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If an archaeologist uncovers a seashell which contains \(60 \%\) of the \({ }^{14} \mathrm{C}\) of a living shell, how old do you estimate that shell, and thus that site, to be? (You may assume the half-life of \({ }^{14} C\) to be 5,568 years.

Short Answer

Expert verified
The shell is approximately 4,118 years old.

Step by step solution

01

Understand the Problem

We are given that a seashell has 60% of the original amount of C. We need to find how old the shell is, using the concept of radioactive decay. The half-life of C is given as 5,568 years.
02

Set Up the Radioactive Decay Formula

The general formula for radioactive decay is \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the remaining quantity of the substance, \( N_0 \) is the initial quantity, \( t \) is time, and \( k \) is the decay constant. We are looking for \( t \).
03

Determine the Decay Constant

The decay constant \( k \) can be calculated using the half-life formula: \( k = \frac{\ln(2)}{\text{half-life}} \). Substituting the half-life, we have \( k = \frac{\ln(2)}{5568} \approx 0.000124 \text{ per year} \).
04

Substitute Known Values into the Decay Formula

We have \( N(t)/N_0 = 0.6 \). Substituting into the decay formula gives: \( 0.6 = e^{-0.000124t} \).
05

Solve for Time \( t \)

First take the natural logarithm of both sides to solve for \( t \): \( \ln(0.6) = -0.000124t \). This leads to \( t = \frac{\ln(0.6)}{-0.000124} \).
06

Calculate the Age

Substituting into the logarithm calculation: \( t \approx \frac{-0.5108}{-0.000124} \approx 4118 \text{ years} \).

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Key Concepts

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

Unraveling Archaeology
Archaeology is the scientific study of human history through the excavation of sites and the analysis of artifacts and other physical remains. When archaeologists uncover ancient objects, they embark on a journey to piece together not only the age of the objects themselves but also the historical eras they pertain to. By achieving this, they can build a comprehensive story of past human activity. Understanding the age of artifacts is crucial, as it allows archaeologists to understand the timeline of settlements, migration, and technological advancements. For example, analyzing a seashell found at a site can provide clues about the diet, environment, and trade of earlier communities. Beyond mere excavation, archaeology uses various dating techniques to pinpoint ages accurately. As we transition into the next sections, you'll see how carbon dating, a common technique, plays a significant role in this scientific discipline.
Decoding Carbon Dating
Carbon dating, or radiocarbon dating, is a technique that archaeologists use to determine the age of organic materials. This method is based on the radioactive decay of 14 C, a naturally occurring isotope of carbon. Essentially, living organisms continuously exchange carbon with the environment in the form of carbon dioxide. Once they die, this exchange stops, but the 14 C in their remains continues to decay at a constant rate. Given enough time, the amount of 14 C decreases predictably thanks to its known half-life. By comparing the ratio of remaining 14 C in a sample to what one would expect in a living organism, scientists can infer the time that has elapsed since the organism's death. Carbon dating isn't reserved just for seashells. It applies to any organic material up to about 50,000 years old. The technique has revolutionized archaeology, providing more accurate timelines for examining the human past. Understanding how to calculate ages using carbon dating is a fundamental skill for anyone delving into historical analysis.
Demystifying Half-Life Calculation
Half-life is a crucial concept in the field of radioactive dating. It represents the time required for half of the radioactive nuclei in a sample to decay. For carbon-14 (14C), this time is approximately 5,568 years. This essentially means that every 5,568 years, half of a sample's original 14C atoms will have decayed into a different element.To calculate the age of an object based on carbon dating, scientists use the decay formula \(N(t) = N_0 e^{-kt}\), where:
  • \(N(t)\) is the current amount of 14C
  • \(N_0\) is the original amount when the organism was alive
  • \(k\) is the decay constant, determined using the half-life (\(k = \frac{\ln(2)}{\text{half-life}}\))
Once the formula is set up with the known values, scientists can solve for \(t\), representing the time since the organism's death, providing a chronological anchor for archaeological findings. By understanding half-life and using simple math, we can unravel mysteries of the past, from the age of seashells to ancient human settlements.

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Most popular questions from this chapter

(From Borelli and Coleman (1996).) Olduvai Gorge, in Kenya, cuts through volcanic flows, tuff (volcanic ash), and sedimentary deposits. It is the site of bones and artefacts of early hominids, considered by some to be precursors of man. In 1959, Mary and Louis Leakey uncovered a fossil hominid skull and primitive stone tools of obviously great age, older by far than any hominid remains found up to that time. Carbon-14 dating methods being inappropriate for a specimen of that age and nature, dating had to be based on the ages of the underlying and overlying volcanic strata. The method used was that of potassium-argon decay. The potassium-argon clock is an accumulation clock, in contrast to the \({ }^{14} \mathrm{C}\) dating method. The potassium-argon method depends on measuring the accumulation of 'daughter' argon atoms, which are decay products of radioactive potassium atoms. Specifically, potassium- \(40\left({ }^{40} \mathrm{~K}\right)\) decays to argon \(\left({ }^{40} \mathrm{Ar}\right)\) and to Calcium- 40 \(\left({ }^{40} \mathrm{Ca}\right)\) by the branching cascade illustrated below in Figure 2.17. Potassium decays to calcium by emitting a \(\beta\) particle (i.e. an electron). Some of the potassium atoms, however, decay to argon by capturing an extra-nuclear electron and emitting a \(\gamma\) particle. The rate equations for this decay process may be written in terms of \(K(t), A(t)\) and \(C(t)\), the potassium, argon and calcium in the sample of rock: $$ \begin{aligned} &K^{\prime}=-\left(k_{1}+k_{2}\right) K \\ &A^{\prime}=k_{1} K \\ &C^{\prime}=k_{2} K \end{aligned} $$ where $$ k_{1}=5.76 \times 10^{-11} \text { year }^{-1}, \quad k_{2}=4.85 \times 10^{-10} \text { year }^{-1} \text { . } $$ (a) Solve the system to find \(K(t), A(t)\) and \(C(t)\) in terms of \(k_{1}, k_{2}\), and \(k_{3}=k_{1}+k_{2}\), using the initial conditions \(K(0)=k_{0}, A(0)=C(0)=0 .\) (b) Show that \(K(t)+A(t)+C(t)=k_{0}\) for all \(t \geq 0 .\) Why would this be the case? (c) Show that \(K(t) \rightarrow 0, A(t) \rightarrow k_{1} k_{0} / k_{3}\) and \(C(t) \rightarrow k_{2} k_{0} / k_{3}\) as \(t \rightarrow \infty\). (d) The age of the volcanic strata is the current value of the time variable \(t\) because the potassiumargon clock started when the volcanic material was laid down. This age is estimated by measuring the ratio of argon to potassium in a sample. Show that this ratio is $$ \frac{A}{K}=\frac{k_{1}}{k_{3}}\left(e^{k_{3} t}-1\right) $$ (e) Now show that the age of the sample in years is $$ \frac{1}{k_{3}} \ell \mathrm{n}\left[\left(\frac{k_{3} A}{k_{1} K}\right)+1\right] $$ (f) When the actual measurements were made at the University of California at Berkeley, the age of the volcanic material (and thus the age of the bones) was estimated to be \(1.75\) million years. What was the value of the measured ratio \(A / K ?\)

Consider the differential equations $$t \frac{d x}{d t}=x, \quad x\left(t_{0}\right)=x_{0},$$ and $$ y^{2} \frac{d x}{d y}+x y=2 y^{2}+1, \quad x\left(y_{0}\right)=x_{0} $$ Put each equation into normal form and then use the integrating factor technique to find the solutions. Establish whether these solutions are unique, and which part of each solution is a response to the initial data and which part a response to the input or forcing.

In Section \(2.7\), we also developed a model to describe the levels of antihistamine and decongestant in a patient taking a course of cold pills: $$ \begin{aligned} &\frac{d x}{d t}=I-k_{1} x, \quad x(0)=0, \\ &\frac{d y}{d t}=k_{1} x-k_{2} y, \quad y(0)=0 . \end{aligned} $$ Here \(k_{1}\) and \(k_{2}\) describe rates at which the drugs move between the two sequential compartments (the GI-tract and the bloodstream) and \(I\) denotes the amount of drug released into the GI-tract in each time step. The levels of the drug in the GI-tract and bloodstream are \(x\) and \(y\), respectively. By solving the equations sequentially show that the solution is $$ x(t)=\frac{I}{k_{1}}\left(1-e^{-k_{1} t}\right), \quad y(t)=\frac{I}{k_{2}}\left[1-\frac{1}{k_{2}-k_{1}}\left(k_{2} e^{-k_{1} t}-k_{1} e^{-k_{2} t}\right)\right] . $$

North American lake system. Consider the American system of two lakes: Lake Erie feeding into Lake Ontario. What is of interest is how the pollution concentrations change in the lakes over time. You may assume the volume in each lake to remain constant and that Lake Erie is the only source of pollution for Lake Ontario. (a) Write a differential equation describing the concentration of pollution in each of the two lakes, using the variables \(V\) for volume, \(F\) for flow, \(c(t)\) for concentration at time \(t\) and subscripts 1 for Lake Erie and 2 for Lake Ontario. (b) Suppose that only unpolluted water flows into Lake Erie. How does this change the model proposed? (c) Solve the system of equations to get expressions for the pollution concentrations \(c_{1}(t)\) and \(c_{2}(t)\) (d) Set \(T_{1}=V_{1} / F_{1}\) and \(T_{2}=V_{2} / F_{2}\), and then \(T_{1}=k T_{2}\) for some constant \(k\) as \(V\) and \(F\) are constants in the model. Substitute this into the equation describing pollution levels in Lake Ontario to eliminate \(T_{1}\). Then show that, with the initial conditions \(c_{1,0}\) and \(c_{2,0}\), the solution to the differential equation for Lake Ontario is $$ c_{2}(t)=\frac{k}{k-1} c_{1,0}\left(e^{-t /\left(k T_{2}\right)}-e^{-t / T_{2}}\right)+c_{2,0} e^{t / T_{2}} $$ (One way of finding the solution would be to use an integrating factor. See Appendix A.4.) (e) Compare the effects of \(c_{1}(0)\) and \(c_{2}(0)\) on the solution \(c_{2}(t)\) over time.

Each of the following differential equations has only one equilibrium solution. Find that equilibrium solution and determine if it is stable or unstable? (a) \(\frac{d y}{d t}=y-1\). (b) \(\frac{d C}{d t}=\frac{F}{V} c_{i}-\frac{F}{V} C\), where \(F, V, c_{i}\) are positive constants.

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