Problem 41
Question
A 10.00 -g plant fossil from an archaeological site is found to have \(\mathrm{a}^{14} \mathrm{C}\) activity of 3094 disintegrations over a period of ten hours. A living plant is found to have a \({ }^{14} \mathrm{C}\) activity of 9207 disintegrations over the same period of time for an equivalent amount of sample with respect to the total contents of carbon. Given that the half-life of \({ }^{14} \mathrm{C}\) is 5715 years, how old is the plant fossil?
Step-by-Step Solution
Verified Answer
The plant fossil is approximately 8703 years old.
1Step 1: Calculate the Current Carbon-14 Activity Rate
Let's first calculate the rates of disintegration per hour for both the fossil and the living plant. We know the timespan is 10 hours. So, for the fossil, the rate is \( \frac{3094}{10} = 309.4 \) disintegrations per hour. Similarly, for the living plant, the rate is \( \frac{9207}{10} = 920.7 \) disintegrations per hour.
2Step 2: Find the Fraction of Carbon-14 Remained in Fossil
The fraction of carbon-14 that remains in the fossil compared to a living plant can be calculated by the ratio of their activity rates: \( f = \frac{309.4}{920.7} \approx 0.336 \). This fraction represents the remaining carbon-14 in the fossil compared to a current living plant.
3Step 3: Use Formula to Calculate the Age of the Fossil
The formula that relates the fraction of remaining carbon-14 to the age is \( f = \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \), where \( t \) is the age of the fossil and \( T_{1/2} = 5715 \) years is the half-life. Rearrange to find \( t \):\[ t = T_{1/2} \cdot \frac{\log(f)}{\log(0.5)} \].Substitute the known values: \( t = 5715 \cdot \frac{\log(0.336)}{\log(0.5)} \approx 8703 \) years.
Key Concepts
Half-lifeRadioactive decayArchaeological datingDisintegration rate
Half-life
The concept of half-life is central to understanding radioactive decay processes, including carbon-14 dating. The half-life is the amount of time it takes for half of the radioactive atoms in a sample to decay. For Carbon-14, this period is 5,715 years.
Half-life is crucial because it provides a predictable way to determine the age of objects, particularly in archaeology. By knowing how long half of the Carbon-14 in a sample takes to decay, scientists can calculate how long a given sample has been decaying.
Half-life is crucial because it provides a predictable way to determine the age of objects, particularly in archaeology. By knowing how long half of the Carbon-14 in a sample takes to decay, scientists can calculate how long a given sample has been decaying.
- Formula: \( T_{1/2} = 5715 \) years for Carbon-14
- Applications: Helps calculate how many half-lives have passed, aiding in determining sample age
Radioactive decay
Radioactive decay is the process through which unstable atomic nuclei lose energy by emitting radiation. Carbon-14 undergoes radioactive decay by changing into nitrogen over time. This decay process follows an exponential pattern, which means the rate of decay decreases over time.
Carbon-14 decay is particularly useful in dating because it is continuous and measurable. The decay results in fewer Carbon-14 nuclei over the years, which can be quantitatively analyzed to determine how many years have passed since the death of the organism.
Carbon-14 decay is particularly useful in dating because it is continuous and measurable. The decay results in fewer Carbon-14 nuclei over the years, which can be quantitatively analyzed to determine how many years have passed since the death of the organism.
- Emits beta radiation
- Transforms to a stable form (Nitrogen-14)
- Used to estimate age in organic materials
Archaeological dating
Archaeological dating with Carbon-14 allows scientists to determine the age of ancient organic materials. The fundamental idea relies on the fact that living organisms maintain a consistent level of Carbon-14. Once the organism dies, it stops absorbing Carbon-14, and the existing isotopes begin to decay.
By measuring the current level of Carbon-14 compared to expected levels in a living organism of the same type, scientists can estimate the time since death. This methodology is widely used in archaeology to date fossils and artifacts.
By measuring the current level of Carbon-14 compared to expected levels in a living organism of the same type, scientists can estimate the time since death. This methodology is widely used in archaeology to date fossils and artifacts.
- Provides a timeline for historical events and ancient biology
- Helps understand the development of human culture and technology
- Accurate for dating within tens of thousands of years
Disintegration rate
The disintegration rate refers to how quickly the Carbon-14 atoms in a sample are breaking down over time. Measured in disintegrations per unit time, this rate provides insight into how much Carbon-14 remains in a sample.
In the given exercise, comparing the disintegration rate of the fossil with that of a current living plant helps determine how much of the original Carbon-14 remains.
In the given exercise, comparing the disintegration rate of the fossil with that of a current living plant helps determine how much of the original Carbon-14 remains.
- Calculated as disintegrations per minute/hour
- Provides the fraction of remaining radioactive material
- Indicative of the sample's age
Other exercises in this chapter
Problem 36
It takes 180 minutes for a 200 -mg sample of an unknown radioactive substance to decay to \(112 \mathrm{mg}\). What is the halflife of this substance?
View solution Problem 38
How much time is required for a 5.00-g sample of \({ }^{233} \mathrm{~Pa}\) to decay to \(0.625 \mathrm{~g}\) if the half-life for the beta decay of \({ }^{233}
View solution Problem 42
A wooden artifact from an Indian temple has a \({ }^{14} \mathrm{C}\) activity of 42 counts per minute as compared with an activity of 58.2 counts per minute fo
View solution Problem 43
Phosphorus-32 is commonly used in nuclear medicine for the identification of malignant tumors. It decays to sulphur-32 with a half-life of 14.29 days. If a pati
View solution