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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
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
practice
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.

band of stability
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

types of decay
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

nuclear equations
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

transmutations
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

practice1
Practice

Complete the following nuclear equations.

  • 73Ga  73Ge + _____
  • 192Pt  188Os + ____
  • 205Bi  205Pb + ____
  • 241Cm + ____  241Am
  • ______ + 42He  24397Bk + 10n
  • 24998Cf + _____  260105Db + 410n
practice2
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?

slide10

Half - Life

  • The time it takes for ½ of a radioactive material to decay
slide11

Nuclear Fission

  • The splitting of nucleus into smaller nuclei
  • Takes place in power plants
slide12

Nuclear Fusion

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