Chapter 5: Problem 16
The half-life of \({ }^{40} \mathrm{~K}\) is \(13 \cdot 10^{9}\) years. The abundance of this isotope in naturally occurring potassium is \(1 \%\) that of the stable isotope \({ }^{39} \mathrm{~K}\). How many years ago were they equally abundant?
Short Answer
Expert verified
Approximately 8.64 billion years ago.
Step by step solution
01
Understanding the Half-Life Concept
The half-life of an isotope is the time required for half of the radioactive atoms to decay. Given, the half-life of \(^{40}\mathrm{K}\) is \(13 \times 10^9\) years,this means every \(13 \times 10^9\) years, the quantity of \(^{40}\mathrm{K}\) halves.
02
Initial Conditions
Initially, the radioactive \(^{40}\mathrm{K}\) is \(1\%\) of the stable \(^{39}\mathrm{K}\). So the initial abundance ratio of \(^{40}\mathrm{K}\) \ to \(^{39}\mathrm{K}\) is \(0.01:1\). We need to find out when they were in equal abundance, meaning the ratio \(1:1\).
03
Using Exponential Decay
The decay of \(^{40}\mathrm{K}\) follows an exponential function, \( N(t) = N_0 \, e^{-kt} \), where \(N(t)\) is the remaining quantity at time \(t\), \(N_0\) is the initial quantity, and \(k\) is the decay constant. The decay constant \(k\) can be calculated using \( k = \frac{\ln(2)}{T_{1/2}} \). Here, the half-life \(T_{1/2} = 13 \times 10^9 \) years.
04
Calculating the Decay Constant
Calculate the decay constant, \( k = \frac{\ln(2)}{13 \times 10^9} \approx 5.33 \times 10^{-11} \text{ years}^{-1} \).
05
Equating Abundances
Set \( N(t) = 1 \) for the ratio to be \(1:1\), so start by \(0.01 e^{-kt} = 1\).Rearrange to solve for \(t\), \(e^{-kt} = 100\).Then, take the natural logarithm of both sides to get \(-kt = \ln(100)\).
06
Solving for Time
Solve for \(t\): \( t = -\frac{\ln(100)}{k} \). Substitute the value of the decay constant, \( t \approx -\frac{4.605}{5.33 \times 10^{-11}} \approx 8.64 \times 10^9 \text{ years} \).This result means that roughly \(8.64 \times 10^9\) years ago, they were in equal abundance.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Half-Life
Half-life is a fundamental concept when studying radioactive decay. It represents the time it takes for half of the radioactive atoms in a sample to decay. For example, the half-life of the isotope potassium-40, represented as \(^{40}\mathrm{K}\), is a staggering \(13 \times 10^9\) years. This long period helps explain why certain isotopes are still present on Earth.
Understanding half-life helps scientists dating ancient rocks or archaeological finds by measuring the remaining radioactive isotopes compared to stable ones. In the case of \(^{40}\mathrm{K}\), this means that if you start with a specific number of atoms, after \(13 \times 10^9\) years, only half of them will remain.
Key points include:
Understanding half-life helps scientists dating ancient rocks or archaeological finds by measuring the remaining radioactive isotopes compared to stable ones. In the case of \(^{40}\mathrm{K}\), this means that if you start with a specific number of atoms, after \(13 \times 10^9\) years, only half of them will remain.
Key points include:
- Half-life is constant and intrinsic to each isotope.
- It provides a measure of the speed of decay—long half-lives indicate slow decay.
- Knowing the half-life allows prediction of how long an isotope will remain detectable or effective.
Exponential Decay
Exponential decay describes processes where the amount of something decreases at a rate proportional to its current value. In radioactive decay, such as with \(^{40}\mathrm{K}\), the decay can be modeled mathematically using an equation: \( N(t) = N_0 \cdot e^{-kt} \). Here, \(N(t)\) is the amount remaining at time \(t\), \(N_0\) is the initial amount, and \(k\) is the decay constant.
The decay constant \(k\) gives the rate of decay, and it is related to half-life through the formula \( k = \frac{\ln(2)}{T_{1/2}} \). For \(^{40}\mathrm{K}\), \(k \approx 5.33 \times 10^{-11} \text{ years}^{-1} \).
Some overarching characteristics:
The decay constant \(k\) gives the rate of decay, and it is related to half-life through the formula \( k = \frac{\ln(2)}{T_{1/2}} \). For \(^{40}\mathrm{K}\), \(k \approx 5.33 \times 10^{-11} \text{ years}^{-1} \).
Some overarching characteristics:
- Exponential decay models processes where the rate of change is proportional to the quantity.
- The decay constant reflects how rapidly the substance decays.
- This model helps predict how much of a substance remains over time, especially useful in radioactive decay studies.
Potassium Isotopes
Potassium, a common element, has several isotopes, including the radioactive \(^{40}\mathrm{K}\) and the stable \(^{39}\mathrm{K}\). The abundance of potassium isotopes and their decay provide insights into geological and biological processes.
Here are a few interesting facts:
Here are a few interesting facts:
- \(^{40}\mathrm{K}\) makes up about 0.01% of naturally occurring potassium.
- The decay of \(^{40}\mathrm{K}\) into other elements like calcium and argon helps date rocks. This is because scientists can measure the amount of these decay products to estimate the age of a sample.
- The more prevalent \(^{39}\mathrm{K}\) is stable and does not undergo radioactive decay.