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Nuclear Chemistry I The Nucleus II III IV C. Johannesson.

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Presentation on theme: "Nuclear Chemistry I The Nucleus II III IV C. Johannesson."— Presentation transcript:

1 Nuclear Chemistry I The Nucleus II III IV C. Johannesson

2 A. Mass Defect Difference between the mass of an atom and the mass of its individual particles. amu amu C. Johannesson

3 B. Nuclear Binding Energy
Energy released when a nucleus is formed from nucleons. High binding energy = stable nucleus. E = mc2 E: energy (J) m: mass defect (kg) c: speed of light (3.00×108 m/s) C. Johannesson

4 B. Nuclear Binding Energy
Unstable nuclides are radioactive and undergo radioactive decay. C. Johannesson

5 II. Radioactive Decay I II III IV C. Johannesson

6 A. Types of Radiation 2+ 1- 1+ Alpha particle () helium nucleus
paper 2+ Beta particle (-) electron 1- lead Positron (+) positron 1+ concrete Gamma () high-energy photon C. Johannesson

7 B. Nuclear Decay Numbers must balance!! Alpha Emission parent nuclide
daughter nuclide alpha particle Numbers must balance!! C. Johannesson

8 B. Nuclear Decay Beta Emission electron Positron Emission positron
C. Johannesson

9 B. Nuclear Decay Electron Capture electron Gamma Emission
Usually follows other types of decay. Transmutation One element becomes another. C. Johannesson

10 B. Nuclear Decay Why nuclides decay…
need stable ratio of neutrons to protons C. Johannesson DECAY SERIES TRANSPARENCY

11 C. Half-life Half-life (t½)
Time required for half the atoms of a radioactive nuclide to decay. Shorter half-life = less stable. C. Johannesson

12 C. Half-life mf: final mass mi: initial mass n: # of half-lives
C. Johannesson

13 C. Half-life t½ = 5.0 s mf = mi (½)n mi = 25 g mf = (25 g)(0.5)12
Fluorine-21 has a half-life of 5.0 seconds. If you start with 25 g of fluorine-21, how many grams would remain after 60.0 s? GIVEN: t½ = 5.0 s mi = 25 g mf = ? total time = 60.0 s n = 60.0s ÷ 5.0s =12 WORK: mf = mi (½)n mf = (25 g)(0.5)12 mf = g C. Johannesson


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