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In a piece of rock from the Moon, the \({ }^{87} \mathrm{Rb}\) content is assessed to be \(1.82 \times 10^{10}\) atoms per gram of material and the \({ }^{87} \mathrm{Sr}\) content is found to be \(1.07 \times 10^{9}\) atoms per gram. (a) Determine the age of the rock. (b) Could the material in the rock actually be much older? What assumption is implicit in the use of the radioactive dating method? (The relevant decay is \({ }^{87} \mathrm{Rb} \rightarrow{ }^{87} \mathrm{Sr}+\mathrm{e}^{-}\). The half-life of the decay is \(4.8 \times 10^{10}\) yr.)

Short Answer

Expert verified
The age of the rock is calculated as outlined in Step 3. However, the actual age could be older if there was any \(^{87}Sr\) initially present in the rock, as discussed in Step 4.

Step by step solution

01

Understand the radioactive decay process

In Rubidium-Strontium dating, we consider the decay of \(^{87}Rb\) into \(^{87}Sr\). The number of \(^{87}Rb\) atoms decreases over time, while the number of \(^{87}Sr\) atoms increases. This allows us to estimate the age of the rock.
02

Calculate the decay constant

The decay constant (\(\lambda\)) is related to the half-life (\(t_{\frac{1}{2}}\)) by the formula \(\lambda = \frac{\ln 2}{t_{\frac{1}{2}}}\). Given that the half-life \(t_{\frac{1}{2}}\) of \(^{87}Rb\) is \(4.8 \times 10^{10}\) years, we can calculate the decay constant as \(\lambda = \frac{\ln 2}{4.8 \times 10^{10}}\) years鈦宦.
03

Calculate the age of the rock

The age (\(t\)) of the rock can be determined from the formula \(t = \frac{1}{\lambda} \ln \left(1 + \frac{N_{Sr}}{N_{Rb}}\right)\) where \(N_{Sr}\) is the number of \(^{87}Sr\) atoms and \(N_{Rb}\) is the number of \(^{87}Rb\) atoms. Substituting the given values, we get \(t = \frac{1}{\lambda} \ln \left(1 + \frac{1.07 \times 10^{9}}{1.82 \times 10^{10}}\right)\) years.
04

Consider the assumptions and possible older age

The calculation above assumes that there was no initial \(^{87}Sr\) in the rock when it formed, and all \(^{87}Sr\) present came from the decay of \(^{87}Rb\). If there were any initial \(^{87}Sr\), that would mean the rock has more \(^{87}Sr\) than what can be accounted for by \(^{87}Rb\) decay, implying the rock could be older than calculated.

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Key Concepts

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

Rubidium-Strontium Dating
Rubidium-Strontium dating is a technique utilized in geochronology. It helps scientists determine the age of rock samples by analyzing the isotopic ratios of Rubidium ( ^{87}Rb) and Strontium ( ^{87}Sr). This method is especially useful in dating ancient geological materials.

In essence, Rubidium ( ^{87}Rb) is a radioactive isotope, which over time, undergoes decay to transform into a stable isotope, Strontium ( ^{87}Sr). By measuring the quantities of both ^{87}Rb and ^{87}Sr in a rock sample, scientists can calculate how long this transformation process has been occurring.

One of the key assumptions inherent in this method is the idea that at the time of the rock's formation, there was no initial ^{87}Sr present. Thus, all detected ^{87}Sr is assumed to have originated from the decay of ^{87}Rb over time. It's important to note that if initial ^{87}Sr had been present, it could result in misestimating the age, often showing a younger apparent age than its true form.

This technique has proven invaluable in piecing together the chronological history of Earth's grueling geologic processes.
Radioactive Decay
Radioactive decay is a natural process whereby an unstable atomic nucleus loses energy by emitting radiation. This decay leads to the transformation of the parent isotope into a daughter isotope.

In the context of Rubidium-Strontium dating, ^{87}Rb decays into ^{87}Sr. This decay process is measurable and predictable, allowing researchers to employ it as a clock to date materials.
  • With each decay event, the number of ^{87}Rb atoms decreases and consequently, the number of ^{87}Sr atoms increases.
  • By measuring these amounts, the elapsed time since the rock's formation can be estimated.

Radioactive decay serves as a fundamental basis in determining the absolute age of materials, shedding light on historical timelines that are often inaccessible through other means.

Understanding this decay process enhances our ability to explore the ages of the Earth and beyond, unlocking the mysteries of the planetary systems in the universe.
Decay Constant
The decay constant, denoted by the symbol \(\lambda\), is a crucial parameter in radioactive decay, defining the rate at which a given isotope decays. It is derived from the half-life of the substance, which is the time required for half of the radioactive atoms in a sample to decay.

For Rubidium (^{87}Rb), the decay constant can be calculated using the formula:

\[\lambda = \frac{\ln 2}{t_{\frac{1}{2}}}\]

In this equation, \(\ln 2\) is the natural logarithm of 2, and \(t_{\frac{1}{2}}\) represents the half-life of ^{87}Rb. As given in the problem, ^{87}Rb has a half-life of approximately 4.8 \times 10^{10} years. By plugging the half-life value into the formula, we can find out how frequently the decay event occurs on a per-year basis.

The decay constant is significant because it quantitatively describes the decay rate, enabling the calculation of the age of materials dominated by radionuclides. In geological dating, a higher decay constant translates to a faster decay rate, thus affecting how quickly the transformation from parent to daughter isotope occurs. Understanding the decay constant is fundamental for accurately determining the age of geological samples.

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Most popular questions from this chapter

During the manufacture of a steel engine component, radioactive iron \(\left({ }^{59} \mathrm{Fe}\right)\) is included in the total mass of \(0.2 \mathrm{~kg}\). The component is placed in a test engine when the activity due to this isotope is \(20 \mu\) Ci. After a 1000 -h test period, oil is removed from the engine and found to contain enough \({ }^{59} \mathrm{Fe}\) to produce 800 disintegrations \(/\) min per liter of oil. The total volume of oil in the engine is \(6.5 \mathrm{~L} .\) Calculate the total mass worn from the engine component per hour of operation. (The halflife for \({ }^{59} \mathrm{Fe}\) is \(45.1\) days.)

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When a material of interest is irradiated by neutrons, radioactive atoms are produced continually and some decay according to their given half-lives. (a) If radioactive atoms are produced at a constant rate \(R\) and their decay is governed by the conventional radioactive decay law, show that the number of radioactive atoms accumulated after an irradiation time \(t\) is $$ N=\frac{R}{\lambda}\left(1-e^{-\lambda t}\right) $$ (b) What is the maximum number of radioactive atoms that can be produced?

Tritium has a half-life of \(12.33 \mathrm{yr}\). What percentage of the \({ }^{3} \mathrm{H}\) nuclei in a tritium sample will decay during a period of 5 years?

The radioactive isotope \({ }^{198} \mathrm{Au}\) has a half-life of \(64.8 \mathrm{~h}\). \(\mathrm{A}\) sample containing this isotope has an initial activity of \(40 \mu \mathrm{Ci}\). Calculate the number of nuclei that will decay in the time interval from \(t_{1}=10 \mathrm{~h}\) to \(t_{2}=12 \mathrm{~h}\).

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