# differential equations

A certain moon rock was found to contain equal numbers of potassium and argon atoms. Assume that all the argon is the result of radioactive decay of potassium (half life $1.28 \times 10^9$) and that one of every 9 potassium atom disintegrations yields an argon atom. What is the age of the rock measured from the time it contained only potassium.

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This question doesn't seem to make sense; what happens to the other 8 out of every 9 potassium decays? Unless we treat them as disappearing completely, or decaying into another potassium isotope, there will be a third substance in the rock. –  Zev Chonoles Aug 17 '11 at 19:10
@Zev: I don't think that affects the relative quantity of potassium and argon atoms, does it? The question only states that the rock contains equal numbers of potassium and argon atoms, not that, say, 50% of all the atoms are argon. –  Rahul Aug 17 '11 at 19:20
Aha, you're correct, Rahul. I'm doing terribly today it seems... –  Zev Chonoles Aug 17 '11 at 19:21
Ordinary potassium contains a very rare radioactive isotope. The other isotopes are not radioactive. It is utterly impossible for the ordinary mix of potassium isotopes to end up half potassium and half argon. One has to assume that by potassium the author means the radioactive version only. –  André Nicolas Aug 17 '11 at 19:22

After Rahul's comment above, I understand the question now.

After $t$ years, exactly $\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$ of the potassium will remain, and exactly $\frac{1}{9}$ of the decays will have produced an argon atom. Thus, we have that $$\text{argon}=\frac{1}{9}\cdot\text{ decayed potassium}$$ and $$\text{decayed potassium}=\text{original potassium} - \text{remaining potassium}$$ so $$\text{decayed potassium}=\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)\cdot\text{ original potassium},$$ so $$\text{argon}=\frac{1}{9}\cdot\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)\cdot\text{ original potassium}$$ Now we set this equal to the remaining amount of potassium, which is $$\text{remaining potassium }=\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\cdot\text{ original potassium}$$ This gives the equation $$\frac{1}{9}\cdot\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)=\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$$ in which we want to solve for $t$. This is just simple manipulations now.

We have $$\frac{1}{9}\cdot\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)=\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$$ which becomes $$\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}=\frac{1}{10}$$ so $$\frac{t}{1.28\times 10^9}\times\ln\left(\frac{1}{2}\right)=\ln\left(\frac{1}{10}\right)$$ so $$t=(1.28\times10^9)\frac{\ln\left(\frac{1}{10}\right)}{\ln\left(\frac{1}{2}\right)}=4.252\ldots\times 10^9$$

(see here)

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Thank u for ur reply but I am still confused because the answer is 1.25 billion years. –  rupa Aug 17 '11 at 20:36
I don't think that answer is correct, unless I've made a manipulation error somewhere. –  Zev Chonoles Aug 17 '11 at 20:52
@ruchi, 1.25 billion years can't be right. That's a little less than one half-life, so just about one-half of the original potassium will have decayed, so half the original potassium will be left; but of the half that decayed, only one-ninth will have become argon, so there will be about 9 times as much potassium as argon after 1.25 billion years. –  Gerry Myerson Aug 18 '11 at 1:51