Potassium-40 decays to argon-40 with a half-life of \(1.27 \times 10^{9}\) yr. What is the age of a rock in which the mass ratio of \({ }^{40} \mathrm{Ar}\) to \({ }^{40} \mathrm{~K}\) is \(4.2 ?\)

Short Answer

Expert verified
The age of the rock is approximately \(2.64 \times 10^9\) years.

Step by step solution

01

Understand the radioactive decay process

In this decay process, Potassium-40 (K-40) decays to Argon-40 (Ar-40). The half-life of this process is given as \(1.27 \times 10^9\) years. Half-life is the time taken for half of the radioactive substance to decay. In other words, it indicates how fast the substance is decaying.
02

Write down the decay equation

For the radioactive decay process, we can use the decay equation: \[ N(t) = N_0 \times (1/2)^{\frac{t}{\text{half-life}}}, \] where \(N(t)\) is the amount of undecayed substance remaining at time \(t\), \(N_0\) is the initial amount of the substance, and "half-life" is the half-life of the substance.
03

Set up the ratio equation

We are given the mass ratio of the Argon-40 to Potassium-40 in the rock as 4.2. Let's denote the initial amount of K-40 as \(N_0^\text{K}\) and the initial amount of Ar-40 as \(N_0^\text{Ar}\). Then, after time \(t\), we have: \(N^\text{Ar}(t) = N_0^\text{Ar} + \left( N_0^\text{K} - N^\text{K}(t) \right)\) We can replace the value of \(N^\text{K}(t)\) by using the decay equation: \(N^\text{Ar}(t) = N_0^\text{Ar} + \left( N_0^\text{K} - N_0^\text{K} \times (1/2)^{\frac{t}{\text{half-life}}} \right)\) Now, we can use the given mass ratio of Ar-40 to K-40 as: \(\frac{N^\text{Ar}(t)}{N^\text{K}(t)} = 4.2\)
04

Solve for the rock's age

Rearrange the ratio equation to solve for \(t\): \(4.2 = \frac{N_0^\text{Ar} + \left( N_0^\text{K} - N_0^\text{K} \times (1/2)^{\frac{t}{\text{half-life}}} \right)}{N_0^\text{K} \times (1/2)^{\frac{t}{\text{half-life}}}}\) \(4.2 = \frac{N_0^\text{Ar} - \left( N_0^\text{K} - N_0^\text{K} \times (1/2)^{\frac{t}{\text{half-life}}} \right)}{N_0^\text{K} \times (1/2)^{\frac{t}{\text{half-life}}}}\) We can cancel out the initial amounts of K-40 and Ar-40, as they are constants, and solve for the time \(t\): \(4.2 = \frac{1 - (1/2)^{\frac{t}{\text{half-life}}}}{(1/2)^{\frac{t}{\text{half-life}}}}\) Now, solve for \(t\) using the given half-life \(1.27 \times 10^9\text{ yr}\): \(4.2 = \frac{1 - (1/2)^{\frac{t}{(1.27 \times 10^9 \text{ yr})}}}{(1/2)^{\frac{t}{(1.27 \times 10^9 \text{ yr})}}}\) \(t \approx 2.64 \times 10^9 \text{ years}\) So, the age of the rock is approximately \(2.64 \times 10^9\) years.

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Key Concepts

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

Half-Life Calculation
Understanding the concept of half-life is crucial for calculating the age of geological samples and understanding radioactive decay. In simplest terms, the half-life of a radioactive element is the time it takes for half of a given amount of the substance to transform into another element via radioactive decay.

For example, if we start with 100 grams of a radioactive isotope with a half-life of 1 year, we would expect to have 50 grams of that isotope remaining after 1 year.

To calculate the number of half-lives that have passed for a given time period, we use the formula:
\[\begin{equation}\text{Number of half-lives} = \frac{t}{\text{half-life}}\end{equation}\]
where \(t\) is the time elapsed, and \(\text{half-life}\) is the duration of the half-life. Since each half-life reduces the quantity by half, after \(n\) half-lives, the fraction of the starting amount that remains is \((1/2)^n\).
Mass Ratio of Isotopes
The mass ratio of isotopes in a rock sample can tell us a great deal about its age. This is especially true when considering parent-daughter isotope systems, where one isotope decays into another over time. In the context of potassium-40 decaying to argon-40, the ratio of the masses of these isotopes reflects the amount of radioactive decay that has occurred.

Initially, a sample contains only the parent isotope (potassium-40 in our example). As time passes, potassium-40 decays into argon-40. This increases the relative amount of argon-40 and decreases the relative amount of potassium-40. Measuring the current ratio between the two can indicate how many half-lives have elapsed, allowing scientists to calculate the sample's age.

The calculation involves taking the current measured ratio and applying the decay formula:
\[\begin{equation}\frac{N^{\text{Ar}}(t)}{N^{\text{K}}(t)} = \text{measured ratio}\end{equation}\]
where \(N^{\text{Ar}}(t)\) is the amount of the daughter isotope (argon-40) and \(N^{\text{K}}(t)\) is the amount of the parent isotope (potassium-40) at time \(t\). By substituting in the half-life of the parent isotope and rearranging, we can solve for \(t\), the time since the rock formed.
Dating Rocks Using Isotopes
Radioactive isotopes are invaluable tools in geological dating, a process known as radiometric dating. When a rock forms, it often contains a known ratio of isotopes. As the isotopes decay over time, the ratio changes in a predictable way, based on their half-lives.

Dating a rock involves measuring the ratio of the remaining parent isotope to the produced daughter isotope and calculating the time needed to reach this ratio. This method assumes that there was no daughter isotope present when the rock formed, and that it has remained a closed system since then—meaning that neither parent nor daughter isotopes have been added to or removed from the rock since its formation.

The age of the rock is then calculated using the decay equation we discussed earlier, considering the current isotope ratio. This is how the age of the rock in our exercise was determined to be approximately \(2.64 \times 10^9\) years based on the decay of potassium-40 to argon-40 and the measured mass ratio.

Importance of Cross-Checking

To improve accuracy, it's often best to measure different isotopic systems in the same rock sample. This cross-checking can reveal whether the rock has remained a closed system and provide a more accurate estimate of its formation age.

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Most popular questions from this chapter

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min} .\) (a) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left({ }^{235} \mathrm{U}\right.\) nuclear mass, \(234.9935 \mathrm{amu} ;{ }^{141} \mathrm{Ba}\) nuclear mass, 140.8833 amu; \({ }^{92} \mathrm{Kr}\) nuclear mass, 91.9021 amu \()\) is taken as typical of that occurring in a nuclear reactor, what mass of uranium- 235 is required to equal \(0.10 \%\) of the solar energy that falls on Earth in 1.0 day?

What is meant by enriched uranium? How is enriched uranium different from natural uranium?

Suppose you had a detection device that could count every decay event from a radioactive sample of plutonium- \(239\left(t_{1 / 2}\right.\) is 24,000 yr). How many counts per second would you obtain from a sample containing \(0.385 \mathrm{~g}\) of plutonium- \(239 ?\)

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One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: (a) \({ }_{20}^{40} \mathrm{Ca}\) and \({ }_{20}^{45} \mathrm{Ca},(\mathbf{b}){ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C},(\mathrm{c})\) lead- 206 and thorium- \(230 .\) Explain your choice in each case.

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