Carbon Dating

1. Dec 31, 2011

Fixxxer125

1. The problem statement, all variables and given/known data
The equilibrium concentration of C14/C12 ≈ 1.5×10−12
A 2.5 g sample of Carbon from an old tree fragment has an activity of 4.57 decays per minute. How old is the sample?
2. Relevant equations
I'm pretty sure I use A(t)=λN(0)*exp(-tλ)
but I am having trouble calculating N(0)! The amount of C14 c.f. C12 is so tiny I don't know how to manipulate the ratio to get the mass of C14, and thus the number of nucli, in the sample. Thanks!

2. Dec 31, 2011

vela

Staff Emeritus
I don't see why the smallness of the ratio should present any difficulty.

3. Dec 31, 2011

Fixxxer125

Well I was trying to calculate the amount of C14 present by dividing 2.5g by (1 + 1.5×10−12), am I doing this the wrong way?thanks

4. Dec 31, 2011

vela

Staff Emeritus
Yes, that's wrong. You have

C14/C12 ≈ 1.5×10−12

so

C14 ≈ 1.5×10−12 C12

5. Dec 31, 2011

Fixxxer125

Ah yes I see that but then I was having trouble working out the actual amount of C12 that there was 1.5×10−12 of C14 to. surely there is something like 0.00000000001g of C14 in the sample?

6. Dec 31, 2011

vela

Staff Emeritus
How did you come up with that?

7. Dec 31, 2011

Fixxxer125

Well I thought for every 1g of C12 there is 1.5×10−12g of C14? so if there are 2.5g in total basically all of this must be C12?

8. Dec 31, 2011

vela

Staff Emeritus
Yes, it's virtually all C12, but the amount of C14 you calculated is wrong.

9. Dec 31, 2011

Fixxxer125

Could you give me a bit of help on doing the ratio please?Thanks!

10. Jan 1, 2012

ehild

Originally the ratio of C14 to C12 was the same as the equilibrium ratio. But C14 decayed in the isolated sample.
The original amount of C14 was 2.5 x1.5-12 gram. It means quite a lot of atoms, you can not take it zero!

You need lambda to solve the equation for t. Do you know it?

ehild

11. Jan 1, 2012

Fixxxer125

Oh yea, I know T1/2 so can work it out, many thanks!