of a chemical element All isotopes of a given element have the same # of protons Differ in the number of neutrons. Hydrogen. Tritium. Deuterium. Isotopes. General Form: A X Z Chemical symbol for the element: X Atomic # (# of Protons): Z Protons + Neutrons: A.

latika
Download Presentation

Isotopes

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. Isotopes • Variants of a chemical element • All isotopes of a given element have the same # of protons • Differ in the number of neutrons Hydrogen Tritium Deuterium

  2. Isotopes • General Form: AXZ • Chemical symbol for the element: X • Atomic # (# of Protons): Z • Protons + Neutrons: A Deuterium Tritium Hydrogen 1H1 3H1 2H1

  3. Radioactive Decay • Alpha Decay • 238U92 => 234Th90 + 4He2 • Beta Decay • Electron Emission (β-) • 3H1 => 3He2 + 0e-1 • 234Th90 => 234Pa91 + 0e-1 • Electron Capture • 40K19 + 0e-1 => 40Ar18 + hv • Postitron Emission (β+) • 40K19 => 40Ar18 + 0e+1

  4. Trinity Test • Trinity Test Video

  5. Early History of the Bomb • 1931—Crockroft & Walton split the atom • 1932—Chadwick discovers the neutron (Nobel Prize 1953 • 1934—Joliot & Curies bombard a target to produce new elements using α particles α Alpha Particle Thin Foil 7Li 11B

  6. Early History of the Bomb • Fermi repeats bombarding experiment with neutrons and finds: • Uranium produces several radioactive by products. α β γ Neutron N β Uranium γ

  7. Uranium-235 Fission • 1N0 + 235U92236U92 144Ba56 + 89Kr36 + 3N + ENERGY • Protons + Neutrons  1 + 235 = 236 • Protons  0 + 92 = 92 • Protons + Neutrons  144 + 89 + 3 = 236 • Protons  56 + 36 = 92

  8. Early History of the Bomb • 1938—Hahn & Stassmann prove that Fermi observed fission and published 12/22/1938. • Fermi Wins Nobel Prize. • 1939—Frisch and Meitner describe fission and the potential for large amounts of energy to be released. • The Question? Are neutrons liberated in the process??

  9. Early History of the Bomb • 1939—Leo Szilard confirms that neutrons are produced and an explosive chain reaction is possible.

  10. Early History of the Bomb • 1939—April 22. Letter in Nature by Joliot confirmed that excess neutrons are produced and a chain reaction is confirmed. • World War II begins.

  11. Fission of Uranium • Mass of a Neutron = 1.008 u

  12. Fission of Uranium • Mass of a Neutron = 1.008 u • Mass of 235U = 235.044 u

  13. Fission of Uranium • Mass of a Neutron = 1.008 u • Mass of 235U = 235.044 u • Mass of 144Ba56 = 143.923 u

  14. Fission of Uranium • Mass of a Neutron = 1.008 u • Mass of 235U = 235.044 u • Mass of 144Ba56 = 143.923 u • Mass of 89Kr36 = 88.918 u

  15. Fission of Uranium • Mass of a Neutron = 1.008 u • Mass of 235U = 235.044 u • Mass of 144Ba56 = 143.923 u • Mass of 89Kr36 = 88.918 u • 1N0 + 235U92236U92 144Ba56 + 89Kr36 + 3N + ENERGY

  16. Fission of Uranium • Mass of a Neutron = 1.008 u • Mass of 235U = 235.044 u • Mass of 144Ba56 = 143.923 u • Mass of 89Kr36 = 88.918 u • 1N0 + 235U92236U92 144Ba56 + 89Kr36 + 3N + ENERGY • 1.008 + 235.044 = 236.052

  17. Fission of Uranium • Mass of a Neutron = 1.008 u • Mass of 235U = 235.044 u • Mass of 144Ba56 = 143.923 u • Mass of 89Kr36 = 88.918 u • 1N0 + 235U92236U92 144Ba56 + 89Kr36 + 3N + ENERGY • 1.008 + 235.044 = 236.052 • 143.923 + 88.918 + 3(1.008) = 235.865

  18. Fission of Uranium • Mass of a Neutron = 1.008 u • Mass of 235U = 235.044 u • Mass of 144Ba56 = 143.923 u • Mass of 89Kr36 = 88.918 u • 1N0 + 235U92236U92 144Ba56 + 89Kr36 + 3N + ENERGY • 1.008 + 235.044 = 236.052 • 143.923 + 88.918 + 3(1.008) = 235.865 • Δ M = 0.187 u

  19. Fission of Uranium • Δ M = 0.187 u • 1 u = 1.66 x 10-27 kg

  20. Fission of Uranium • Δ M = 0.187 u • 1 u = 1.66 x 10-27 kg • E = mc2

  21. Fission of Uranium • Δ M = 0.187 u • 1 u = 1.66 x 10-27 kg • E = mc2 • E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2

  22. Fission of Uranium • Δ M = 0.187 u • 1 u = 1.66 x 10-27 kg • E = mc2 • E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2 • E = 2.775 * 10-11 Joules

  23. Fission of Uranium • Δ M = 0.187 u • 1 u = 1.66 x 10-27 kg • E = mc2 • E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2 • E = 2.775 * 10-11 Joules Note: A unit we like to use is the electron volt (eV). This is the energy an electron will gain as it moves across an electric potential difference of one volt. 1eV = 1.602 x 10-19 Joule

  24. Fission of Uranium • Δ M = 0.187 u • 1 u = 1.66 x 10-27 kg • E = mc2 • E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2 • E = 2.775 * 10-11 Joules Note: A unit we like to use is the electron volt (eV). This is the energy an electron will gain as it moves across an electric potential difference of one volt. 1eV = 1.602 x 10-19 Joule 173. x 106eV 1.73 x 108eV 2.775 x 10-11 J x 1 eV = = 1.602 x 10-19 J

  25. Fission of Uranium • Δ M = 0.187 u • 1 u = 1.66 x 10-27 kg • E = mc2 • E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2 • E = 2.775 * 10-11 Joules Note: A unit we like to use is the electron volt (eV). This is the energy an electron will gain as it moves across an electric potential difference of one volt. 1eV = 1.602 x 10-19 Joule 173. x 106eV 1.73 x 108eV 2.775 x 10-11 J x 1 eV = = 173 MeV 1.602 x 10-19 J =

  26. So What’s the Big Deal?? • If 2.775 * 10-11 Joules of energy are released in one fission • What if 1 Kg of Uranium 235 fissions? • How much energy is released?

  27. So What’s the Big Deal?? • If 2.775 * 10-11 Joules of energy are released in one fission • What if 1 Kg of Uranium 235 fissions? • How much energy is released? 1 mole 235U92 = 235.044g 1 Kg = 1000 g

  28. So What’s the Big Deal?? • If 2.775 * 10-11 Joules of energy are released in one fission • What if 1 Kg of Uranium 235 fissions? • How much energy is released? 1 mole 235U92 = 235.044g 1 Kg = 1000 g 1000 g235U92x 1 mole 235U92 = 4.25 Moles 235U92 235.044 g235U92

  29. The Big Deal 4.25 Moles 235U92 x 6.022 x 1023 atoms = 2.56*1024 Atoms 235U92 mole

  30. The Big Deal 4.25 Moles 235U92 x 6.022 x 1023 atoms = 2.56*1024 Atoms 235U92 mole If 1 Kg of 235U92 fissions, we get: 2.775 X 10-11 Joule x 2.56*1024 Atoms 235U92 = 7.10* 1013 Joules Atom

  31. The Big Deal 4.25 Moles 235U92 x 6.022 x 1023 atoms = 2.56*1024 Atoms 235U92 mole If 1 Kg of 235U92 fissions, we get: 2.775 X 10-11 Joule x 2.56*1024 Atoms 235U92 = 7.10* 1013 Joules Atom As a comparison: 1 TON of TNT 4.184* 109 Joules

  32. The Big Deal 4.25 Moles 235U92 x 6.022 x 1023 atoms = 2.56*1024 Atoms 235U92 mole If 1 Kg of 235U92 fissions, we get: 2.775 X 10-11 Joule x 2.56*1024 Atoms 235U92 = 7.10* 1013 Joules Atom As a comparison: 1 TON of TNT 4.184* 109 Joules 7.10* 1013 Joules = 1.7* 104 Tons of TNT 1 Kg of 235U92 4.184* 109 Joules 17,000 Tons of TNT

  33. Fission Videos • Nuclear Fission • Chain Reaction

  34. Fission Cross Sections

  35. The Problem Fission occurs in about 10-8 seconds 80 Generations pass in 0.8 microseconds It takes less than a millionth of a second to fission a kg of 235U

  36. The Solution Use a gun to shoot the slug in Little Boy Gun Type Weapon Equivalent to 20,000 Tons of TNT

  37. Fat Man Bomb

  38. Fat Man Bomb

  39. Homework • Given the Equation: • 1N0 + 235U92236U92 144Ba56 + 89Kr36 + 3N + ENERGY • What percentage of the mass is converted to energy?

More Related