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A lunar rock was analyzed for argon by mass spectrometry and for potassium by atomic absorption. The results of these analyses showed that the sample contained \(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1}\) of argon and \(0.083 \%\) of potassium. The half-life of potassium- 40 is \(1.248 \times\) \(10^{9} \mathrm{y} \cdot\) Calculate the age of the lunar rock.

Short Answer

Expert verified
The age of the lunar rock is approximately \(3.5 \times 10^{9}\) years.

Step by step solution

01

Calculate the number of moles of argon

First calculate the number of moles of argon in the lunar rock. Recall that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters or \(2.24 \times 10^{4} \mathrm{mL} \). Therefore, if the lunar rock contains \(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1}\) of argon, the number of moles of argon (\(n_{Ar}\)) per gram of lunar rock can be calculated as \(n_{Ar} = \frac{(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1})}{(2.24 \times 10^{4} \mathrm{mL / mol})} = 1.35 \times 10^{-9} \mathrm{mol} \mathrm{g}^{-1} \)
02

Calculate the number of moles of potassium-40

Next calculate the number of moles of potassium-40 in the lunar rock. The total potassium content is given as \(0.083 \%\). Given natural isotopic abundance of potassium-40 is 0.0117\%, the amount of potassium-40 (\(n_{K}\)) per gram of lunar rock can be calculated as \(n_{K} = 0.083 \times 0.0117 \times \frac{1}{100} = 9.71 \times 10^{-5} \mathrm{mol} \mathrm{g}^{-1} \)
03

Calculate the age of the rock

Use the ratio of the number of moles of argon to the number of moles of potassium-40 to calculate the age of the rock. The ratio of argon-40 to potassium-40 is given by (\(1.35 \times 10^{-9}\) / \(9.71 \times 10^{-5}\)) = \(1.39 \times 10^{-5}\). Use the equation for radioactive decay which is \(N_t = N_0 \times \frac{1}{2}^{\left(\frac{t}{t_{1/2}}\right)}\). Rearranging gives \(t = t_{1/2} \times \frac{log(\frac{N_0}{N_t})}{log(2)} = 1.248 \times 10^{9} years \times \frac{log(1 / 1.39 \times 10^{-5})}{log(2)} = 3.5 \times 10^{9} years\)

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Key Concepts

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

Potassium-argon dating
Potassium-argon dating is a method used to determine the age of rocks and minerals. It relies on measuring the levels of potassium-40 and argon-40 in a sample. Potassium-40 is a naturally occurring isotope that undergoes radioactive decay. Over time, it decays into argon-40, a stable gas. This decay process allows scientists to calculate the age of a rock by comparing the ratio of potassium-40 to argon-40. The more argon-40 present, the older the rock is.
  • This technique is particularly useful for dating igneous and volcanic rocks.
  • It helps us understand geological events and the history of planetary bodies like the moon.
For lunar rocks, this method provides insight into the age and formation history of the moon's surface. By analyzing a sample’s argon and potassium content, scientists gain a clearer picture of its age.
Radioactive decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. Over time, isotopes like potassium-40 lose particles or energy and transform into more stable forms. This process is vital in dating rocks and minerals as it provides a 'clock' that scientists can use to measure time. At the heart of the calculation is the half-life, which is the time it takes for half of the radioactive atoms in a sample to decay into another form. For potassium-40, this half-life is approximately 1.248 billion years. The decay chain for potassium-40 includes both argon-40 and calcium-40, but for dating purposes, only argon-40 is considered. The predictable rate of this decay helps scientists determine the age of rocks.
Mass spectrometry analysis
Mass spectrometry is a technique used to measure the mass-to-charge ratio of ions. In the context of age determination of lunar rocks, it helps in analyzing the quantity of argon-40. This technique involves ionizing chemical species and sorting the ions based on their mass and charge.
  • The method aids in detecting and quantifying isotopes present in a sample with high precision.
  • Mass spectrometry is crucial for accurate age dating, as it precisely measures the tiny amounts of argon-40 generated from decay.
By using mass spectrometry, scientists can ensure that the calculations related to the age of lunar rocks are based on reliable data, thereby giving them confidence in their findings.
Atomic absorption spectroscopy
Atomic absorption spectroscopy is utilized to measure the concentration of elements, such as potassium in a sample. This analytical technique involves measuring the absorption of light by vaporized atoms. When a sample is heated, its atoms absorb light at specific frequencies. For lunar rocks, this method detects the presence and concentration of potassium-40. It involves:
  • Vaporizing the sample to make it suitable for measurement.
  • Detecting how much light is absorbed by potassium atoms, indicating their concentration.
By knowing the amount of potassium in a sample, scientists can calculate the total amount of potassium-40 present, which is necessary for determining the age of the rock. This process complements mass spectrometry to provide a comprehensive analysis of lunar 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).

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Radioactive decay and mass spectrometry are often used to date rocks after they have cooled from a magma. \(^{87} \mathrm{Rb}\) has a half-life of \(4.8 \times 10^{10}\) years and follows the radioactive decay $$^{87} \mathrm{Rb} \longrightarrow^{87} \mathrm{Sr}+\beta^{-}$$ A rock was dated by assaying the product of this decay. The mass spectrum of a homogenized sample of rock showed the \(^{87} \mathrm{Sr} /^{86} \mathrm{Sr}\) ratio to be \(2.25 .\) Assume that the original \(^{87} \mathrm{Sr} /^{86} \mathrm{Sr}\) ratio was 0.700 when the rock cooled. Chemical analysis of the rock gave \(15.5 \mathrm{ppm}\) Sr and 265.4 ppm \(\mathrm{Rb},\) using the average atomic masses from a periodic table. The other isotope ratios were \(^{86} \mathrm{Sr} /^{88} \mathrm{Sr}=\) 0.119 and \(^{84} \mathrm{Sr} /^{88} \mathrm{Sr}=0.007 .\) The isotopic ratio for \(^{87} \mathrm{Rb} /^{85} \mathrm{Rb}\) is 0.330. The isotopic masses are as follows:Calculate the following: (a) the average atomic mass of Sr in the rock (b) the original concentration of \(\mathrm{Rb}\) in the rock in \(\mathrm{ppm}\) (c) the percentage of rubidium- 87 decayed in the rock (d) the time since the rock cooled.

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