Radioactive decay question. W(t) = W(0)^(-0.000122t) where W(t) is the number of Carbon atoms left after "t" years. Assume W(0) = 6x10^10. Assume we cannot detect C if there are less than 10^3 atoms. What is the oldest fossil we can date?

Guest Sep 14, 2014

#1**+5 **

Set W(t) = 10^{3} and solve for t.

10^{3} = 6*10^{10}*e^{-0.000122t} You didn't have an "e" in your expression, but I suspect it should be there, so I've included it.

Divide both sides by 6*10^{10}

10^{-7}/6 = e^{-0.000122t}

Take logs (natural log) of both sides:

ln(10^{-7}/6) = -0.000122t

Divide both sides by -0.000122

t = -ln(10^{-7}/6)/0.000122 = (7*ln(10)+ln(6))/0.000122

$${\mathtt{t}} = {\frac{\left({\mathtt{7}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{10}}\right)}{\mathtt{\,\small\textbf+\,}}{ln}{\left({\mathtt{6}}\right)}\right)}{{\mathtt{0.000\: \!122}}}} \Rightarrow {\mathtt{t}} = {\mathtt{146\,802.091\: \!149\: \!068\: \!645\: \!811}}$$

t ≈ 146802 years

Alan Sep 14, 2014

#1**+5 **

Best Answer

Set W(t) = 10^{3} and solve for t.

10^{3} = 6*10^{10}*e^{-0.000122t} You didn't have an "e" in your expression, but I suspect it should be there, so I've included it.

Divide both sides by 6*10^{10}

10^{-7}/6 = e^{-0.000122t}

Take logs (natural log) of both sides:

ln(10^{-7}/6) = -0.000122t

Divide both sides by -0.000122

t = -ln(10^{-7}/6)/0.000122 = (7*ln(10)+ln(6))/0.000122

$${\mathtt{t}} = {\frac{\left({\mathtt{7}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{10}}\right)}{\mathtt{\,\small\textbf+\,}}{ln}{\left({\mathtt{6}}\right)}\right)}{{\mathtt{0.000\: \!122}}}} \Rightarrow {\mathtt{t}} = {\mathtt{146\,802.091\: \!149\: \!068\: \!645\: \!811}}$$

t ≈ 146802 years

Alan Sep 14, 2014