/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 The isotope \({ }^{40} \mathrm{~... [FREE SOLUTION] | 91Ó°ÊÓ

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The isotope \({ }^{40} \mathrm{~K}\) can decay to either \({ }^{40} \mathrm{Ca}\) or \({ }^{40} \mathrm{Ar} ;\) assume both decays have a half-life of \(1.26 \times 10^{9} \mathrm{y}\). The ratio of the Ca produced to the Ar produced is \(8.54 / 1=8.54 .\) A sample originally had only \({ }^{40} \mathrm{~K}\). It now has equal amounts of \({ }^{40} \mathrm{~K}\) and \({ }^{40} \mathrm{Ar}\); that is, the ratio of \(\mathrm{K}\) to \(\mathrm{Ar}\) is \(1 / 1=1 .\) How old is the sample? (Hint: Work this like other radioactive-dating problems, except that this decay has two products.)

Short Answer

Expert verified
The sample is approximately 1.26 billion years old.

Step by step solution

01

Understand the Problem

The problem involves a radioactive decay of \(^{40}K\) to \(^{40}Ca\) or\(^{40}Ar\). Given is the decay half-life and that the current amounts of \(^{40}K\) and \(^{40}Ar\) are equal. We need to determine the age of the sample.
02

Use the Decay Formula

We use the decay formula \(N = N_0 e^{-\lambda t}\),where\(N\) is the number of radioactive atoms at any given time,\(N_0\) is the initial quantity of radioactive atoms, and \(\lambda\) is the decay constant.Here, \(N/N_0 = 1/2\) because \(^{40}Ar\) originally came from \(^{40}K\).The decay constant, \(\lambda\), can be calculated as\( \lambda = \frac{\ln(2)}{t_{1/2}} \).
03

Calculate Decay Constant

The decay constant is given by\(\lambda = \frac{\ln(2)}{1.26 \times 10^9} = \frac{0.693}{1.26 \times 10^9} \text{ y}^{-1}\).
04

Use Decay Equation to Find Time

With\(N/N_0 = \frac{1}{2}\),solve for\(t\)using\(\frac{N}{N_0} = e^{-\lambda t}\)or\(t = \frac{\ln(\frac{N_0}{N})}{\lambda}\).Since\(N = N_0 / 2\) from the condition \(K/Ar = 1/1\), the time \(t\) can be found using the formula\[t = \frac{\ln(2)}{\lambda} = t_{1/2}\].
05

Calculate the Age of the Sample

Substitute \(\lambda = \frac{0.693}{1.26 \times 10^9}\) back into the equation to find:\[t \approx t_{1/2} = 1.26 \times 10^9 \text{ years}\].Given the ratio of \(Ca\) to \(Ar\) is 8.54, only a single half-life is needed since \(K\) to \(Ar\) is already\(1/1\).

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Key Concepts

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

Isotope Decay
In the realm of nuclear physics, isotopes play a fascinating role due to their ability to undergo radioactive decay. An isotope is a variant of a particular chemical element, which, while sharing the same number of protons, has different numbers of neutrons, leading to differing atomic masses. Over time, some isotopes are unstable and will "decay" or transform into other isotopes or elements. This process is referred to as isotope decay.
When the isotope \({ }^{40}K\) decays, it can turn into either \({ }^{40}Ca\) or \({ }^{40}Ar\). Such transformation is essential in the calculations for determining the age of geological samples. As decay progresses, the number of original isotopes decreases, while the number of decay products increases. This transformation proceeds in a predictable manner, allowing scientists to use it as a time gauge.
The consistent rate of decay follows an exponential decay model. Understanding this model is crucial when working with radioactive materials, as it helps predict how long a sample has been undergoing decay. Scientists use this knowledge for various purposes, such as understanding the age of rocks and meteorites.
Half-Life Calculation
The concept of half-life is central to understanding how radioactive decay helps us calculate the age of materials. The half-life is the time taken for half the amount of a radioactive isotope in a sample to decay. It indicates the stability and speed of decay of an isotope.
For \({ }^{40}K\), the half-life is \(1.26 \times 10^{9} \, \text{years}\). This means that after approximately 1.26 billion years, half of the original \({ }^{40}K\) isotopes will have decayed into \({ }^{40}Ca\) and \({ }^{40}Ar\).
In this exercise, with equal amounts of \({ }^{40}K\) and \({ }^{40}Ar\), it implies that one half-life has passed. Scientists can use the half-life to calculate the time since the sample had only pure \({ }^{40}K\), which in this case, is indeed \( ext{1.26 billion years}\). This calculation assumes a constant decay rate which is a property of the isotope.
Radioactive Dating
Radioactive dating, also called radiometric dating, is a technique used to date materials such as rocks or carbon, based on knowledge of their radioactive isotopes. This method is possible because these isotope decays occur at a consistent rate over time, making them a "clock" of sorts for scientists to interpret.
By measuring the ratio of the original radioactive isotope to its decay products, such as \({ }^{40}K\) to \({ }^{40}Ar\), geologists can ascertain the age of the sample. In this case, the method helped determine that our sample is \( ext{1.26 billion years}\) old.
The precision of radioactive dating makes it indispensable in studying Earth's history. It provides a way to quantify the age of geological formations and can be used to trace the evolutionary history of Earth, as well as for archaeological purposes. Understanding the principles behind radiometric dating allows us to piece together significant events in the history of our planet.

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Most popular questions from this chapter

A \({ }^{7} \mathrm{Li}\) nucleus with a kinetic energy of \(3.00 \mathrm{MeV}\) is sent toward a \({ }^{232} \mathrm{Th}\) nucleus. What is the least center-to- center separation between the two nuclei, assuming that the (more massive) \({ }^{232} \mathrm{Th}\) nucleus does not move?

The nuclide \({ }^{14} \mathrm{C}\) contains (a) how many protons and (b) how many neutrons?

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?

Calculate the mass of a sample of (initially pure) \({ }^{40} \mathrm{~K}\) that has an initial decay rate of \(1.70 \times 10^{5}\) disintegrations/s. The isotope has a half-life of \(1.28 \times 10^{9} \mathrm{y}\).

Under certain rare circumstances, a nucleus can decay by emitting a particle more massive than an alpha particle. Consider the decays $${ }^{223} \mathrm{Ra} \rightarrow{ }^{209} \mathrm{~Pb}+{ }^{14} \mathrm{C} \quad \text { and } \quad{ }^{223} \mathrm{Ra} \rightarrow{ }^{219} \mathrm{Rn}+{ }^{4} \mathrm{He}$$ Calculate the \(Q\) value for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is \(30.0 \mathrm{MeV}\). What is the barrier height for \({ }^{14} \mathrm{C}\) emission? (Be careful about the nuclear radii.) The needed atomic masses are $$ \begin{aligned} &\begin{array}{llll} { }^{223} \mathrm{Ra} & 223.01850 \mathrm{u} & { }^{14} \mathrm{C} & 14.00324 \mathrm{u} \end{array}\\\ &{ }^{209} \mathrm{~Pb} \quad 208.98107 \mathrm{u} \quad{ }^{4} \mathrm{He} \quad 4.00260 \mathrm{u}\\\ &{ }^{219} \mathrm{Rn} \quad 219.00948 \mathrm{u} \end{aligned} $$

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