1. Oct 30, 2005

### Iscariot

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?

2. Oct 30, 2005

### big man

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 $$\lambda=ln(2)/T_1_/_2$$

Last edited: Oct 30, 2005
3. Oct 30, 2005

### Iscariot

Thanks a lot!!

Last edited: Oct 30, 2005