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