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- 1. Uses of Nuclear Chemistry 18.3
- 2. 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.
- 3. Half-Life
- 4. Half-Life The amount of time it takes for ½ of an unstable element to decay.
- 5. 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?
- 6. 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.
- 7. 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
- 8. 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
- 9. 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. 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
- 11. Radioactive Dating
- 12. 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)
- 13. 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)
- 14. 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
- 15. 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
- 16. 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
- 17. 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
- 18. 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
- 19. ½ Life Example
- 20. 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)
- 21. 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
- 22. 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
- 23. 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
- 24. 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
- 25. 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
- 26. 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
- 27. 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
- 28. Homework Half-life problems
- 29. 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

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