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:
  • \(N(t)\) 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:
  • \[ \tau = \frac{5730}{\ln(2)} \approx 8267.679 \text{ years}\]
This conversion is necessary because it allows us to switch between the concepts of half-life and mean lifetime to make calculations more manageable.
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:
  • 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 step-by-step calculation shows how we can determine the age of an archaeological specimen by understanding and implementing each part of the exponential decay 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.

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