Isotopes Variants of a chemical element

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Presentation transcript:

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

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

Radioactive Decay Beta Decay Alpha Decay Electron Emission (β-) 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

Trinity Test Trinity Test Video

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

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

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

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??

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

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.

Fission of Uranium Mass of a Neutron = 1.008 u

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

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

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

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

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

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

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

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

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

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

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

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

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 2.775 x 10-11 J x 1 eV 1.73 x 108 eV 173. x 106 eV = = 1.602 x 10-19 J

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 2.775 x 10-11 J x 1 eV 1.73 x 108 eV 173. x 106 eV = = 1.602 x 10-19 J = 173 MeV

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?

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

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 g 235U92 x 1 mole 235U92 = 4.25 Moles 235U92 235.044 g 235U92

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

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

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

The Big Deal 17,000 Tons of TNT 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

Fission Videos Nuclear Fission Chain Reaction

Fission Cross Sections

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

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

Fat Man Bomb

Fat Man Bomb

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