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Section 13.6 Carbon Dating and Half-Life

Subsection 13.6.1 Application: Carbon Dating

Carbon dating is a method for estimating how long ago a living thing died. Using it, we can estimate how long ago animals like mammoths lived, or ancient humans, without having actual records of them.

Example 13.6.1.

All living things contain carbon, and there are different types of carbon.
  • One is called carbon-12, which is the β€œnormal” type of carbon.
  • Another is carbon-14, which is a radioactive type of carbon (called an isotope), which decays over time into a different element.
When a plant or animal is alive, it always has roughly the same amount of carbon-14. However, when it dies, it stops taking in new carbon-14 (or any carbon at all). Over time, its amount of carbon-14 decreases. It turns out that it decreases exponentially.
To measure the rate of exponential decay, instead of using a percentage, we use its half-life, which is the time it takes for the amount to decay to half its initial amount.
Carbon-14 has a half-life of 5730 years. This means that:
This means if we start with 100% of the carbon-14, then the percent of carbon-14 remaining after \(t\) years is given by,
\begin{gather*} P(t) = 100 \brac{\frac{1}{2}}^{\frac{t}{5730}} \end{gather*}
  • The growth factor is \(\frac{1}{2}\) because we want to multiply the amount by \(\frac{1}{2}\)
  • \(\frac{t}{5730}\) represents the number of half-lives in \(t\) years. For example, if \(t=5730\text{,}\) that is \(\frac{5730}{5730}=1\) half-lives.
The idea behind carbon dating is that if we measure how much carbon-14 the dead organism has, and also know how much it had when alive, we can estimate how long the carbon-14 has been decaying for, and therefore how long ago it died.

Checkpoint 13.6.2.

Carbon-14 has a half-life of 5730 years. Find the approximate age of a dead organism that has 20% of its original carbon-14 in its tissues.
Hint.
\(A=A_0 \brac{\frac{1}{2}}^{t/5730}\text{,}\) find \(t\) when \(A=0.2A_0\)
Answer.
Approximately 13308 years

Remark 13.6.3.

Carbon dating works for organisms that are up to about 60000 years old, as anything older than that has too little carbon-14 to measure. It was developed in 1949 by Willard Libby (1908-1980), American chemist, and it revolutionized archaeology and paleontology.

Subsection 13.6.2 Application: Radioactive Decay

Many elements have radioactive isotopes, which are versions of the element which are radioactive. These radioactive isotopes decay over time into regular, non-radioactive elements. They decay continuously at a constant percentage, and so they can be modelled using exponential decay.
The half-life of a substance is the time required for a decaying amount to decay to half its initial amount.

Remark 13.6.4.

Half-life is analogous to doubling time for exponential growth.
\begin{align*} \boxed{A = A_0 \brac{\frac{1}{2}}^{t/T} \quad \longrightarrow \begin{cases} A & \text{amount after time } t \\ A_0 & \text{initial amount (at } t = 0\text{)} \\ t & \text{time that has passed} \\ T & \text{half-life (time to reduce to half)} \end{cases}} \end{align*}
  • \(\frac{1}{2}\) represents the growth factor (or decay factor, since it is exponential decay).
  • \(\frac{t}{T}\) represents the number of times that factor is applied (in other words, the number of half-lives in time \(t\)).

Example 13.6.5.

If a radioactive substance starts with 100g, after 1 half-life it will decay to 50g. After another half-life, that 50g will decay to 25g (half of 50g), and so on.
# of half-lives Amount (g)
0 100
1 50
2 25
3 12.5
4 6.25
5 3.125
6 1.5625

Checkpoint 13.6.6.

Checkpoint 13.6.7.

Checkpoint 13.6.8.

Americium (Am) is a radioactive element that is used in household smoke detectors. Am-241 has a half-life of approximately 432 years. The average smoke detector contains 0.28 \(\mu\text{g}\) of Am-241. Find an equation for the amount \(A\) of Am-241 in a smoke detector after \(t\) years.
Answer.
\(A=0.28\brac{\frac{1}{2}}^{\frac{t}{432}}\)

Checkpoint 13.6.9.

Radioactive argon-39 has a half-life of 275 years. If we initially have 84 grams of argon-39, how much remains after 1600 years?
Hint.
Use \(A=84\brac{\frac{1}{2}}^{\frac{t}{275}}\text{,}\) and \(t=1600\)
Answer.
Approximately 1.49 grams

Checkpoint 13.6.10.

Radioactive phosphorus-32 is used to study liver function. It has a half-life of 14.3 days. If a small amount of phosphorus-32 is injected into a person’s body, find how long it will take for the level of radiation to drop to 5% of its original value.
Hint.
\(A=A_0\brac{\frac{1}{2}}^{t/14.3}\text{,}\) find \(t\) when \(A=0.05 A_0\)
Answer.
\(t = \frac{14.3\log{0.05}}{\log{\brac{\frac{1}{2}}}} \approx 61.8\)

Checkpoint 13.6.11.

How long would it take 70% of a radioactive substance to decay if its half-life was 20 hours?
Hint.
\(A=A_0\brac{\frac{1}{2}}^{t/20}\text{,}\) find \(t\) when \(A=0.3 A_0\)
Answer.
\(t=\frac{20\log(0.30)}{\log\brac{\frac{1}{2}}}\approx 34.7\)

Checkpoint 13.6.12.

A radioactive substance is produced from nuclear fallout. If 300 g of this substance decays to 120 g in 30 years, what is the half-life of this substance?
Hint.
\(A=300\brac{\frac{1}{2}}^{t/T}\text{,}\) and when \(t=30\text{,}\) \(A=120\text{,}\) and solve for \(T\)
Answer.
\(T=\frac{30\log\brac{\frac{1}{2}}}{\log\brac{\frac{2}{5}}} \approx 22.7\)

Checkpoint 13.6.13.

The half-life of an element is 15.9 hours. A medical clinic purchases a 50 gram sample of this element. How long until there would be only 12.5 grams left? Solve without using logs.
Hint.
50g to 12.5g is 2 half-lives
Answer.
31.8 hours

Checkpoint 13.6.14.

An element has a half-life of 30 days. How long would it take until 20% of the original amount remained?
Hint.
\(A=A_0\brac{\frac{1}{2}}^{t/30}\text{,}\) find \(t\) when \(A=0.2A_0\)
Answer.
\(t = \frac{30\log\brac{\frac{1}{5}}}{\log\brac{\frac{1}{2}}} \approx 69.7\)

Checkpoint 13.6.15.

(a)
Write a function to relate the amount remaining, \(A\text{,}\) in milligrams, to the time, \(t\text{,}\) in days.
Answer.
\(A(t)=72\brac{\frac{1}{2}}^{\frac{t}{10}}\)
(d)
How long, to the nearest day, will it take for there to be 0.07 mg of the initial sample remaining?
Answer.
\(t=\frac{10\log\brac{\frac{0.07}{72}}}{\log\brac{\frac{1}{2}}}\approx 100\)