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What is the age of a piece of volcanic rock that has a mass ratio of argon- 40 to potassium- 40 of \(2.9 ?\) The half-life of potassium-40 by \(\beta\) decay is \(1.248 \times 10^{9} \mathrm{y}\) and by electron capture \(t_{1 / 2}=1.4 \times 10^{9} \mathrm{y}\).

Short Answer

Expert verified
Plug the calculated decay constant and the provided mass ratio into the elapsed time equation to find the age of the volcanic rock. Please execute the calculations to get the final result.

Step by step solution

01

Understand the Problem

This is a problem concerning radiometric age dating. We are given a mass ratio of argon-40 to potassium-40 in a piece of volcanic rock λ=2.9. We are also given two possible half-lives of potassium-40.
02

Formulate the Decay Rate Equation

Let's utilize the radioactive decay rate equation: \( t = \frac{1}{λ} \cdot \ln(1 + \frac{N}{N_{0}}) \), where t is the elapsed time, N is the number of atoms currently present, N_{0} is the number of atoms initially present, and \(λ\) is the decay constant calculated from the half-life.
03

Calculate Decay Constant

Using the given half-life value, the decay constant \(\( λ \) can be calculated using the formula: \( λ = \frac{0.693}{t_{\frac{1}{2}}} \). Since the half-life estimation is uncertain, let's take an average of the two given values, Hence, \( λ = \frac{0.693}{((1.248 + 1.4) / 2) * 10^{9} } \). Calculating this will give us the decay constant.
04

Substitute Values to Find Elapsed Time

Substitute the given mass ratio 2.9 (= N / N_{0}) and calculated decay constant into the decay rate equation to find the elapsed time. This will give us the age of the volcanic rock.

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Key Concepts

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

Radioactive Decay Rate Equation
Understanding the radioactive decay rate equation is vital for calculating the age of geological samples. This equation reflects the relationship between the number of radioactive nuclei remaining in a sample and the time that has passed. In mathematical terms, the equation is expressed as:
\[ t = \frac{1}{\text{(lambda)}} \times \text{ln} \left(1 + \frac{N}{N_0}\right) \]
where:
  • t is the time that has elapsed,
  • N is the current number of radioactive atoms,
  • N_0 is the initial number of radioactive atoms, and
  • (lambda)(lambda) is the decay constant.

The decay constant ((lambda)) is a key factor in this calculation and is derived from the half-life of the radioactive isotope. To solve radiometric dating problems, we estimate the time elapsed since the formation of the rock by using this decay rate equation, essentially allowing us to 'read' the geological clock of a sample.
Potassium-40 Half-Life
The concept of half-life is central to the science of radiometric age dating. It describes the time required for half the quantity of a radioactive isotope to decay. For the particular case of potassium-40, or K40, it has a significant half-life of
\[ 1.248 \times 10^9 \text{ years} \]
by (beta) decay, and \[ 1.4 \times 10^9 \text{ years} \]
by electron capture. This long half-life makes potassium-40 a valuable isotope for dating ancient geological events.
To make our age determination as accurate as possible, we average the two possible half-lives provided for potassium-40 when calculating the decay constant, considering both modes of decay that the isotope undergoes. It's this reasoned approach that enhances the reliability of the age estimate for the volcanic rock in question.
Argon-40 to Potassium-40 Mass Ratio
The mass ratio of argon-40 to potassium-40 in a sample is a crucial aspect of radiometric age dating. Argon-40 is a noble gas that is created as a result of the radioactive decay of potassium-40. In a closed system, where no argon is lost to the atmosphere, the accumulation of argon-40 can provide us with important insights into the sample's age.
In your exercise, the provided ratio is
\[ \text{Argon-40} : \text{Potassium-40} = 2.9 \]
This ratio essentially indicates that for every atom of potassium-40 present, there are about 2.9 atoms of argon-40 that have been produced through radioactive decay. By employing the radioactive decay rate equation, this ratio allows us to determine the elapsed time since the volcanic rock solidified and ceased exchanging gases with the environment. Understanding these ratios and how they apply to the decay rate equation is integral to unlocking the history recorded within rocks and other materials.

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Most popular questions from this chapter

A certain shale deposit containing \(0.006 \%\) U by mass is being considered for use as a potential fuel in a breeder reactor. Assuming a density of \(2.5 \mathrm{g} / \mathrm{cm}^{3},\) how much energy could be released from \(1.00 \times 10^{3} \mathrm{cm}^{3}\) of this material? Assume a fission energy of \(3.20 \times 10^{-11} \mathrm{J}\) per fission event (that is, per U atom).

Write nuclear equations to represent (a) the decay of \(^{214} \mathrm{Ra}\) by \(\alpha\) -particle emission (b) the decay of \(^{205}\) At by positron emission (c) the decay of \(^{212} \mathrm{Fr}\) by electron capture (d) the reaction of two deuterium nuclei (deuterons) to produce a nucleus of \(\frac{3}{2} \mathrm{He}\). (e) the production of \({243}_{97} \mathrm{Bk}\) get by the \(\alpha\) -particle bombardment of\({241}_{95} \mathrm{Am}\) (f) a nuclear reaction in which thorium-232 is bombarded with \(\alpha\) particles, producing a new nuclide and four neutrons.

Write a plausible equation for the decay of tritium, 3 \(\mathrm{H}\), the radioactive isotope of hydrogen. 1 \(\textrm{ }\).

Which member of the following pairs of nuclides would you expect to be most abundant in natural sources: (a) \(_{10}^{20} \mathrm{Ne}\) or \(_{10}^{22} \mathrm{Ne} ;\) (b) \(_{8}^{17} \mathrm{O}\) or \(_{8}^{18} \mathrm{O} ;\) (c) \(_{3}^{6} \mathrm{Li}\) or \(_{3}^{7} \mathrm{Li} ?\) Explain your reasoning.

A sample containing \(_{88}^{224} \mathrm{Ra},\) which decays by \(\alpha\) -particle emission, disintegrates at the following rate, expressed as disintegrations per minute or counts per minute \((\mathrm{cpm}): t=0,1000 \mathrm{cpm} ; t=1 \mathrm{h}\) \(992 \mathrm{cpm} ; t=10 \mathrm{h}, 924 \mathrm{cpm} ; t=100 \mathrm{h}, 452 \mathrm{cpm}\) \(t=250 \mathrm{h}, 138 \mathrm{cpm} .\) What is the half-life of this nuclide?

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