Anthropologists can estimate the age of a bone or other sample of organic matter by its carbon- 14 content. The carbon- 14 in a living organism is constant until the organism dies, after which carbon-14 decays with first- order kinetics and a half-life of 5730 years. Suppose a bone from an ancient human contains \(19.5 \%\) of the C-14 found in living organisms. How old is the bone?

Short Answer

Expert verified
The bone is approximately 17,190 years old.

Step by step solution

01

Understand the Concept of Half-Life

Half-life is the time required for half of a quantity of a radioactive isotope to decay. In first-order kinetics, the decay rate is directly proportional to the quantity of the remaining substance. The half-life of Carbon-14 is given as 5730 years.
02

Use the Decay Formula

To estimate the age of the bone, apply the first-order decay formula which is \( N(t) = N_0 (1/2)^{t/T} \), where \( N(t) \) is the remaining amount of C-14 at time 't', \( N_0 \) is the initial amount of C-14, 't' is the time that has passed, and 'T' is the half-life of the substance. Here, \( N(t)/N_0 = 19.5\% \).
03

Calculate the Age of the Bone

Plugging in the values, you get \( 0.195 = (1/2)^{t/5730} \). To find 't', take the logarithm of both sides: \( \ln(0.195) = \ln((1/2)^{t/5730}) \). Solving for 't' yields \( t = 5730 \times \frac{\ln(0.195)}{\ln(1/2)} \).
04

Solve for 't'

Using the formula \( t = 5730 \times \frac{\ln(0.195)}{\ln(1/2)} \), compute the value of 't' to find the age of the bone.

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Key Concepts

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

Radioactive Isotope Decay
At the heart of carbon-14 dating is the principle of radioactive isotope decay. Radioactive isotopes, such as carbon-14, are unstable forms of elements that release energy by emitting radiation, transforming into a different element over time. This decay occurs predictably and at a fixed rate, characterized by what we call a half-life. For instance, carbon-14, a key player in archeological dating, transforms into nitrogen-14 as it decays.

It's important to note that once an organism dies, it no longer exchanges carbon with its environment, making the carbon-14 it contains at the time of death a closed system. Over time, the carbon-14 decays while the carbon-12, a stable isotope, remains unchanged. This predictable decay rate of carbon-14 allows scientists to calculate the age of formerly living materials by measuring the remaining carbon-14.
Half-Life Calculation
The half-life of a radioactive isotope is the time it takes for half of the atoms in a sample to decay. It's a pivotal concept in understanding radioactive decay as it directly influences age determination of an archeological sample. The half-life remains constant regardless of the initial amount of the isotope or the size of the sample.

For carbon-14, the half-life is approximately 5730 years. When we know this value, we can estimate the time that has elapsed since the death of the organism by measuring how much carbon-14 remains in comparison to the stable carbon-12. With each passing half-life, the amount of carbon-14 is reduced by half, forming a time marker for scientists to use in their calculations.
First-Order Kinetics
First-order kinetics is a term that describes a scenario where the rate of reaction is directly proportional to the concentration of a single reactant. In the context of radioactive decay, and specifically carbon-14 dating, this means that the rate at which carbon-14 decays is proportional to the amount of carbon-14 present in the sample at any given time.

Using mathematical language, the first-order decay formula is written as \( N(t) = N_0 (1/2)^{t/T} \), where \( N(t) \) is the amount of the substance at time 't', \( N_0 \) is the original amount, 't' is the elapsed time, and 'T' is the half-life of the substance. This relationship allows us to predict how much of the radioisotope will remain after a certain period has passed, which is instrumental in calculating the age of archeological samples.
Archaeological Age Determination
Carbon-14 dating is a powerful tool used in archaeology to determine the age of organic artifacts. It relies on the principles of radioactive isotope decay and first-order kinetics as discussed earlier. The process begins by measuring the ratio of carbon-14 to carbon-12 in the artifact. By comparing this ratio to the expected ratio in a living organism, scientists can determine how many half-lives have passed since the death of the organism.

In the provided exercise, with a residual carbon-14 content of 19.5%, we can apply the decay formula and calculate that \( t = 5730 \times \frac{\ln(0.195)}{\ln(1/2)} \) to find the time elapsed since the organism's death. Using these calculations, the age of organic materials, such as bone, can be estimated with a fair degree of accuracy. Carbon-14 dating thus serves as a window into the past, helping us to piece together the timeline of human history and prehistory.

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Most popular questions from this chapter

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and has a rate constant of \(1.42 \times 10^{-4} \mathrm{~s}^{-1}\) at a certain temperature. a. What is the half-life for this reaction? b. How long will it take for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(25 \%\) of its initial concentration? c. If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.00 \mathrm{M}\), how long will it take for the concentration to decrease to \(0.78 \mathrm{M} ?\) d. If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(0.150 \mathrm{M},\) what is the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) after \(2.00 \times 10^{2} \mathrm{~s}\) ? After \(5.00 \times 10^{2} \mathrm{~s} ?\)

Ethyl chloride vapor decomposes by the first-order reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{HCl} $$ The activation energy is \(249 \mathrm{~kJ} / \mathrm{mol}\), and the frequency factor is \(1.6 \times 10^{14} \mathrm{~s}^{-1} .\) Find the value of the rate constant at \(710 \mathrm{~K}\) What fraction of the ethyl chloride decomposes in 15 minutes at this temperature? Find the temperature at which the rate of the reaction would be twice as fast.

Consider this overall reaction, which is experimentally observed to be second order in AB and zero order in C: $$ \mathrm{AB}+\mathrm{C} \longrightarrow \mathrm{A}+\mathrm{BC} $$ Is the following mechanism valid for this reaction? $$ \begin{array}{ll} \mathrm{AB}+\mathrm{AB} \longrightarrow \mathrm{AB}_{2}+\mathrm{A} & \text { Slow } \\ \mathrm{AB}_{2}+\mathrm{C} \longrightarrow \mathrm{AB}+\mathrm{BC} & \text { Fast } \end{array} $$

The evaporation of a 120 -nm film of \(n\) -pentane from a single crystal of aluminum oxide is zero order with a rate constant of \(1.92 \times 10^{13} \mathrm{molecules} / \mathrm{cm}^{2} \cdot \mathrm{s}\) at \(120 \mathrm{~K}\) a. If the initial surface coverage is \(8.9 \times 10^{16}\) molecules \(/ \mathrm{cm}^{2}\), how long will it take for one-half of the film to evaporate? b. What fraction of the film is left after 10 s? Assume the same initial coverage as in part a.

Explain the difference between a normal chemical equation for a chemical reaction and the mechanism of that reaction.

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