20
Absolute Dating Notes and Practice

Absolute Dating Notes and Practice. Directions: Use the following presentation to complete the notes sheet

Embed Size (px)

Citation preview

Absolute Dating Notes and Practice

Directions:Use the following presentation to complete the notes sheet.

Absolute DatingAbsolute dating involves methods to determine the age of an event or object in a specific unit of time.

To determine the absolute age of fossils and rocks, scientists analyze isotopes of radioactive elements.

Isotopes- atoms of the same element with the same number of protons but different numbers of neutrons.

Unstable IsotopesRadioactive elements are unstable. They decay, change, into differentelements over time.

During radioactive decay, the unstable parent isotope decays into stabledaughter product.

Radioactive Decay

The half-life of an element is the time it takes for half of the material you started with to decay. Remember, it doesn’t matter howmuch you start with. After 1 half-life, half of it will have decayed.Each element has its own half-life

Each element decays into a new elementC14 decays into N14 while U238 decays into Pb206 (lead), etc.

The half-life of each element is constant. It’s like a clock keeping perfect time.

Over time, your sample will have progressively more daughter material andless parent material. The total amount of matter will not change.

Demonstration Time

The grid below represents a quantity of C14. Each time you click,one half-life goes by. Try it! C14 – blue (parent) N14 – red (daughter)

As we begin notice that no timehas gone by and that 100% of thematerial is C14

Half

lives

% C14 %N14 Ratio of

C14 to N14

0 100% 0% no ratio

Age = 0 half lives (5700 x 0 = 0 yrs)

The grid below represents a quantity of C14. Each time you click,one half-life goes by. Try it! C14 – blue N14 - red

Half

lives

% C14 %N14 Ratio of

C14 to N14

0 100% 0% no ratio

1 50% 50% 1:1

After 1 half-life (5700 years), 50% ofthe C14 has decayed into N14. The ratioof C14 to N14 is 1:1. There are equalamounts of the 2 elements.

Age = 1 half lives (5730 x 1 = 5730 yrs)

The grid below represents a quantity of C14. Each time you click,one half-life goes by. Try it! C14 – blue N14 - red

Half

lives

% C14 %N14 Ratio of

C14 to N14

0 100% 0% no ratio

1 50% 50% 1:1

2 25% 75% 1:3

Now 2 half-lives have gone by for a totalof 11,400 years. Half of the C14 that waspresent at the end of half-life #1 has nowdecayed to N14. Notice the C:N ratio. Itwill be useful later.Age = 2 half lives (5730 x 2 = 11,460 yrs)

The grid below represents a quantity of C14. Each time you click,one half-life goes by. Try it! C14 – blue N14 - red

Half

lives

% C14 %N14 Ratio of

C14 to N14

0 100% 0% no ratio

1 50% 50% 1:1

2 25% 75% 1:3

3 12.5% 87.5% 1:7

After 3 half-lives (17,100 years) only12.5% of the original C14 remains. Foreach half-life period half of the materialpresent decays. And again, notice the ratio, 1:7Age = 3 half lives (5730 x 3 = 17,190 yrs)

C14 – blue N14 - red How can we find the age of asample without knowing howmuch C14 was in it to begin with?

1) Send the sample to a lab which will determine the C14 : N14

ratio. 2) Use the ratio to determine how many half lives have gone by since the sample formed.Remember, 1:1 ratio = 1 half life 1:3 ratio = 2 half lives 1:7 ratio = 3 half lives

In the example above, the ratio is 1:3.

3) Look up the half life for that isotope and multiply that value times the number of half lives determined by the ratio.

If the sample has a ratio of 1:3 that means it is 2 half lives old. If the halflife of C14 is 5,730 years then the sample is 2 x 5,730 or 11,460 years old.

C14 has a short half life and can only be used on organic (once living) material.To date an ancient rock we use the uranium – lead method (U238 : Pb206).

Here is our sample. Remember we have no ideahow much U238 was in the rock originally but allwe need is the U:Pb ratio in the rock today. Thiscan be obtained by standard laboratory techniques.

As you can see the U:Pb ratio is 1:1. From whatwe saw earlier a 1:1 ratio means that 1 half lifehas passed.Rock Sample

Now all we have to do is see what the half-life for U238 is.

Try the next one on your own.............orto review the previous frames click here.

1 half-life = 4.5 x 109 years (4.5 billion), so the rock is 4.5 billionyears old.

Element X (Blue) decays into Element Y (red)The half life of element X is2000 years.How old is our sample?

If you said that the sample was 8,000 years old, you understandradioactive dating.

If you’re unsure and want anexplanation just click.

See if this helps:1 HL = 1:1 ratio2 HL = 1:3 3 HL = 1:74 HL = 1:15

Element X (blue) Element Y (red)How old is our sample?

We know that the sample wasoriginally 100% element X. There arethree questions:First: What is the X:Y ratio now?Second: How many half-lives had togo by to reach this ratio?Third: How many years does this number of half-lives represent?

2) As seen in the list on the previous slide, 4 half-lives must go by in order to reach a 1:15 ratio.

3) Since the half life of element X is 2,000 years, four half-lives would be 4 x 2,000 or 8,000 years. This is the age of the sample.

1) There is 1 blue square and 15 red squares. Count them. This is a 1:15 ratio.

Try it:"What is the half-life of this element?"

Just remember that at the endof one half-life, 50% of theelement will remain. Find 50%on the vertical axis, Follow theblue line over to the red curveand drop straight down to findthe answer:

The half-life of this element is 1 million years.

Try It:"What percent of the materialoriginally present will remainafter 2 million years?"

Find 2 million years on thebottom, horizontal axis. Thenfollow the green line up to the red curve. Go to the left andfind the answer.

After 2 million years 25% of the original materialwill remain.

How to Solve Half Life Problems

• Read over the steps for how to solve half life problems on the front of the WS.

• Use these process to solve the questions on the back of the worksheet.

• When you are done, check your answers on the last slide of this presentation.

Key to Half Life Problems

1. 0.125g

2. 25 mg

3. 0.625 g

4. 6 years

5. 75 minutes

6. 0.0625 g would remain after 15 minutes. You would need to order about 100 g.

Directions

Go to http://www.glencoe.com/sites/common_assets/science/virtual_labs/E18/E18.html

Follow the directions on the left side of the screen to simulate radioactive decay.