# Carbon-14 and age of a sample

1. Sep 1, 2015

### Rectifier

The problem
The isotope $^{14}C$ has a half-life of $5730$ years. The ratio between the amount of $^{14}C$ nuclei and $^{12}C$ nuclei in a living matter is $1.3 \cdot 10 ^{-12}$. When an organism dies the amount of $^{14}C$ atoms starts dropping and thus the ratio decreases.

a) What is the activity of $^{14}C$ -nuclei in a sample from a living matter that contains 1g Carbon.
b) Assume that an archaeological finding (from a dead animal) that contains 200g carbon has an activity of 400 decays per minute. How old is the finding?

This problem was translated from Swedish. Sorry for my weak English ;(

The attempt
a)
The activity is given by:
$N=N_0 \cdot e^{-\lambda t}$
$\lambda = \frac{ln(2)}{T_{\frac{1}{2}}}$
$T_{\frac{1}{2}}=5730 \ years = 2.091 \cdot 10^6 \ days$

There is 1g of carbon. That means that
$m_{^{14}C} + m_{^{12}C} = 1$

the ration of $^{14}C$ atoms and $^{12}C$ atoms in a living matter is $1.3 \cdot 10 ^{-12}$ thus

$\frac{ ^{14}C n }{ ^{12}C n} = 1.3 \cdot 10 ^{-12}$

I am not entirely sure whether I should have n (amount of mols) or another unit. But I guess that it does not matter since units cancel out in the division.

To calculate N i need N_0 which is initial activity.

The activity N halves then t=half-time

Therefore:
$\frac{1}{2} = N_0e^{- \lambda t} \\ \frac{1}{2} = N_0e^{- \frac{ln(2)}{t_{\frac{1}{2}}} t}$ and for $t = t_{\frac{1}{2}}$ $N_0 = 1$

Aaaand I am stuck :,(

2. Sep 1, 2015

### DEvens

Nearly there, nearly there. First, C12 means that 12 grams of atoms is one mole.

The extra weight of the tiny fraction of C14 atoms will make essentially no difference to the total number of carbon atoms.

At the half life you have half the original atoms left, as you indicate. You nearly have $\lambda$ calculated, you just have to get your algebra straight.

$\frac{1}{2} = \exp ( -\lambda t_{\frac{1}{2}})$

The activity is the rate of change of C14's at any given time. That is the derivative with respect to time of the number $N(t)$. And you have the equation for that as soon as you calculate $\lambda$.

3. Sep 2, 2015

### Rectifier

I am sorry but I am a bit lost in your advice. Should I find the derivative of $N=N_0 \cdot e^{-\lambda t}$ with respect to t? But what is $\frac{1}{2} = e^ { -\lambda t_{\frac{1}{2}}}$ then? Is it what remain when I set $t= t_{\frac{1}{2}}$?

4. Sep 2, 2015

### haruspex

Yes, take the derivative of N(t) to find the activity at time t.
$\frac{1}{2} = e^ { -\lambda T_{\frac{1}{2}}}$ Is equivalent to your equation $\lambda=\frac{\log 2}{T_{\frac 12}}$. Not sure why DEvens mentioned it.

5. Sep 2, 2015

### Rectifier

Here is another attempt:
$N(t)=N_0 \cdot e^{-\lambda t}$
$N'(t)= -N_0 \lambda e^{-\lambda t}$
And since $N'(0) = N_0$
$N'(0)=N_0= -N_0 \lambda e^{-\lambda 0} \\1= - \lambda e^{-\lambda 0} \\ 1= - \lambda \\ \lambda = -1$

This cant be right, can it?

EDIT:
I am quite sure that $N'(0) \neq N_0$. I think that $N_0$ should be calculated by using the ration of nuclei.

Last edited: Sep 2, 2015
6. Sep 2, 2015

### Rectifier

Here is yet another try:
a)
Here is another attempt:
$N(t)=N_0 \cdot e^{-\lambda t}$
$N'(t)= -N_0 \lambda e^{-\lambda t}$
$N'(0)= -N_0 \lambda$

I need the value of $\lambda$ and $N_0$
$\lambda = \frac{ln(2)}{half-life} \\ \lambda = \frac{ln(2)}{5730}$

12g of carbon is one mole
$12u = 12*1,660538921 \cdot 10^{-24}$
1g of carbon has therefore

Number of $^{12}C$ atoms in one gram is therefore $\frac{1}{12 \cdot 1,660538921 \cdot 10^{-24} }$ (not entirely sure though it seems like this is the amount of carbon atoms in 12g)

$N_{_{14}C} = N_{_{12}C} \cdot 1.3*10^{-12} \\ N_{_{14}C} = \frac{1}{12 \cdot 1,660538921 \cdot 10^{-24} } \cdot 1.3 \cdot 10^{-12} =$

Does this look right?

Last edited: Sep 2, 2015
7. Sep 2, 2015

### Rectifier

I made a lot of edits in the post above. Sorry about that.

8. Sep 2, 2015

### haruspex

Yes, that all looks right.

9. Sep 2, 2015

### Rectifier

Thank you!

So that means that:
$N_{_{14}C} = N_{_{12}C} \cdot 1.3 \cdot 10^{-12} \\ N_{_{14}C} = \frac{1}{12 \cdot 1,6605... \cdot 10^{-24} } \cdot 1.3 \cdot 10^{-12} = 6.523986... \cdot 10^10 \\ N_0= 6.523986 \cdot 10^{10}$

and therefore
$N'(0)= -N_0 \lambda \\ N'(0)= -6.523986 \cdot 10^{10} \frac{ln(2)}{5730} \\ N'(0) = -7.891942... \cdot 10^6$ Shouldn't this one be positive? hmm...
2 is the least significant figures in the problem and therefore the answer is $N'(0) = -7.9 \cdot 10^6$

Does this look right?

10. Sep 2, 2015

### haruspex

Yes, except that the activity will be -N'(t). N' is the rate of gain of 14C, while the activity is the rate of its loss.

11. Sep 2, 2015

### Rectifier

Thank you so much!

I tried b) too.

So here is what I get:
mass = 200
activity = 400 decays / minute or also = 210240000 dacays / year

Strategy:
I want to calculate the new initial activity $N_0$ of the animal that has 200g carbon in it. To calculate the age of the animal I want to find t when $N'(t)=210240000$ (3).

I derive the same function as above once again:
$N(t)=N_0 \cdot e^{-\lambda t}$
$N'(t)= -N_0 \lambda e^{-\lambda t}$ (1)
$N'(0)= -N_0 \lambda = N'(0)= -N_0 \frac{ln(2)}{5730}$ (2)

12g of C-12 atoms is one mole.
$12u = 12*1,660538921 \cdot 10^{-24}$
Amount of C-12 atoms in 200g sample is therefore
$N_{C-12}=\frac{200}{12 \cdot 1,660538921 \cdot 10^{-24} }$

When the animal was alive it had $N_{C-14}$ atoms. It is its the initial activity sine C-14 nuclei start to decay when it dies. It means that we can use the ratio in a)

$N_0=N_{_{14}C} = N_{_{12}C} \cdot 1.3 \cdot 10^{-12} \\ N_{_{14}C} = \frac{200}{12 \cdot1.660538921 \cdot 10^{-24} } \cdot 1.3 \cdot 10^{-12} =1.08733106627... \cdot 10^{12}$

$N_0 = 1.08733106627... \cdot 10^{12}$

(1) gives:
$N'(t)= -N_0 \lambda e^{-\lambda t}$
$N'(t)= -1.08733106627... \cdot 10^{12} \frac{ln(2)}{5730} e^{-\frac{ln(2)}{5730} t}$

(3) gives
$210240000= -1.08733106627... \cdot 10^{12} \frac{ln(2)}{5730} e^{-\frac{ln(2)}{5730} t} \\ t = -3877.028729217900$
I guess that the animal is 3877 years old then. Not sure if how I should round that though.

When I removed the minus in fron of $N_0$ there were no real solutions.

12. Sep 2, 2015

### haruspex

As I wrote, N' is by definition the rate of gain of C-14 atoms, so is always negative. So your eqn (3) should read N'(t)= - 210240000 year-1.
t should come out positive. Your value for the initial number of C-14 atoms is too low. I get more like 2 1013.

13. Sep 2, 2015

### Rectifier

Oh I see. I will be back shortly. I will try to find the error. Thank you for your help!

14. Sep 2, 2015

### Rectifier

I input that in wolfram:
N_{_{14}C} = \frac{200}{12 \cdot1.660538921 \cdot 10^{-24} } \cdot 1.3 \cdot 10^{-12}
and get
$1.30... \cdot 10^{13}$ could that be right?

15. Sep 2, 2015

### haruspex

Yes. My mental calculation was very approximate, and was a bit of an overestimate.

16. Sep 2, 2015

### Rectifier

Thank you once again!

I will recalculate b) now.

17. Sep 2, 2015

### Rectifier

$N=N_{_{14}C} = \frac{200}{12 \cdot1.660538921 \cdot 10^{-24} } \cdot 1.3 \cdot 10^{-12}=1.30 \cdot 10^{13}$

$N'(t)= -N_0 \lambda e^{-\lambda t}$
$N'(t)= -1.30 \cdot 10^{13} \frac{ln(2)}{5730} e^{-\frac{ln(2)}{5730} t}$

(3) gives
$-210240000= -1.30 \cdot 10^{13} \frac{ln(2)}{5730} e^{-\frac{ln(2)}{5730} t} \\ t = 16634.4$

That animal is 16634.4 old then.

Here is wolfram input on the last stage:
-210240000= -1.30 \cdot 10^{13} \frac{log(e,2)}{5730} e^{-\frac{log(e,2)}{5730} x}

18. Sep 2, 2015

### haruspex

I haven't checked the calculation, but it seems reasonable. But you should not specify so many decimal digits. It overstates the accuracy..

19. Sep 2, 2015

### Rectifier

How many digits should I have? There are 2 significant digits in the problem then I guess it should be 17000 years (not 17000.0)

20. Sep 2, 2015

Yes.