Chapter 18: Problem 28
An archaeological specimen was analyzed and found to be emitting only \(25 \%\) as much \(C-14\) radiation per gram of carbon as newly cut wood. How old is this specimen?
Short Answer
Expert verified
Approximately 16535 years old.
Step by step solution
01
- Understanding Radioactive Decay
Carbon-14 (\textsuperscript{14}C) undergoes radioactive decay over time. The decay can be modeled by the equation \(N(t) = N_0 e^{-\frac{t}{\tau}}\), where \(N(t)\) is the amount of \textsuperscript{14}C at time \(t\), \(N_0\) is the initial amount of \textsuperscript{14}C, \(\tau\) is the mean lifetime of \textsuperscript{14}C, and \(t\) is the time that has passed.
02
- Half-Life Understanding
The half-life of \textsuperscript{14}C, the time taken for half of the initial amount of \textsuperscript{14}C to decay, is 5730 years. The relation between half-life (\(T_{1/2}\)) and mean lifetime (\(\tau\)) is given by: \[\tau = \frac{T_{1/2}}{\ln(2)}\].
03
- Calculate Mean Lifetime
Using the half-life of 5730 years, calculate the mean lifetime \(\tau\): \[\tau = \frac{5730}{\ln(2)} \approx 8267.679 \, \text{years}\].
04
- Set Up the Decay Equation
Insert the known values into the decay equation. Given that the specimen has 25% of the original \textsuperscript{14}C contents: \[ 0.25N_0 = N_0 e^{-\frac{t}{\tau}}\].
05
- Simplify the Equation
Divide both sides by \(N_0\): \[0.25 = e^{-\frac{t}{\tau}}\].
06
- Solve for Time
Take the natural logarithm of both sides to solve for \(t\): \[\ln(0.25) = -\frac{t}{\tau}\].
07
- Final Calculation
Isolate \(t\) and calculate: \[t = -\tau \cdot \ln(0.25)\]. Substitute \(\tau \approx 8267.679\) years: \[t = -8267.679 \, \text{years} \cdot \ln(0.25) \approx 16535.357 \, \text{years}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Carbon-14 decay
Carbon-14 (also written as \textsuperscript{14}C) is a radioactive isotope of carbon. It naturally occurs in small amounts in the atmosphere and is absorbed by living organisms. When a living organism dies, it stops absorbing Carbon-14, and the existing \textsuperscript{14}C begins to decay. This radioactive decay process allows scientists to estimate the age of archaeological specimens through radioactive decay dating, which is the basis of Carbon-14 dating.
Over time, the Carbon-14 nuclei break down and emit radiation in the form of beta particles. Despite this decay, the amount of \textsuperscript{14}C in the atmosphere remains relatively constant because it is continuously produced by cosmic rays. This steady production allows for an effective dating method for once-living materials.
To model this decay, we use the exponential decay equation: \[N(t) = N_0 e^{-\frac{t}{\tau}}\]\, where:
Over time, the Carbon-14 nuclei break down and emit radiation in the form of beta particles. Despite this decay, the amount of \textsuperscript{14}C in the atmosphere remains relatively constant because it is continuously produced by cosmic rays. This steady production allows for an effective dating method for once-living materials.
To model this decay, we use the exponential decay equation: \[N(t) = N_0 e^{-\frac{t}{\tau}}\]\, where:
- \(N(t)\) is the amount of \textsuperscript{14}C remaining at time \(t\)
- \(N_0\) is the initial amount of \textsuperscript{14}C when the organism died
- \(\tau\) is the mean lifetime of \textsuperscript{14}C
- \(t\) is the time that has elapsed since the organism died
Calculating the Half-Life
The concept of half-life is essential in understanding radioactive decay. The half-life (\textsuperscript{T_{1/2}}) is the time required for half of the radioactive atoms in a sample to decay. For Carbon-14, the half-life is known to be 5730 years, with which we can make age determinations of archaeological finds.
There is a relationship between the half-life (\textsuperscript{T_{1/2}}) and the mean lifetime (\textsuperscript{\tau}}) of a radioactive isotope, expressed by the equation: \[\tau = \frac{T_{1/2}}{\ln(2)}\]\ where\(\ln(2)\) is the natural logarithm of 2.
Using this equation, we can calculate the mean lifetime of Carbon-14:
There is a relationship between the half-life (\textsuperscript{T_{1/2}}) and the mean lifetime (\textsuperscript{\tau}}) of a radioactive isotope, expressed by the equation: \[\tau = \frac{T_{1/2}}{\ln(2)}\]\ where\(\ln(2)\) is the natural logarithm of 2.
Using this equation, we can calculate the mean lifetime of Carbon-14:
- \[ \tau = \frac{5730}{\ln(2)} \approx 8267.679 \text{ years}\]
Exponential Decay Equation in Action
The exponential decay equation is at the core of radioactive dating techniques. To determine the age of an archaeological specimen using Carbon-14 dating, we use the equation:
This approach underscores the importance of each concept: the initial amount of \textsuperscript{14}C (\textsuperscript{N_0}), the remaining amount (\textsuperscript{N(t)}), and the mean lifetime (\textsuperscript{\tau}}) within the exponential decay framework.
- Given: The specimen has 25% of the original \( N_0 \) left, we set up the equation: \[ 0.25 N_0 = N_0 e^{-\frac{t}{\tau}}\]
- Simplify by dividing both sides by \( N_0 \): \[ 0.25 = e^{-\frac{t}{\tau}}\]
- Take the natural logarithm of both sides: \[ \ln(0.25) = -\frac{t}{\tau}\]
- Solve for \( t \): \[ t = -\tau \cdot \ln(0.25) \]
- Substitute \( \tau \approx 8267.679 \text{ years}\): \[ t = - 8267.679 \text{ years} \cdot \ln(0.25) \approx 16535.357 \text{ years}\]
This approach underscores the importance of each concept: the initial amount of \textsuperscript{14}C (\textsuperscript{N_0}), the remaining amount (\textsuperscript{N(t)}), and the mean lifetime (\textsuperscript{\tau}}) within the exponential decay framework.