2.
Radioactive Dating
ïƒ˜ Finding the age of an object based on
the amount of radioactive isotopes in it.
ïƒ˜ Example:
ïƒ˜Carbon-14 has a half life of 5715 years.
ïƒ˜Carbon-12 & Carbon-14 amounts remain
constant while an object is living.
ïƒ˜When an organic artifact is found the
Carbon-14 can be compared to the
Carbon-12.
4.
Half-Life
ïƒ˜ The amount of time it takes for Â½ of an
unstable element to decay.
5.
Half-Life
ïƒ˜ The amount of time it takes for Â½ of an
unstable element to decay.
ïƒ˜ Example: carbon-14 half life is 5715. If a
80mg sample decays over 22860 years, how
much will remain?
6.
Half-Life
ïƒ˜ The amount of time it takes for Â½ of an
unstable element to decay.
ïƒ˜ Example: carbon-14 half life is 5715. If a
80mg sample decays over 22860 years, how
much will remain?
ïƒ˜ 22860/5715 = 4 half lives.
7.
Half-Life
ïƒ˜ The amount of time it takes for Â½ of an
unstable element to decay.
ïƒ˜ Example: carbon-14 half life is 5715. If a
80mg sample decays over 22860 years, how
much will remain?
ïƒ˜ 22860/5715 = 4 half lives.
ïƒ˜ 80/2=40mg 1st half life
8.
Half-Life
ïƒ˜ The amount of time it takes for Â½ of an
unstable element to decay.
ïƒ˜ Example: carbon-14 half life is 5715. If a
80mg sample decays over 22860 years, how
much will remain?
ïƒ˜ 22860/5715 = 4 half lives.
ïƒ˜ 80/2=40mg 1st half life
ïƒ˜ 40/2=20mg 2nd half-life
9.
Half-Life
ïƒ˜ The amount of time it takes for Â½ of an
unstable element to decay.
ïƒ˜ Example: carbon-14 half life is 5715. If a
80mg sample decays over 22860 years, how
much will remain?
ïƒ˜ 22860/5715 = 4 half lives.
ïƒ˜ 80/2=40mg 1st half life
ïƒ˜ 40/2=20mg 2nd half-life
ïƒ˜ 20/2=10mg 3rd half-life
10.
Half-Life
ïƒ˜ The amount of time it takes for Â½ of an
unstable element to decay.
ïƒ˜ Example: carbon-14 half life is 5715. If a
80mg sample decays over 22860 years, how
much will remain?
ïƒ˜ 22860/5715 = 4 half lives.
ïƒ˜ 80/2=40mg 1st half life
ïƒ˜ 40/2=20mg 2nd half-life
ïƒ˜ 20/2=10mg 3rd half-life
ïƒ˜ 10/2=5mg 4th half-life
12.
Radioactive Dating
ïƒ˜ 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
ïƒ˜ 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
ïƒ˜ An object was found with 1/8 the ratio of
carbon-14 to carbon-12. How old is it? (one
half-life = 5715 years)
ïƒ˜ 1 divided by 2 = Â½ 1st half-life
15.
Radioactive Dating
ïƒ˜ An object was found with 1/8 the ratio of
carbon-14 to carbon-12. How old is it? (one
half-life = 5715 years)
ïƒ˜ 1 divided by 2 = Â½ 1st half-life
ïƒ˜ Â½ divided by 2 = Â¼ 2nd half-life
16.
Radioactive Dating
ïƒ˜ An object was found with 1/8 the ratio of
carbon-14 to carbon-12. How old is it? (one
half-life = 5715 years)
ïƒ˜ 1 divided by 2 = Â½ 1st half-life
ïƒ˜ Â½ divided by 2 = Â¼ 2nd half-life
ïƒ˜ Â¼ divided by 2 = 1/8 3rd half-life
17.
Radioactive Dating
ïƒ˜ An object was found with 1/8 the ratio of
carbon-14 to carbon-12. How old is it? (one
half-life = 5715 years)
ïƒ˜ 1 divided by 2 = Â½ 1st half-life
ïƒ˜ Â½ divided by 2 = Â¼ 2nd half-life
ïƒ˜ Â¼ divided by 2 = 1/8 3rd half-life
18.
Radioactive Dating
ïƒ˜ An object was found with 1/8 the ratio of
carbon-14 to carbon-12. How old is it? (one
half-life = 5715 years)
ïƒ˜ 1 divided by 2 = Â½ 1st half-life
ïƒ˜ Â½ divided by 2 = Â¼ 2nd half-life
ïƒ˜ Â¼ divided by 2 = 1/8 3rd half-life
ïƒ˜ 3 x 5715 = 17145 years
20.
Â½ Life Example
ïƒ˜ 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.
Â½ Life Example
ïƒ˜ 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)
ïƒ˜ 3.84/1.28 = 3 half lives
22.
Â½ Life Example
ïƒ˜ 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)
ïƒ˜ 3.84/1.28 = 3 half lives
ïƒ˜ 4.3 x 2 = 8.6 1st half-life
23.
Â½ Life Example
ïƒ˜ 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)
ïƒ˜ 3.84/1.28 = 3 half lives
ïƒ˜ 4.3 x 2 = 8.6 1st half-life
ïƒ˜ 8.6 x 2 = 17.2 2nd half-life
24.
Â½ Life Example
ïƒ˜ 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)
ïƒ˜ 3.84/1.28 = 3 half lives
ïƒ˜ 4.3 x 2 = 8.6 1st half-life
ïƒ˜ 8.6 x 2 = 17.2 2nd half-life
ïƒ˜ 17.2 x 2 = 34.4 3rd half-life
25.
Â½ Life Example
ïƒ˜ 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)
ïƒ˜ 3.84/1.28 = 3 half lives
ïƒ˜ 4.3 x 2 = 8.6 1st half-life
ïƒ˜ 8.6 x 2 = 17.2 2nd half-life
ïƒ˜ 17.2 x 2 = 34.4 3rd half-life
ïƒ˜ Answer = 34.4mg
26.
Uses of Nuclear Chemistry
ïƒ˜ Smoke detectors: use Americium-241
ïƒ˜ Art forgeries: found through neutron
activation analysis.
ïƒ˜ Medical: to diagnose and treat disease.
ïƒ˜X-rays
ïƒ˜Thallium-201 intravenously
ïƒ˜PET scans
27.
Radiation Exposure
ïƒ˜ 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
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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