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Nuclear Chemistry

Nuclear Chemistry. Deals with the nucleus of the atom (protons & neutrons) Matter is changed into energy by the breaking up of the nucleus of an atom Mass defect – The difference in the calculated mass of an atom and the measured mass of an atom Mass of neutron = 1.67493x10 -27 kg

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Nuclear Chemistry

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  1. Nuclear Chemistry • Deals with the nucleus of the atom (protons & neutrons) • Matter is changed into energy by the breaking up of the nucleus of an atom • Mass defect – The difference in the calculated mass of an atom and the measured mass of an atom • Mass of neutron = 1.67493x10-27kg • Mass of proton = 1.67262x10-27 kg • Nuclear Binding Energy – amount of energy needed to break up the nucleus = a measure of the stability of the nucleus • E = mc2

  2. Practice Calculate the binding energy of for carbon -12 (atomic mass 12.000) and uranium-235 (atomic mass 235.0439). The atomic mass of 11H is 1.00782 amu and the mass of a neutron is 1.00866 amu.

  3. Band of stability Looks at the proton/neutron ratio • The preferred ratio of p+/n0 is 1:1 • Atoms with low atomic #s have close to that ratio • Larger atoms (atomic #s 84+) do not so they are radioactive Atoms with a high p+/n0 - emit- particles Atoms with a low p+/n0 - +emission or e-capture Atomic #s 84+ -  emission

  4. Types of Decay Emission(given off) •  - alpha – 42He Weakest type 2. - - beta particle – 0-1e 3.  - gamma ray – energy Strongest type 4. + - positron – 0+1e Capture(taken in) 1. 0-1e – electron is taken in

  5. Nuclear Equations • Equations written to represent the splitting that takes place in the nucleus (called a transmutation) • The sum of the atomic #s of products = sum the atomic #s of reactants • The sum of the mass #s of products = sum of the mass #s of the reactants 116C  115B + 01 116C + 0-1e  115B 146C  147N + 0-1 19278Pt  18876Os + 42He

  6. Transmutations • The splitting of the nucleus • Larger atoms split spontaneously b/c they are radioactive • Others can be induced (forced) to occur by hitting a nucleus with a particle (nuclear fission) 23592U + 10n  9336Kr + 14056Ba + 310n Shortened notation = 23592U (10n, 310n) 9336Kr, 14056Ba

  7. Practice Complete the following nuclear equations. • 73Ga  73Ge + _____ • 192Pt  188Os + ____ • 205Bi  205Pb + ____ • 241Cm + ____  241Am • ______ + 42He  24397Bk + 10n • 24998Cf + _____  260105Db + 410n

  8. Decay Series

  9. Practice One type of commerical smoke detector contains a minute amount of radioactive americium-241 which decays by  particle production. The  particles ionize molecules ini the air, allowing it to conduct an electric current. When smoke particles enter, the conductivity of the air is changed and the alarm buzzes. • Write the equation for the decary of americium-241 by  particle production. b. The complete decay of 241Am involves successively , , , , , , , , , , , and . What is the final stable nucleus produced in this decay series?

  10. Half - Life • The time it takes for ½ of a radioactive material to decay

  11. Nuclear Fission • The splitting of nucleus into smaller nuclei • Takes place in power plants

  12. Nuclear Fusion • Smaller nuclei combining to form a large one • HUGE amounts of energy involved • Sun, Hydrogen bomb

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