Question

Devise an exponential decay function that fits the following data; then answer the accompanying questions. Be sure to identify the reference point \((t=0)\) and units of time. The half-life of \(\mathrm{C}-14\) is about 5730 yr. a. Archaeologists find a piece of cloth painted with organic dyes. Analysis of the dye in the cloth shows that only \(77 \%\) of the \(\mathrm{C}-14\) originally in the dye remains. When was the cloth painted? b. A well-preserved piece of wood found at an archaeological site has \(6.2 \%\) of the \(\mathrm{C}-14\) that it had when it was alive. Estimate when the wood was cut.

Short answer

Question: In an archaeological site, two samples were found: a cloth with a pigment containing 77% of its original C-14 content and a piece of wood containing 6.2% of its original C-14 content. Based on the C-14 half-life of 5730 years, determine the approximate age of each artifact. Answer: The cloth was painted approximately 1417.9 years ago and the wood was cut about 14241.7 years ago.

Step-by-step solution

Formulate the exponential decay function with the given half-life

With the half-life of \(\mathrm{C}-14\) given as 5730 years, we can plug this into the exponential decay function: $$N(t) = N_0 \cdot \left(\frac{1}{2}\right)^\frac{t}{5730}$$

Solve for \(t\) in part a

We're given that \(77\%\) of the \(\mathrm{C}-14\) is remaining in the dye, which means \(N(t) = 0.77N_0\). Now we need to find the time \(t\) when this occurs: $$0.77N_0 = N_0 \cdot \left(\frac{1}{2}\right)^\frac{t}{5730}$$ Divide both sides by \(N_0\) to eliminate it: $$0.77 = \left(\frac{1}{2}\right)^\frac{t}{5730}$$

Use logarithms to solve for \(t\)

To isolate \(t\), we will use logarithms. Take the logarithm of both sides of the equation with base \(2\): $$\log_2(0.77) = \frac{t}{5730}$$ Multiply both sides by \(5730\) to get \(t\): $$t = 5730 \cdot \log_2(0.77)$$ Calculate the value of \(t\): $$t \approx 1417.9$$ So, the cloth was painted about 1417.9 years ago.

Solve for \(t\) in part b

We're given that \(6.2\%\) of the \(\mathrm{C}-14\) is remaining in the wood, which means \(N(t) = 0.062N_0\). Now we need to find the time \(t\) when this occurs: $$0.062N_0 = N_0 \cdot \left(\frac{1}{2}\right)^\frac{t}{5730}$$ Divide both sides by \(N_0\) to eliminate it: $$0.062 = \left(\frac{1}{2}\right)^\frac{t}{5730}$$

Use logarithms again to solve for \(t\)

Take the logarithm of both sides of the equation with base \(2\): $$\log_2(0.062) = \frac{t}{5730}$$ Multiply both sides by \(5730\) to get \(t\): $$t = 5730 \cdot \log_2(0.062)$$ Calculate the value of \(t\): $$t \approx 14241.7$$ So, the wood was cut about 14241.7 years ago.

Key concepts

Understanding Half-Life

One of the core concepts in understanding exponential decay, especially in radiocarbon dating, is the notion of half-life. A half-life is the time required for a quantity to reduce to half its initial value. For instance, with Carbon-14 ( C-14 ), which is a radioactive isotope used in dating organic material, the half-life is about 5730 years.
This means that every 5730 years, only half of the original C-14 remains. The formula used to describe this decay is: \( N(t) = N_0 \cdot \left(\frac{1}{2}\right)^\frac{t}{\text{half-life}} \) , where \( N_0 \) is the original quantity of the substance.
Half-life is important because it allows scientists to measure the passage of time in a consistent and predictable manner, using the decrease in C-14 to date ancient organic materials.

Carbon-14 Dating

Carbon-14 dating, or radiocarbon dating, is a method used by archaeologists and other scientists to determine the age of organic materials. This technique is based on the decay of C-14 , a naturally occurring radioactive isotope of carbon. Since living organisms continually exchange carbon with their environment, they maintain a constant level of C-14 .
When an organism dies, it stops absorbing carbon, and the C-14 present begins to decay. By measuring how much C-14 remains in a sample compared to how much is expected from a living organism (\( N_0 \)), the time since death, or age, can be calculated.

• The standard equation used is \( N(t) = N_0 \cdot \left(\frac{1}{2}\right)^\frac{t}{5730} \), showing the relationship between remaining C-14 and time.
• This method is highly effective and generally accurate for materials up to about 50,000 years old.

Logarithms in Exponential Decay

Logarithms are an essential tool in solving exponential decay problems, like those involving C-14 dating. When we know the amount of a substance left and need to find how long the decay took, logarithms help us solve for time \( t \).
For example, if \( 77\% \) of C-14 is remaining, we would set up the equation \( 0.77 = \left(\frac{1}{2}\right)^\frac{t}{5730} \). To isolate \( t \), logarithms are used:

• Take the logarithm (base 2) of both sides: \( \log_2(0.77) = \frac{t}{5730} \).
• Solve for \( t \) by multiplying both sides by 5730: \( t = 5730 \cdot \log_2(0.77) \).

Logarithms transform the exponential equation to a linear form that is much simpler to solve, making complex decay processes manageable and understandable.