/*! 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 76 Potassium- 40 decays by the two ... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 76

Potassium- 40 decays by the two different paths, shown below. $$ \begin{aligned} &{ }_{19}^{40} \mathrm{~K} \rightarrow{ }_{20}^{40} \mathrm{Ca}+{ }_{-1}^{0} \mathrm{e} \quad(89.3 \%) \\ &{ }_{19}^{40} \mathrm{~K} \stackrel{\mathrm{EC}}{\longrightarrow}{ }_{18}^{41} \mathrm{Ar} \quad(10.7 \%) \end{aligned} $$ where EC stands for electron capture. The overall half-life for the decay of \({ }_{19}^{40} \mathrm{~K}\) is \(1.248 \times 10^{9}\) years. The potassium- 40 to argon- 40 reaction is thought to be the source of argon in the earth's atmosphere. Potassium-argon dating is used in geology and archaeology to date sedimentary rocks. Estimate the age of sedimentary rocks with a \({ }^{40} \mathrm{Ar}\) -to- \({ }^{-4} \mathrm{~K}\) ratio of \(0.0102\).

Short Answer

Expert verified
The estimated age of the sedimentary rocks is approximately 1.25 billion years.

Step by step solution

01

Understand the Problem

The exercise involves determining the age of sedimentary rocks using the potassium-argon dating method. We are given the decay paths of potassium-40 and the overall half-life, and we need to find the age based on the argon-40 to potassium-40 ratio.
02

Decay Paths Analysis

Understand the two decay modes of potassium-40: one to calcium-40 via beta decay, with 89.3% probability, and another to argon-40 via electron capture with 10.7% probability. Argon-40 formation contributes to the argon-40 present in the rock.
03

Calculate the Effective Decay Constant

The overall decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by \( \lambda = \frac{\ln(2)}{t_{1/2}} \). With \( t_{1/2} = 1.248 \times 10^9 \) years, compute \( \lambda = \frac{\ln(2)}{1.248 \times 10^9} \).
04

Adjust for Argon Formation Probability

Since only 10.7% of decays form argon-40, calculate the decay constant specifically for argon formation: \( \lambda_{Ar} = 0.107 \lambda \).
05

Apply Decay Formula to Find Rock Age

Using the formula \( [^{40}\mathrm{Ar}] = [^{40}\mathrm{K}]_0 \times (1 - e^{-\lambda_{Ar}t}) \), substitute \( \frac{^{40}\mathrm{Ar}}{^{40}\mathrm{K}} \) ratio of 0.0102 to get \((0.0102) = (1 - e^{-0.107\lambda t}) \). Solve for \( t \).
06

Solve for Time \( t \)

Rearrange the equation to solve for time \( t \): \( t = \frac{-1}{0.107\lambda} \ln(1-0.0102) \). Substitute \( \lambda \) to compute \( t \).
07

Final Computation

Finally, execute the computation to find \( t \), the age of the rock, which results in approximately 1.25 billion years.

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Key Concepts

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

Radioactive Decay
Radioactive decay is a fascinating process where unstable atomic nuclei transform into more stable nuclei over time. This happens by releasing particles or radiation. In the context of potassium-40, it has two primary decay paths: - **Beta decay** to calcium-40, with an occurrence probability of 89.3%. - **Electron capture** to argon-40, occurring 10.7% of the time.
In both cases, the original potassium-40 nucleus is losing energy and changing into a different element. Understanding these decay paths is crucial for estimating the age of geological samples, such as rocks, accurately.
Half-Life Calculation
Half-life is essentially the time required for half of a sample's radioactive nuclei to decay. It's a constant and helps describe the rate of decay for any radioactive substance. For potassium-40, the half-life is long, about 1.248 billion years, reflecting the slow process of its decay.To find the decay constant \( \lambda \), use the formula:\[\lambda = \frac{\ln(2)}{t_{1/2}}\]where \( t_{1/2} \) is the half-life. With this decay constant, scientists adjust for specific decay paths, like calculating \( \lambda_{Ar} \) for the formation of argon-40, which is crucial to the potassium-argon dating method.
Geological Dating Methods
Geological dating methods like potassium-argon dating are essential for understanding Earth's history. They help geologists and archaeologists determine the age of rocks and fossils. This method is particularly useful for dating volcanic rocks.The idea relies on measuring the ratio of argon-40 to potassium-40 in a rock sample. Given the known half-life and the decay probabilities of potassium-40, scientists can backtrack to estimate how long the decay has been occurring, thus dating the rock. For instance, a measured \( ^{40} \text{Ar} \) to \( ^{40} \text{K} \) ratio of 0.0102 can be used with decay equations to estimate the age of sedimentary rocks, giving insight into processes millions to billions of years ago.

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Most popular questions from this chapter

A \(1.00-\mathrm{mL}\) sample of blood is withdrawn from an animal, and the red blood cells are labeled with phosphorus-32 ( \(t_{1 j}=14.28\) days). The activity of this sample is 50000 disintegrations \(\cdot \mathrm{min}^{-1}\). The sample is then reinjected into the animal. A few hours later, another \(1.00\) -mL sample is withdrawn, and its activity is \(10.0\) disintegrations \(\cdot \mathrm{min}^{-1} .\) Determine the volume of blood in the animal. Assume that phosphorus-32 is uniformly distributed throughout the blood and that the activity due to phosphorus- 32 remains constant during the experiment. By using similar methods, it has been found that the human body contains about \(75 \mathrm{~mL}\) of blood per kilogram of body weight.

Suggest an experimental method for measuring the rate of the reactions described by the following chemical equations: (a) \(\mathrm{I}_{2}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+2 \mathrm{I}^{-}(a q)\) yellow \(\quad\) colorless colorless colorless (b) \(4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)\)

You order a sample of \(\mathrm{Na}_{5} \mathrm{PO}_{4}\) containing the radioisotope phosphorus-32 \(\left(t_{y / 2}=14.28\right.\) days). If the shipment is delayed in transit for two weeks, how much of the original activity will remain when you receive the sample?

Determine the rate law for the reaction described by $$ \mathrm{NO}(g)+\mathrm{H}_{2}(g) \rightarrow \text { products } $$ from the initial-rate data tabulated below: \begin{tabular}{ccc} \hline Initial pressure of \(P_{\mathrm{H}_{2}} /\) Torr & Initial pressure of \(P_{\mathrm{No}} /\) Torr & Initial rate of reaction/Torr \(\cdot \mathrm{s}^{-1}\) \\ \hline 400 & 159 & 34 \\ 400 & 300 & 125 \\ 289 & 400 & 160 \\ 205 & 400 & 110 \\ 147 & 400 & 79 \\ \hline \end{tabular} Calculate the value of the rate constant for this reaction.

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by $$ \mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g) $$ The rate law for this reaction is $$ \text { rate of reaction }=k\left[\mathrm{O}_{3}\right][\mathrm{NO}] $$ Given that \(k=2.99 \times 10^{6} \mathrm{M}^{-1} \cdot \mathrm{s}^{-1}\) at \(310 \mathrm{~K}\), calculate the initial reaction rate when \(\left[\mathrm{O}_{3}\right]\) and \([\mathrm{NO}]\) remain essentially constant at the values \(\left[\mathrm{O}_{3}\right]_{0}=2.0 \times 10^{-6} \mathrm{M}\) and \([\mathrm{NO}]_{0}=6.0 \times 10^{-5} \mathrm{M}\), owing to continuous production from separate sources. Calculate the number of moles of \(\mathrm{NO}_{2}(g)\) produced per hour per liter of air.
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