- #1
Iscariot
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The carbon in living matter contains a fixed proportion of the radioactive isotope carbon-14. The carbon-14 in 1.00g of carbon from living matter has an activity of 0.250Bq. The half-life of carbon-14 is 5730. When a plant dies the proportion of carbon-14 decreases due to radioactive decay. A 1.00g sample of carbon from an ancient boat has an activity of 0.160Bq. Determine the age of the board.
Here's how I solved it...
Original Activity = 0.25Bq
Activity of Sample = 0.16Bq
Then I just calculated what's that as a ratio of the original activity...
0.16/0.25 = 0.64
Then multiplied the half time by this number:
5730 * 0.64 = 3666 years ~ 3700 years
Which is the correct answer. However this seems like a bit of a fluke. Especially since I've got a feeling I should be using this formula:
x = x(original) ^-(lamda)*(time)
Can anyone put my mind at ease, was my answer a fluke or is that a valid method to calculating the answer?
Here's how I solved it...
Original Activity = 0.25Bq
Activity of Sample = 0.16Bq
Then I just calculated what's that as a ratio of the original activity...
0.16/0.25 = 0.64
Then multiplied the half time by this number:
5730 * 0.64 = 3666 years ~ 3700 years
Which is the correct answer. However this seems like a bit of a fluke. Especially since I've got a feeling I should be using this formula:
x = x(original) ^-(lamda)*(time)
Can anyone put my mind at ease, was my answer a fluke or is that a valid method to calculating the answer?