The amount of substance left of a radioactive element of half life, [tex]t_{\frac{1}{2}}[/tex] after a time, t, is given by:
[tex]N(t)=N_0\left( \frac{1}{2} \right)^ \frac{t}{t_{ \frac{1}{2} }}[/tex]
Given that potassium-40 has a half life of approximately 1.25 billion
years.
The number of years it will take for 0.1% of potassium-40 to remain is obtained as follows:
[tex]0.1=100\left( \frac{1}{2} \right)^ \frac{t}{1.25}} \\ \\ \Rightarrow\left( \frac{1}{2} \right)^ \frac{t}{1.25}}=0.001 \\ \\ \Rightarrow\frac{t}{1.25}\ln\left( \frac{1}{2} \right)=\ln(0.001) \\ \\ \Rightarrow \frac{t}{1.25}= \frac{\ln(0.001)}{\ln\left( \frac{1}{2} \right)} =9.966 \\ \\ t=9.966(1.25)=12.5[/tex]
Therefore, the maximum age of a fossil that we could date using 40k is 12.5 billion years.
