# Calculating the age of a radioactive source

1. Apr 30, 2006

### lando45

Hey,

I have been set this question, and I am finding it extremely difficult to begin. So much information is given, I don't know how to start to answer it. Also, some of the terms are confusing...

A living specimen in equilibrium with the atmosphere contains one atom of 14C (half-life = 5730 yr) for every 7.7×10^11 stable carbon atoms. An archeological sample of wood (cellulose, C12 H22 O11) contains 22.2 mg of carbon. When the sample is placed inside a shielded beta counter with 89.0% counting efficiency, 950 counts are accumulated in one week. Assuming that the cosmic-ray flux and the Earth's atmosphere have not changed appreciably since the sample was formed, find the age of the sample.

So could somebody point me in the right direction to start with?

2. May 1, 2006

### lando45

Anyone? I only have another 21 hours to answer it!

3. May 1, 2006

### Hootenanny

Staff Emeritus
What equations do you know which relate number of radioactive atoms, count rate and time?

HINT: Important data which you are given;

-Number of carbon 14 atoms in living tree (inital amount)
-Total mass of carbon in dead speciemen
-Fraction of carbon which is radioactive
-Counts per week (decays per week)
-Efficency of GM tube.

~H

4. May 1, 2006

### daveb

First you'll want to find the activity of C-14 at any time in the past as a function of time (hint: the decay equation). Use this and the total number of carbon atoms in the sample to find out when the ratio of C-14 to C-12 equals the given ratio.

5. May 1, 2006

### lando45

OK, well I was discussing this with a friend and this is what we came up with:

Using
N= No*e^(-tLambda)
We need: N, No, Lambda, to find 't'.
Lambda is given by Ln(2)/Half Life
This yields N as A/Lambda
A is given by counts recorded, divided by efficiency (as a decimal <1), then divided by (7*24*60*60), ie, a week in seconds.
We therefore have N.
To find No (this is the bit which I’m unsure of)…
We take moles = mass / atomic mass, ie: 0.021/12.
Then times that by Avogadro’s number to get the number of Carbon 12 atoms currently present in the sample.
I find this step to be a leap of knowledge, so tell me if it’s wrong:
The data says “One in [number] carbon atoms are C14” – so would we divide the number found by (mass)/(atomic mass)*Avogadro by 1/(figure given)?

If this yields No, then t would be given by
Ln(No/N) /Lambda. But that’s wrong.

Where is that going wrong?

6. May 1, 2006

### Hootenanny

Staff Emeritus
No you should multiply by 1+efficencey (i.e. 1.89).

I'm note sure what you've done here, the mass of the same is 22.2mg therefore the number of moles is given by 0.0222/12.

Also for your final formula I believe you have made a mistake.

$$N = N_{0}e^{-\lambda t}$$
$$\ln\frac{N}{N_{0}} = -\lambda t$$
$$t = -\frac{1}{\lambda}\ln{\frac{N}{N_{0}}$$

~H

7. May 1, 2006

### lando45

I don't understand why I should be multiplying by 1 + efficiency? Surely counts recorded = 0.89 x real counts?

And also, the reason a couple of the values are different is because that was the data in my friends question, we worked it out using his values.

8. May 1, 2006

### Hootenanny

Staff Emeritus
My apologies, you are correct. However, take note of the error in your rearranged formula, I think that's where you slipped up.

~H

9. May 1, 2006

### lando45

Oh right OK. You're saying I should use:

t = (-1/lambda) * [ln (N/No)]

But if rearranged, that's the same as what I used:

t = [ln (No/N)] / lambda

Isn't it?

10. May 1, 2006

### Hootenanny

Staff Emeritus
Yes, it is. So why do you believe your answer to be wrong? Just to check have you converted your halflife into seconds?

~H

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