Finding Original Carbon nuclei from given sample.

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Homework Statement


A small animal bone fragment found in an archaelogical site has a carbon mass of 155g. When the animal was alive, the ratio of radioactive 146C to the stable 126C was 1.31×10-12. What was the number of 146C nuclei found in the sample when the animal was alive?


Homework Equations


None given, but I would assume:
N=N0e-λt


The Attempt at a Solution


Not too sure where to start so I got the decay constant, λ by using half life of Carbon14, 5730 Years
0.5=e-λ(5730)
λ=1.21×10-4

Then I solved the for the number of years since the animal was alive by plugging everything back into the original equation, assuming N/N0 = 1.31×10-12

t=-ln(1.31×10-12)/-1.21×10-4 = 226180 Years

Up until here I don't think I did anything wrong, but here is where I am unsure of what to do.

I tried using the same formula to solve for N0, but this time using the given 155g.

155=N0e-(1.2110-4)(226180)
N0=1.18×1014g

I don't think in doing the right thing here. Can anyone give me some guidance?

Thanks!
 

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Hello rcubed,

Welcome to Physics Forums!
rcubed said:

Homework Statement


A small animal bone fragment found in an archaelogical site has a carbon mass of 155g. When the animal was alive, the ratio of radioactive 146C to the stable 126C was 1.31×10-12. What was the number of 146C nuclei found in the sample when the animal was alive?

Homework Equations


None given, but I would assume:
N=N0e-?t

The Attempt at a Solution


Not too sure where to start so I got the decay constant, ? by using half life of Carbon14, 5730 Years
0.5=e-?(5730)
?=1.21×10-4
Okay, you've found that
[tex]\frac{N}{N_0} = e^{-1.21 \times 10^4 \ t}[/tex]

Although I don't think that helps for this problem. (Maybe it does later in a different part of the problem not listed in the above statement).
Then I solved the for the number of years since the animal was alive by plugging everything back into the original equation, assuming N/N0 = 1.31×10-12

t=-ln(1.31×10-12)/-1.21×10-4 = 226180 Years
No, wait. You're using the 1.31×10-12 out of context. It is *not* the ratio of the final amount of 14C to the original amount of 14C.

As a matter of fact, you don't even know what the final amount of 14C is. That information is not given in the problem statement. And since the age of the animal is not given either, you can't even calculate it (at least not without additional information).

Can anyone give me some guidance?

Thanks!
All you're trying to find is the original amount of 14C (when the animal was alive). For this particular problem, don't worry about how much of the sample is 14C today, or even how old the sample is. There's not enough information given anyway.

So here is what we know. There is 155 g sample of carbon. When the animal was alive, the fraction of carbon (in terms of the ratio of the number of nuclei) that was 14C was 1.31×10-12 (Technically that number is the ratio of 14C to 12C, but it's also approximately the same ratio as the number of 14C to total). So when the animal was alive, how much of that sample was 14C?

[Hint: You might want to start by determining how many carbon nuclei (primarily 12C) are in 155 g of carbon. Then, since you know what fraction of that that was 14C, determine the number of 14C nuclei].

-------------
Edit:

By the way, I am presently interpreting the 1.31×10-12 ratio as the ratio of number of nuclei. If instead it is a ratio of masses or weights, then my advice needs to be modified to take that into account (i.e. 14C is heavier than 12C, which needs consideration).
 
Last edited:
^Thanks for the reply!

So its just simply finding the number of 12C Molecules first, then using the ratio to find the 14C?
155g/12gmol-1*NA = 7.78e24

7.78e24*1.32e-12 = 1.026e13 ?

Is that it?
 
rcubed said:
^Thanks for the reply!

So its just simply finding the number of 12C Molecules first, then using the ratio to find the 14C?
155g/12gmol-1*NA = 7.78e24

7.78e24*1.32e-12 = 1.026e13 ?

Is that it?
It looks like the right idea to me. :approve: But I think the original problem statement said that the ratio was 1.31 x 10-12 (you used 1.32 x 10-12).