Problem 40 - Chemistry: The Central Science

Question

A wooden artifact from a Chinese temple has \(\mathrm{a}^{14} \mathrm{C}\) activity of \(38.0\) counts per minute as compared with an activity of \(58.2\) counts per minute for a standard of zero age. From the half-life for \({ }^{14} \mathrm{C}\) decay, \(5715 \mathrm{yr}\), determine the age of the artifact.

Step-by-Step Solution

Answer

The age of the artifact is approximately \(3151\) years.

Step 1: Calculate the activity ratio

First, we need to find the ratio of the artifact's activity to the zero-age standard's activity. This can be calculated by dividing the artifact's activity by the zero-age standard's activity: \[\frac{A_{artifact}}{A_{standard}}=\frac{38.0}{58.2}\]

Step 2: Evaluate the activity ratio

Next, determine the value of the activity ratio by dividing the two given activities: \[\frac{38.0}{58.2} \approx 0.6526\]

Step 3: Use the decay equation

We will now use the radioactive decay equation. For an artifact and the zero-age standard, the equation is given by: \[\frac{N}{N_0} = \frac{A_{artifact}}{A_{standard}} = \frac{38.0}{58.2} = e^{-\lambda t}\] where \(N\) is the current amount of \({}^{14}\mathrm{C}\), \(N_0\) is the initial amount of ${}^{14}\mathrm{C}\), \(\lambda\) is the decay constant, and \(t\) is the time in years.

Step 4: Express decay constant in terms of half-life

We're given the half-life of ${}^{14}\mathrm{C} = 5715\,\mathrm{yr}\), and the formula connecting half-life (\(t_{1/2}\)) and decay constant (\(\lambda\)) is given by: \[t_{1/2}=\frac{\ln 2}{\lambda}\] Now, isolate \(\lambda\): \[\lambda = \frac{\ln 2}{t_{1/2}}\]

Step 5: Calculate the decay constant

Next, we substitute the given half-life value (\(5715\,\mathrm{yr}\)) into the decay constant equation: \[\lambda = \frac{\ln 2}{5715} \approx 1.21 \times 10^{-4}\,\mathrm{yr}^{-1}\]

Step 6: Solve for the age of the artifact

Now, substitute the decay constant and the activity ratio into the radioactive decay equation and solve for \(t\): \[e^{-1.21 \times 10^{-4} \times t} = 0.6526\] Next, take the natural logarithm of both sides: \[-1.21 \times 10^{-4} \times t = \ln 0.6526\] Finally, solve for \(t\): \[t = \frac{\ln 0.6526}{-1.21 \times 10^{-4}} \approx 3151 \,\mathrm{yr}\]

Step 7: Final answer

The age of the artifact is approximately \(3151\) years.

Key Concepts

Radioactive Decay

Radioactive decay is a fundamental process through which unstable atomic nuclei transform into more stable forms, releasing energy in the form of radiation. It occurs naturally and can be harnessed in various ways, such as in medical applications, archeology, and determining the age of geological samples.

Each radioactive isotope, such as Carbon-14 (used in radiocarbon dating), decays at a fixed, known rate that is characterized by its half-life. During decay, particles such as alpha particles, beta particles, or gamma rays are emitted, until a stable, non-radioactive isotope is formed.

Understanding radioactive decay is essential for interpreting radiocarbon dating results. By comparing the remaining amount of a radioactive isotope in a sample to a standard of known activity, scientists can estimate the time that has elapsed since the death of an organism, a principle which enables the dating of ancient artifacts and fossils.

Half-Life Calculation

Understanding Half-Life

The half-life of a radioactive isotope is the time required for half of the substance to decay. It is a measure of the rate at which a radioactive material will disappear from a sample. Knowing the half-life is crucial for calculating the age of objects, as demonstrated in the textbook exercise for dating a wooden artifact. In the case of Carbon-14, the half-life is approximately 5,715 years.

An artifact's half-life carries direct implications for radiocarbon dating. It allows us to model the decay over time and, by using the half-life formula \( t_{1/2} = \frac{\ln 2}{\lambda} \), we can connect the decay constant (\(\lambda\)) directly back to the half-life. This relationship is pivotal for scientists to estimate the age of once-living materials and is a cornerstone in the field of archeology.

Exponential Decay Equation

Applying the Exponential Decay Equation

The exponential decay equation describes how the quantity of a substance decreases over time. In our radiocarbon dating example, the equation \( \frac{N}{N_0} = e^{-\lambda t} \) expresses the ratio of the remaining amount of Carbon-14 (\(N\)) to the initial amount at zero time (\(N_0\)). The \(e^{-\lambda t}\) term illustrates the exponential nature of decay, where \(\lambda\) is the decay constant and \(t\) is the time elapsed.

This equation provides a mathematical method for determining the time that has passed since an organism has died and stopped exchanging Carbon-14 with the environment. Solving for the time \(t\) involves isolating \(t\) from the known values or measurements such as activity rates, decay constants, and the natural logarithm function. This equation is an essential tool for archaeologists and geologists for dating artifacts and geological events, allowing them to construct a timeline of historical and prehistorical events.