# Homework Help: 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!

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