Calculating the Age of an Ancient Boat: A Radioactive Decay Problem

AI Thread Summary
The discussion focuses on calculating the age of an ancient boat using carbon-14 decay. The original activity of carbon-14 in living matter is 0.250 Bq, while the sample from the boat shows an activity of 0.160 Bq. A method was proposed that involved calculating the ratio of the sample's activity to the original and multiplying it by the half-life of carbon-14, resulting in an estimated age of approximately 3700 years. However, concerns were raised about the validity of this method, suggesting that a more accurate approach would involve using the exponential decay formula and natural logarithms. The conversation emphasizes the importance of using proper decay equations for accurate age determination.
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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?
 
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Well if you do it your way you aren't really considering the fact that it's an exponential decay.
I would say that you would use the activity equation given by:
A=A_oe^-^\lambda ^t
It's just as easy. Just rearrange it and then take the natural log of both sides to solve for t.
Where \lambda=ln(2)/T_1_/_2
 
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Thanks a lot!
 
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