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18.3 Half-Life
 

18.3 Half-Life

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18.3 Half-Life 18.3 Half-Life Presentation Transcript

  • Uses of Nuclear Chemistry 18.3
  • Radioactive Dating  Finding the age of an object based on the amount of radioactive isotopes in it.  Example: Carbon-14 has a half life of 5715 years. Carbon-12 & Carbon-14 amounts remain constant while an object is living. When an organic artifact is found the Carbon-14 can be compared to the Carbon-12.
  • Half-Life
  • Half-Life  The amount of time it takes for ½ of an unstable element to decay.
  • Half-Life  The amount of time it takes for ½ of an unstable element to decay.  Example: carbon-14 half life is 5715. If a 80mg sample decays over 22860 years, how much will remain?
  • Half-Life  The amount of time it takes for ½ of an unstable element to decay.  Example: carbon-14 half life is 5715. If a 80mg sample decays over 22860 years, how much will remain?  22860/5715 = 4 half lives.
  • Half-Life  The amount of time it takes for ½ of an unstable element to decay.  Example: carbon-14 half life is 5715. If a 80mg sample decays over 22860 years, how much will remain?  22860/5715 = 4 half lives.  80/2=40mg 1st half life
  • Half-Life  The amount of time it takes for ½ of an unstable element to decay.  Example: carbon-14 half life is 5715. If a 80mg sample decays over 22860 years, how much will remain?  22860/5715 = 4 half lives.  80/2=40mg 1st half life  40/2=20mg 2nd half-life
  • Half-Life  The amount of time it takes for ½ of an unstable element to decay.  Example: carbon-14 half life is 5715. If a 80mg sample decays over 22860 years, how much will remain?  22860/5715 = 4 half lives.  80/2=40mg 1st half life  40/2=20mg 2nd half-life  20/2=10mg 3rd half-life
  • Half-Life  The amount of time it takes for ½ of an unstable element to decay.  Example: carbon-14 half life is 5715. If a 80mg sample decays over 22860 years, how much will remain?  22860/5715 = 4 half lives.  80/2=40mg 1st half life  40/2=20mg 2nd half-life  20/2=10mg 3rd half-life  10/2=5mg 4th half-life
  • Radioactive Dating
  • Radioactive Dating  An object was found with 1/8 the ratio of carbon-14 to carbon-12. How old is it? (one half-life = 5715 years)
  • Radioactive Dating  An object was found with 1/8 the ratio of carbon-14 to carbon-12. How old is it? (one half-life = 5715 years)
  • Radioactive Dating  An object was found with 1/8 the ratio of carbon-14 to carbon-12. How old is it? (one half-life = 5715 years)  1 divided by 2 = ½ 1st half-life
  • Radioactive Dating  An object was found with 1/8 the ratio of carbon-14 to carbon-12. How old is it? (one half-life = 5715 years)  1 divided by 2 = ½ 1st half-life  ½ divided by 2 = ¼ 2nd half-life
  • Radioactive Dating  An object was found with 1/8 the ratio of carbon-14 to carbon-12. How old is it? (one half-life = 5715 years)  1 divided by 2 = ½ 1st half-life  ½ divided by 2 = ¼ 2nd half-life  ¼ divided by 2 = 1/8 3rd half-life
  • Radioactive Dating  An object was found with 1/8 the ratio of carbon-14 to carbon-12. How old is it? (one half-life = 5715 years)  1 divided by 2 = ½ 1st half-life  ½ divided by 2 = ¼ 2nd half-life  ¼ divided by 2 = 1/8 3rd half-life
  • Radioactive Dating  An object was found with 1/8 the ratio of carbon-14 to carbon-12. How old is it? (one half-life = 5715 years)  1 divided by 2 = ½ 1st half-life  ½ divided by 2 = ¼ 2nd half-life  ¼ divided by 2 = 1/8 3rd half-life  3 x 5715 = 17145 years
  • ½ Life Example
  • ½ Life Example  A rock has 4.3mg of Potassium-40. The rock was dated 3.84 billion years old. How much Potassium-40 was originally present? (1 half-life = 1.28 billion years)
  • ½ Life Example  A rock has 4.3mg of Potassium-40. The rock was dated 3.84 billion years old. How much Potassium-40 was originally present? (1 half-life = 1.28 billion years)  3.84/1.28 = 3 half lives
  • ½ Life Example  A rock has 4.3mg of Potassium-40. The rock was dated 3.84 billion years old. How much Potassium-40 was originally present? (1 half-life = 1.28 billion years)  3.84/1.28 = 3 half lives  4.3 x 2 = 8.6 1st half-life
  • ½ Life Example  A rock has 4.3mg of Potassium-40. The rock was dated 3.84 billion years old. How much Potassium-40 was originally present? (1 half-life = 1.28 billion years)  3.84/1.28 = 3 half lives  4.3 x 2 = 8.6 1st half-life  8.6 x 2 = 17.2 2nd half-life
  • ½ Life Example  A rock has 4.3mg of Potassium-40. The rock was dated 3.84 billion years old. How much Potassium-40 was originally present? (1 half-life = 1.28 billion years)  3.84/1.28 = 3 half lives  4.3 x 2 = 8.6 1st half-life  8.6 x 2 = 17.2 2nd half-life  17.2 x 2 = 34.4 3rd half-life
  • ½ Life Example  A rock has 4.3mg of Potassium-40. The rock was dated 3.84 billion years old. How much Potassium-40 was originally present? (1 half-life = 1.28 billion years)  3.84/1.28 = 3 half lives  4.3 x 2 = 8.6 1st half-life  8.6 x 2 = 17.2 2nd half-life  17.2 x 2 = 34.4 3rd half-life  Answer = 34.4mg
  • Uses of Nuclear Chemistry  Smoke detectors: use Americium-241  Art forgeries: found through neutron activation analysis.  Medical: to diagnose and treat disease. X-rays Thallium-201 intravenously PET scans
  • Radiation Exposure  rem: the unit used to express biological effects of radiation. 0-25 no effect 25-50 slight decrease in white cell count 50-100 marked decrease in white cell count 100-200 nausea and hair loss 200-500 internal bleeding 500+ death
  • Homework Half-life problems
  • A New Approach All half life problems have 5 parts 1)Length of decay length of decay # of half lives = 2)Length of half life length of half life 3)Initial amount Divide by 2 if finding final amount. Multiply by 2 if finding original amount. 4)Final amount 5)Number of half lives