Students will develop an understanding of how carbon dating works and also learn to solve for the age of an artifact using the carbon dating technique.
Key Equations
\begin{align*}M = M_0 (\frac{1}{2})^{\frac{t}{t_H}}\end{align*}
The amount of mass \begin{align*}M\end{align*} of the substance surviving after an original \begin{align*}\mathit{M}_0\end{align*} \begin{align*}n\end{align*} uclei decay for time \begin{align*}t\end{align*} with a half-life of \begin{align*}\mathit{t}_H\end{align*}.
Example 1
Question: The half-life of \begin{align*}^{239}\end{align*}Pu is \begin{align*}24,119\end{align*} years. You have \begin{align*}31.25\end{align*} micrograms left, and the sample you are studying started with \begin{align*}2000\end{align*} micrograms. How long has this rock been decaying?
Answer: We will use the equation for time and simply plug in the known values. \begin{align*} t=t_H\frac{\ln{\frac{N}{N_0}}}{\ln{\frac{1}{2}}}=24119\mathrm{y}\frac{\ln{\frac{31.25\mu\mathrm{g}}{2000\mu\mathrm{g}}}}{\ln{\frac{1}{2}}}=144,700\mathrm{years} \end{align*}
Watch this Explanation
Time for Practice
- The half-life of \begin{align*}^{239}\mathrm{Pu}\end{align*} is \begin{align*}24,119\end{align*} years. You have \begin{align*}31.25\end{align*} micrograms left, and the sample you are studying started with \begin{align*}2000\end{align*} micrograms. How long has this rock been decaying?
- A certain fossilized plant is \begin{align*}23,000\end{align*} years old. Anthropologist Hwi Kim determines that when the plant died, it contained \begin{align*}0.250 \;\mathrm{g}\end{align*} of radioactive \begin{align*}^{14}\mathrm{C} (t_H = 5730 \;\mathrm{years})\end{align*}. How much should be left now?
- Jaya unearths a guinea pig skeleton from the backyard. She runs a few tests and determines that \begin{align*}99.7946\end{align*}% of the original \begin{align*}^{14}\mathrm{C}\end{align*} is still present in the guinea pig’s bones. The half-life of \begin{align*}^{14}\mathrm{C}\end{align*} is \begin{align*}5730\end{align*} years. When did the guinea pig die?
- You use the carbon dating technique to determine the age of an old skeleton you found in the woods. From the total mass of the skeleton and the knowledge of its molecular makeup you determine that the amount of \begin{align*}^{14}\mathrm{C}\end{align*} it began with was \begin{align*}0.021\end{align*} grams. After some hard work, you measure the current amount of \begin{align*}^{14}\mathrm{C}\end{align*} in the skeleton to be \begin{align*}0.000054\end{align*} grams. How old is this skeleton? Are you famous?
- Micol had in her lab two samples of radioactive isotopes: \begin{align*}^{151}\mathrm{Pm}\end{align*} with a half-life of \begin{align*}1.183\end{align*} days and \begin{align*}^{134}\;\mathrm{Ce}\end{align*} with a half-life of \begin{align*}3.15\end{align*} days. She initially had \begin{align*}100 \;\mathrm{mg}\end{align*} of the former and \begin{align*}50 \;\mathrm{mg}\end{align*} of the latter.
- Do a graph of quantity remaining (vertical axis) vs. time for both isotopes on the same graph.
- Using the graph determine at what time the quantities remaining of both isotopes are exactly equal and what that quantity is.
- Micol can detect no quantities less than \begin{align*}3.00 \;\mathrm{mg}\end{align*}. Again, using the graph, determine how long she will wait until each of the original isotopes will become undetectable.
- The \begin{align*}Pm\end{align*} goes through \begin{align*}\beta-\end{align*} decay and the \begin{align*}Ce\end{align*} decays by means of electron capture. What are the two immediate products of the radioactivity?
- It turns out both of these products are themselves radioactive; the \begin{align*}Pm\end{align*} product goes through \begin{align*}\beta-\end{align*} decay before it becomes stable and the \begin{align*}Ce\end{align*} product goes through \begin{align*}\beta+\end{align*} decay before it reaches a stable isotope. When all is said and done, what will Micol have left in her lab?
Answers to Selected Problems
- \begin{align*}t = 144,700\end{align*} years
- \begin{align*}0.0155 \;\mathrm{g}\end{align*}
- \begin{align*}17\end{align*} years
- \begin{align*}49,000\end{align*} years
- .