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Chapter 42: Problem 77

How many years are needed to reduce the activity of \({ }^{14} \mathrm{C}\) to 0.020 of its original activity? The half-life of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{y}\).

Short Answer

Expert verified
Approximately 28860 years are needed.

Step by step solution

01

Understand the Decay Formula

The formula to describe radioactive decay is given by \( A = A_0 \times (1/2)^{t/T} \), where \( A \) is the remaining activity, \( A_0 \) is the original activity, \( t \) is the time elapsed, and \( T \) is the half-life. In this problem, we need to find \( t \) when \( A = 0.020 \times A_0 \).
02

Set the Equation

Set up the equation using the decay formula: \( 0.020 \times A_0 = A_0 \times (1/2)^{t/5730} \). Simplify by dividing both sides by \( A_0 \), which results in \( 0.020 = (1/2)^{t/5730} \).
03

Take the Logarithm of Both Sides

To solve for \( t \), take the logarithm of both sides: \( \log(0.020) = \log((1/2)^{t/5730}) \). This simplifies to \( \log(0.020) = \frac{t}{5730} \times \log(1/2) \).
04

Solve for t

Rearrange the equation to solve for \( t \): \( t = \frac{5730 \times \log(0.020)}{\log(1/2)} \). Calculate \( \log(0.020) \approx -1.6990 \) and \( \log(1/2) \approx -0.3010 \). Substitute these values into the equation: \( t = \frac{5730 \times -1.6990}{-0.3010} \approx 28860 \) years.

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Key Concepts

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

half-life
In the realm of radioactive decay, the concept of half-life is crucial. Half-life is the time required for a quantity to reduce to half its initial value. It's a natural constant for each radioactive isotope such as carbon-14 ( ^{14}C ).

Every 5730 years, the amount of carbon-14 in any sample halves. This means if you start with a certain amount of carbon-14 today, in 5730 years, you will only have half of that amount left. It's similar to a countdown, where at each half-life period, the available material is halved.

This property is incredibly useful, especially in fields like archeology and geology, because it provides a clock that can help measure the age of objects by evaluating the amount of the radioactive isotope remaining.
carbon-14 dating
Carbon-14 dating is a method used to determine the age of an object containing organic material by measuring the amount of carbon-14 it contains. Living organisms constantly exchange carbon with their environment, maintaining a consistent level of carbon-14.

However, once the organism dies, it no longer takes in carbon-14, and the isotope starts to decay at a known rate, characterized by its half-life of 5730 years. By measuring how much carbon-14 is left in a sample, scientists can infer how long ago the organism died.
  • Archaeologists often use carbon-14 dating to date artifacts of organic origin, such as bone, cloth, wood, and plant fibers.
  • This method has helped to date remains and artifacts as far back as approximately 50,000 years.
Keep in mind that for carbon-14 dating, assumptions about the initial content of ^{14}C and the environment's carbon levels must be considered for accuracy.
logarithmic equations
Logarithmic equations are a powerful tool in solving problems involving exponential growth or decay, such as radioactive decay. These equations include logarithms, which are the inverse operations of exponentiation.

When dealing with decay problems, taking the logarithm of both sides helps isolate the time variable, making it easier to solve for unknowns like time 't' in our original problem. For example, if you know the rate of decay, you can use logarithms to find out how long it will take for the remaining substance to reach a certain level.
  • Natural logs are often used, denoted as ln(x), particularly when working with continuous growth rates.
  • Common logs, denoted as log(x), are also frequently used, especially when solving base 10 problems, such as those involving half-lives.
Logarithmic equations can be intimidating, but they are simply another form of mathematical expression that allows us to powerfully examine exponential relationships.

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Most popular questions from this chapter

The half-life of a radioactive isotope is 140 d. How many days would it take for the decay rate of a sample of this isotope to fall to one-fourth of its initial value?

At the end of World War II, Dutch authorities arrested Dutch artist Hans van Meegeren for treason because, during the war, he had sold a masterpiece painting to the Nazi Hermann Goering. The painting, Christ and His Disciples at Emmaus by Dutch master Johannes Vermeer \((1632-1675),\) had been discovered in 1937 by van Meegeren, after it had been lost for almost 300 years. Soon after the discovery, art experts proclaimed that Emmaus was possibly the best Vermeer ever seen. Selling such a Dutch national treasure to the enemy was unthinkable treason. However, shortly after being imprisoned, van Meegeren suddenly announced that he, not Vermeer, had painted Emmaus. He explained that he had carefully mimicked Vermeer's style, using a 300 -year-old canvas and Vermeer's choice of pigments; he had then signed Vermeer's name to the work and baked the painting to give it an authentically old look. Was van Meegeren lying to avoid a conviction of treason, hoping to be convicted of only the lesser crime of fraud? To art experts, Emmaus certainly looked like a Vermeer but, at the time of van Meegeren's trial in 1947 , there was no scientific way to answer the question. However, in 1968 Bernard Keisch of Carnegie-Mellon University was able to answer the question with newly developed techniques of radioactive analysis. Specifically, he analyzed a small sample of white lead-bearing pigment removed from Emmaus. This pigment is refined from lead ore, in which the lead is produced by a long radioactive decay series that starts with unstable \({ }^{238} \mathrm{U}\) and ends with stable \({ }^{206} \mathrm{~Pb}\). To follow the spirit of Keisch's analysis, focus on the following abbreviated portion of that decay series, in which intermediate, relatively shortlived radionuclides have been omitted: $${ }^{230} \mathrm{Th} \frac{{ }_{75.4 \mathrm{ky}}}{ }^{226} \mathrm{Ra} \frac{{ }_{1.60 \mathrm{ky}}}{ }^{210} \mathrm{~Pb} \frac{{ }_{22.6 \mathrm{ky}}}{ }^{206} \mathrm{~Pb}$$ The longer and more important half-lives in this portion of the decay series are indicated. (a) Show that in a sample of lead ore, the rate at which the number of \({ }^{210} \mathrm{~Pb}\) nuclei changes is given by $$\frac{d N_{210}}{d t}=\lambda_{226} N_{226}-\lambda_{210} N_{210}$$ where \(N_{210}\) and \(N_{226}\) are the numbers of \({ }^{210} \mathrm{~Pb}\) nuclei and \({ }^{226} \mathrm{Ra}\) nuclei in the sample and \(\lambda_{210}\) and \(\lambda_{226}\) are the corresponding disintegration constants. Because the decay series has been active for billions of years and because the half-life of \({ }^{210} \mathrm{~Pb}\) is much less than that of \({ }^{226} \mathrm{Ra}\), the nuclides \({ }^{226} \mathrm{Ra}\) and \({ }^{210} \mathrm{~Pb}\) are in equilibrium; that is, the numbers of these nuclides (and thus their concentrations) in the sample do not change. (b) What is the ratio \(R_{226} / R_{210}\) of the activities of these nuclides in the sample of lead ore? (c) What is the ratio \(N_{226} / N_{210}\) of their numbers? When lead pigment is refined from the ore, most of the \(226 \mathrm{Ra}\) is eliminated. Assume that only \(1.00 \%\) remains. Just after the pigment is produced, what are the ratios (d) \(R_{226} / R_{210}\) and (e) \(N_{226} / N_{210} ?\) Keisch realized that with time the ratio \(R_{226} / R_{210}\) of the pigment would gradually change from the value in freshly refined pigment back to the value in the ore, as equilibrium between the \({ }^{210} \mathrm{~Pb}\) and the remaining \({ }^{226} \mathrm{Ra}\) is established in the pigment. If Emmaus were painted by Vermeer and the sample of pigment taken from it were 300 years old when examined in \(1968,\) the ratio would be close to the answer of (b). If Emmaus were painted by van Meegeren in the 1930 s and the sample were only about 30 years old, the ratio would be close to the answer of (d). Keisch found a ratio of \(0.09 .\) (f) Is Emmaus a Vermeer?

The isotope \({ }^{238} \mathrm{U}\) decays to \({ }^{206} \mathrm{~Pb}\) with a half-life of \(4.47 \times 10^{9} \mathrm{y}\). Although the decay occurs in many individual steps, the first step has by far the longest half-life; therefore, one can often consider the decay to go directly to lead. That is, $${ }^{238} \mathrm{U} \rightarrow{ }^{206} \mathrm{~Pb}+\text { various decay products. }$$ A rock is found to contain \(4.20 \mathrm{mg}\) of \({ }^{238} \mathrm{U}\) and \(2.135 \mathrm{mg}\) of \({ }^{206} \mathrm{~Pb}\). Assume that the rock contained no lead at formation, so all the lead now present arose from the decay of uranium. How many atoms of (a) \({ }^{238} \mathrm{U}\) and (b) \({ }^{206} \mathrm{~Pb}\) does the rock now contain? (c) How many atoms of \({ }^{238} \mathrm{U}\) did the rock contain at formation? (d) What is the age of the rock?

An \(85 \mathrm{~kg}\) worker at a breeder reactor plant accidentally ingests \(2.5 \mathrm{mg}\) of \({ }^{239} \mathrm{Pu}\) dust. This isotope has a half- life of \(24100 \mathrm{y}\), decaying by alpha decay. The energy of the emitted alpha particles is \(5.2 \mathrm{MeV},\) with an \(\mathrm{RBE}\) factor of \(13 .\) Assume that the plutonium resides in the worker's body for \(12 \mathrm{~h}\) (it is eliminated naturally by the digestive system rather than being absorbed by any of the internal organs) and that \(95 \%\) of the emitted alpha particles are stopped within the body. Calculate (a) the number of plutonium atoms ingested, (b) the number that decay during the \(12 \mathrm{~h},\) (c) the energy absorbed by the body, (d) the resulting physical dose in grays, and (e) the dose equivalent in sieverts.

Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a stable, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings: (a) \({ }^{235} \mathrm{U} \rightarrow{ }^{232} \mathrm{Th}+{ }^{3} \mathrm{He},\) (b) \({ }^{235} \mathrm{U} \rightarrow{ }^{231} \mathrm{Th}+{ }^{4} \mathrm{He}\) (c) \({ }^{235} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+{ }^{5} \mathrm{He}\) The needed atomic masses are $$ \begin{array}{llll} { }^{232} \mathrm{Th} & 232.0381 \mathrm{u} & { }^{3} \mathrm{He} & 3.0160 \mathrm{u} \\ { }^{231} \mathrm{Th} & 231.0363 \mathrm{u} & { }^{4} \mathrm{He} & 4.0026 \mathrm{u} \\ { }^{230} \mathrm{Th} & 230.0331 \mathrm{u} & { }^{5} \mathrm{He} & 5.0122 \mathrm{u} \\ { }^{235} \mathrm{U} & 235.0429 \mathrm{u} & & \end{array} $$
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