# Meteoric Radiation Decay Question Check

1. May 21, 2008

### TFM

1. The problem statement, all variables and given/known data

I have done a question, but I am not sure wether the answer I got is correct:

A meteorite sampleis found to have some 40-Ar trapped in a small volume deep inside it, which also contains exactly the same amount of radioactive 40-K. Assuming that when the meteorite was formed there was no trapped argon, that every 100 40-K nuclei decay into 11 40-Argon nuclei (and 89 40-Ca, which play no further part in this question) and that the half life of 40-K is 1.3x10^9 years, estimate the age of the meteorite in years.

2. Relevant equations

$$\frac{N(^{40}Ar)}{N()^{40}K} = e^{\lambda t_{Formation}} - 1$$

$$t_{1/2} = \frac{ln2}{\lambda}$$

3. The attempt at a solution

Rearranging the formula, I get:

$$\frac{N(^{40}Ar)}{N(^{40}K)} = e^{(\frac{ln2}{t_{1/2}}) t_{Formation}} - 1$$

I considered the ratio to be 50:50, or 1:1, since there was equal numbers. Put in all the variables into the above equation, and I got a nice number of 1.3 billion years. It seems a fair enough answer.

My problem is, I don't seem to have taken into consideration the fact that only 11 of the 100 potassium atoms decay into Argon. and the fact that the time is the half-life shows that this should be the answer when the Potassium only forms argon, since half-life is the time for half the atoms to decay, it means from the half that have decaysed, really, only 11% should be argon.

Have I made a mistake somewhere/used the wrong formula?

TFM

2. May 21, 2008

### tiny-tim

Hi TFM!

The question isn't very clear, but I think you're right: if you start with 200 atoms of K, then after one half-life you have 100 of K, 11 of Ar, and 89 of Ca.

So your $$\frac{N(^{40}Ar)}{N()^{40}K} = e^{\lambda t_{Formation}} - 1$$

should say $$\frac{N(^{40}Ar\,+\,^{40}Ca)}{N()^{40}K} = e^{\lambda t_{Formation}} - 1$$,

and then you use N(Ar)/N(Ca) = 11/89.