Problem 97 - Calculus Single Variable

Question

Radiocarbon Dating Two isotopes of carbon, \({ }^{12} \mathrm{C},\) which is stable, and \({ }^{14} \mathrm{C},\) which decays exponentially with a 5700 -year half-life, are found in a known fixed ratio in living matter. After death, carbon is no longer metabolized, and the amount \(m(t)\) of \({ }^{14} \mathrm{C}\) decreases due to radioactive decay. In the analysis of a sample performed \(T\) years after death, the mass of \({ }^{12} \mathrm{C}\), unchanged since death, can be used to determine the mass \(m_{0}\) of \({ }^{14} \mathrm{C}\) that the sample had at the moment of death. The time \(T\) since death can then be calculated from the law of exponential decay and the measurement of \(m(T)\). Use this information for solving Exercises \(95-98\). Prehistoric cave art has recently been found in a number of new locations. At Chauvet, France, the amount of \({ }^{14} \mathrm{C}\) in some charcoal samples was determined to be \(0.0879 \cdot m_{0}\). Ahout how old are the cave drawings?

Step-by-Step Solution

Answer

The cave drawings are about 18,783 years old.

Step 1: Understand the Problem

We know that after death, the isotope \(^{14}\mathrm{C}\) decays according to an exponential law. Given the remaining percentage of \(^{14}\mathrm{C}\) in a sample compared to its original amount, we need to find the time \(T\) since the death or in this case the creation of charcoal drawings.

Step 2: Determine the Exponential Decay Formula

The exponential decay of \(^{14}\mathrm{C}\) can be described using the formula:\[m(t) = m_0 \cdot e^{-\lambda t}\]where \(\lambda\) is the decay constant, \(m(t)\) is the mass at time \(t\), and \(m_0\) is the initial mass.

Step 3: Calculate the Decay Constant \(\lambda\)

The decay constant \(\lambda\) is related to the half-life by the formula:\[\lambda = \frac{\ln(2)}{\text{half-life}}\]Given that the half-life of \(^{14}\mathrm{C}\) is 5700 years, we calculate:\[\lambda = \frac{\ln(2)}{5700} \approx 1.216 \times 10^{-4} \text{ per year}\]

Step 4: Insert Given Values into the Exponential Decay Formula

We know that after some time \(T\), the amount of \(^{14}\mathrm{C}\) is \(0.0879 \cdot m_0\). Thus:\[0.0879 \cdot m_0 = m_0 \cdot e^{-\lambda T}\]Simplifying gives:\[0.0879 = e^{-\lambda T}\]

Step 5: Solve for Time \(T\)

To find \(T\), take the natural logarithm of both sides:\[\ln(0.0879) = -\lambda T\]Substituting \(\lambda\) from Step 3 gives:\[T = -\frac{\ln(0.0879)}{1.216 \times 10^{-4}}\]Calculating gives approximately:\[T \approx 18783 \text{ years}\]

Key Concepts

Exponential Decay

Exponential decay is a mathematical concept that describes how quantities decrease over time. In the context of radiocarbon dating, this principle is important for understanding how the isotope \(^{14} \mathrm{C}\) diminishes after the death of an organism.
Exponential decay can be represented mathematically by the equation: \[m(t) = m_0 \cdot e^{-\lambda t}\]where:

• \(m(t)\) is the remaining amount of the substance at time \(t\).
• \(m_0\) is the initial amount at the starting time (usually at the time of death for radiocarbon).
• \(e\) is the base of the natural logarithms, approximately equal to 2.718.
• \(\lambda\) is the decay constant (a topic we'll discuss in detail soon).

Exponential decay results in a rapid initial decrease which then slows over time. This makes it very effective for calculating the age of old objects in radiocarbon dating, allowing scientists to estimate how long ago a living organism died by measuring the remaining \(^{14} \mathrm{C}\) in samples.

Half-life

The half-life of a radioactive isotope like \(^{14} \mathrm{C}\) is the time it takes for half of the sample to decay. During each half-life, the amount of the radioactive material reduces by 50%.
This property is pivotal in determining the timeline of decay in radiocarbon dating.

• For \(^{14} \mathrm{C}\), the half-life is approximately 5700 years. This length of time gives scientists a window to measure ages up to about 50,000 years.
• Since half of the substance decays every 5700 years, the concept of half-life helps to easily estimate how much of the original \(^{14} \mathrm{C}\) remains at any given point in time.

Understanding half-life allows us to convert our knowledge of how much \(^{14} \mathrm{C}\) is left into years, which tells us how old a sample, like the Chauvet cave drawings, might be. It is a key step in calculating the age of archaeological findings.

Decay Constant

The decay constant \(\lambda\) is a parameter that quantifies the rate at which a radioactive substance undergoes exponential decay. In other words, it tells us how quickly or slowly the isotope is decaying.
For any radioactive isotope, the decay constant relates to the half-life using the formula: \[\lambda = \frac{\ln(2)}{\text{half-life}}\]where \(\ln(2)\) is the natural logarithm of 2, which is approximately 0.693.
For \(^{14} \mathrm{C}\), with a half-life of 5700 years, \(\lambda\) is calculated as follows:

• \(\lambda = \frac{\ln(2)}{5700} \approx 1.216 \times 10^{-4}\) per year.

This decay constant is critical for inserting into the exponential decay formula to calculate the age of carbon-containing samples. By understanding \(\lambda\), scientists can plug known values into the decay formula to determine how many years have passed since an organism containing carbon atoms, like charcoal, ceased to absorb new carbon.