Problem 355
Question
True or False? If true, prove it. If false, find the true answer. If a relic contains 90\(\%\) as much radiocarbon as new material, can it have come from the time of Christ (approximately 2000 years ago)? Note that the half-life of radiocarbon is 5730 years.
Step-by-Step Solution
Verified Answer
False. The relic is about 868 years old, not from 2000 years ago.
1Step 1: Understanding the Problem
We need to determine if a relic containing 90\(\%\) of the radiocarbon as new material can be from around 2000 years ago, assuming the half-life of radiocarbon is 5730 years. This involves using the decay formula for radiocarbon.
2Step 2: Radiocarbon Decay Formula
The decay of radiocarbon can be modeled by the formula \( N(t) = N_0 \cdot (0.5)^{t/T} \), where \( N(t) \) is the quantity of carbon-14 at time \( t \), \( N_0 \) is the initial quantity, and \( T \) is the half-life (5730 years).
3Step 3: Setting Up the Equation
Since the relic has 90\(\%\) of the original radiocarbon, \( N(t) = 0.9N_0 \). We substitute this into the decay formula: \[ 0.9N_0 = N_0 \cdot (0.5)^{t/5730} \]
4Step 4: Solving for t
Cancel \( N_0 \) on both sides and solve for \( t \): \[ 0.9 = (0.5)^{t/5730} \] Take the natural logarithm of both sides:\[ \ln(0.9) = \frac{t}{5730} \cdot \ln(0.5) \] Solving for \( t \):\[ t = \frac{\ln(0.9)}{\ln(0.5)} \times 5730 \approx 868 \text{ years} \]
5Step 5: Conclusion
The relic is approximately 868 years old based on its radiocarbon content. Since this is much less than 2000 years, the relic cannot be from the time of Christ. Therefore, the statement "If it contains 90\(\%\) radiocarbon, it can be from the time of Christ" is false.
Key Concepts
Half-lifeExponential DecayLogarithms
Half-life
The concept of half-life is central to understanding radiocarbon dating. It refers to the time required for half the quantity of a radioactive isotope in a sample to decay. For radiocarbon (carbon-14), the half-life is approximately 5730 years. This means that after 5730 years, only half of the original carbon-14 remains in a relic or sample.
When an organism dies, it ceases to absorb carbon-14, and the isotope begins to decay at a fixed rate. By measuring the remaining amount of carbon-14 in a sample, scientists can determine how many half-lives have passed and estimate the age of the sample.
In practical applications, knowing the half-life allows researchers to use mathematical models to calculate the decay over time. This is why half-life forms the backbone of radiocarbon dating, helping determine the age of archaeological finds.
When an organism dies, it ceases to absorb carbon-14, and the isotope begins to decay at a fixed rate. By measuring the remaining amount of carbon-14 in a sample, scientists can determine how many half-lives have passed and estimate the age of the sample.
In practical applications, knowing the half-life allows researchers to use mathematical models to calculate the decay over time. This is why half-life forms the backbone of radiocarbon dating, helping determine the age of archaeological finds.
Exponential Decay
Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. In radiocarbon dating, exponential decay helps model the decrease in carbon-14 over time. The formula typically used is \[N(t) = N_0 \cdot (0.5)^{t/T}\]
Where:
Where:
- \(N(t)\) is the amount of carbon-14 at time \(t\)
- \(N_0\) is the initial amount of carbon-14
- \(T\) is the half-life (5730 years for carbon-14)
Logarithms
Logarithms are mathematical functions helpful in solving equations involving exponential relationships, such as the radiocarbon decay equation. When we want to find the time \(t\) it takes for a certain percentage of carbon-14 to decay, logarithms can resolve this.
Consider the equation derived from exponential decay:\[0.9 = (0.5)^{t/5730}\]
To solve for \(t\), take the natural logarithm (ln) of both sides, which allows you to bring the exponent down: \[\ln(0.9) = \frac{t}{5730} \cdot \ln(0.5)\]
Rearrange and solve for \(t\):
\[t = \frac{\ln(0.9)}{\ln(0.5)} \times 5730\]
Using logarithms here simplifies the process of finding \(t\), as exponential decay involves powers, and solving powers directly can be complex without them. Logarithms hence play an important role in the context of radiocarbon dating by aiding in the comparison of currently observed amounts and original quantities, eventually determining the time elapsed.
Consider the equation derived from exponential decay:\[0.9 = (0.5)^{t/5730}\]
To solve for \(t\), take the natural logarithm (ln) of both sides, which allows you to bring the exponent down: \[\ln(0.9) = \frac{t}{5730} \cdot \ln(0.5)\]
Rearrange and solve for \(t\):
\[t = \frac{\ln(0.9)}{\ln(0.5)} \times 5730\]
Using logarithms here simplifies the process of finding \(t\), as exponential decay involves powers, and solving powers directly can be complex without them. Logarithms hence play an important role in the context of radiocarbon dating by aiding in the comparison of currently observed amounts and original quantities, eventually determining the time elapsed.
Other exercises in this chapter
Problem 354
True or False? If true, prove it. If false, find the true answer. How old is a skull that contains one-fifth as much radiocarbon as a modern skull? Note that th
View solution Problem 354
For the following exercises, use \(y=y_{0} e^{k t}\). How old is a skull that contains one-fifth as much radiocarbon as a modern skull? Note that the half-life
View solution Problem 355
For the following exercises, use \(y=y_{0} e^{k t}\). If a relic contains \(90 \%\) as much radiocarbon as new material, can it have come from the time of Chris
View solution Problem 356
For the following exercises, use \(y=y_{0} e^{k t}\). The population of Cairo grew from 5 million to 10 million in 20 years. Use an exponential model to find wh
View solution