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Chapter 29: Problem 36

Carbon-14 dating is used to date a bone found at an archaeological excavation. If the ratio of \(\mathrm{C}-14\) to \(\mathrm{C}-12\) atoms is \(3.25 \times 10^{-13},\) how old is the bone? [Hint: Note that this ratio is one fourth the ratio of \(1.3 \times 10^{-12}\) that is found in a living sample. \(]\)

Short Answer

Expert verified
The age of the bone is approximately 11,460 years.

Step by step solution

01

Understand the Problem

We are given the ratio of Carbon-14 to Carbon-12 in a bone and in a living sample. The ratio in the bone is \(3.25 \times 10^{-13}\) and in the living sample is \(1.3 \times 10^{-12}\). We need to find the age of the bone using Carbon-14 dating.
02

Determine the Half-life of Carbon-14

The half-life of Carbon-14 is approximately 5730 years. This value will be used to calculate the decay over time and estimate the age of the bone.
03

Calculate the Decay Rate and Establish the Equation

The decay of Carbon-14 is exponential in nature, described by the formula: \[N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}\]where \(N\) is the current quantity of \(\mathrm{C}-14\), \(N_0\) is the initial quantity, \(t\) is the time elapsed, and \(t_{1/2}\) is the half-life.
04

Calculate the Initial to Current Ratio

Since the given ratio in the bone is one-fourth of the living sample's ratio, we have:\[\frac{N}{N_0} = \frac{3.25 \times 10^{-13}}{1.3 \times 10^{-12}} = \frac{1}{4}\]
05

Solve for Time (Age of the Bone)

Using the ratio \(\frac{1}{4}\) and the equation from Step 3, set it up as:\[\left( \frac{1}{2} \right)^{\frac{t}{5730}} = \frac{1}{4}\]Taking logarithms yields:\[-\frac{t}{5730} \log(2) = \log \left( \frac{1}{4} \right)\]Solve for \(t\):\[t = \frac{-\log \left( \frac{1}{4} \right)}{\log(2)} \times 5730\]Calculate this to find \(t \approx 11460\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life
The concept of half-life is crucial in understanding how substances decay over time, such as Carbon-14 in radiocarbon dating. Half-life is the time taken for half of the radioactive isotopes in a sample to decay. For Carbon-14, this period is approximately 5730 years.
  • Half-life remains constant for a given isotope and does not change over time.
  • This property allows scientists to predict how much of a sample remains after a certain number of years.
The understanding of half-life is pivotal in Carbon-14 dating, as it provides the basis for calculating the age of artifacts.
Knowing that the amount of Carbon-14 decreases by half every 5730 years is a significant part of how we can determine the age of archaeological finds.
Exponential Decay
Exponential decay describes the process by which quantities decrease at a rate proportional to their current value. This concept is fundamental in Carbon-14 dating to calculate how much Carbon-14 remains over time.
The formula used is:\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
  • Where:
    • \(N\) is the remaining quantity of Carbon-14 at time \(t\).
    • \(N_0\) is the initial quantity of Carbon-14.
    • \(t_{1/2}\) is the half-life of Carbon-14, 5730 years.
  • As time \(t\) increases, \(N\) decreases exponentially, following the half-life rules.
Understanding this concept helps us calculate how many half-lives have passed, thereby enabling us to find out the age of ancient artifacts from archaeological digs.
Archaeological Excavation
Archaeological excavation is the process of systematically uncovering past human activity buried under soil or other materials. This often involves careful excavation or digging up of sites where artifacts or human remains are believed to reside.
  • Carbon-14 dating is a method frequently used in archaeology to date organic materials found at excavation sites.
  • This technique is especially valuable when dating previously living things like bones or wooden objects.
As archaeologists uncover items, understanding the techniques of Carbon-14 dating allows them to determine the age of the items. This helps them reconstruct the timeline of human history and understand significant cultural evolutions. Integrating archaeological findings with Carbon-14 dating offers a scientific basis for categorizing and interpreting historical events.

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Most popular questions from this chapter

How many neutrons are found in a \(^{35} \mathrm{Cl}\) nucleus?
In this problem, you will verify the statement (in Section 29.4) that the \(^{14} \mathrm{C}\) activity in a living sample is \(0.25 \mathrm{Bq}\) per gram of carbon. (a) What is the decay constant \(\lambda\) for \(^{14} \mathrm{C} ?\) (b) How many \(^{14} \mathrm{C}\) atoms are in \(1.00 \mathrm{g}\) of carbon? One mole of carbon atoms has a mass of \(12.011 \mathrm{g},\) and the relative abundance of \(^{14} \mathrm{C}\) is \(1.3 \times 10^{-12} .\) (c) Using your results from parts (a) and (b), calculate the \(^{14} \mathrm{C}\) activity per gram of carbon in a living sample.
\(_{52}^{106} \mathrm{Te}\) is radioactive; it \(\alpha\) decays to $^{102} \mathrm{s}_{0} \mathrm{Sn} .\( " \)_{50}^{102} \mathrm{Sn}$ is itself radioactive and has a half-life of 4.5 s. At \(t=0,\) a sample contains 4.00 mol of \(^{106}\) Te and 1.50 mol of \(^{102}\) so Sn. At \(t=\) \(25 \mu \mathrm{s},\) the sample contains \(3.00 \mathrm{mol}\) of \(^{106} \mathrm{Te}\) and $2.50 \mathrm{mol}\( of \)^{102} \mathrm{s} 0$ Sn. How much 102 50 Sn will there be at \(t=50 \mu \mathrm{s} ?\)

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