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Scientists have estimated that the earth's crust was formed 4.3 billion years ago. The radioactive nuclide \(176 \mathrm{Lu},\) which decays to 176 \(\mathrm{Hf}\) , was used to estimate this age. The half-life of 176 \(\mathrm{Lu}\) is 37 billion years. How are ratios of \(^{176} \mathrm{Lu}\) to 176 \(\mathrm{Hf}\) utilized to date very old rocks?

Short Answer

Expert verified
To date very old rocks, scientists use the radioactive decay of \(^{176}\mathrm{Lu}\) to \(^{176}\mathrm{Hf}\) and their half-life of 37 billion years. They first measure the ratio of \(^{176}\mathrm{Lu}\) to \(^{176}\mathrm{Hf}\) in the rock sample and then utilize the following formula to estimate the age of the rock: \( t = \frac{1}{\lambda} \ln \left(1 + \frac{N_\mathrm{Hf}}{N_\mathrm{Lu}}\right) \), where \(t\) is the age, \(\lambda\) is the decay constant, and \(N_\mathrm{Hf}\) and \(N_\mathrm{Lu}\) are the amounts of \(^{176}\mathrm{Hf}\) and \(^{176}\mathrm{Lu}\) in the sample, respectively. By analyzing these ratios, scientists have estimated that the Earth's crust was formed 4.3 billion years ago.

Step by step solution

01

Understand Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. In this case, \(^{176}\mathrm{Lu}\) is decaying to \(^{176}\mathrm{Hf}\). The time it takes for half of the original radioactive substance to decay is called the half-life. For \(^{176}\mathrm{Lu}\), the half-life is 37 billion years.
02

Calculate the ratio of \(^{176}\mathrm{Lu}\) to \(^{176}\mathrm{Hf}\)

To date a rock using this method, scientists first determine the ratio of \(^{176}\mathrm{Lu}\) to \(^{176}\mathrm{Hf}\) in the rock. This requires measuring the amounts of both isotopes present in the rock sample. The ratio of \(^{176}\mathrm{Lu}\) to \(^{176}\mathrm{Hf}\) will give an indication of how long the decay process has been occurring since the rock's formation.
03

Use half-life to find the age of the rock

With the half-life of \(^{176}\mathrm{Lu}\) and the ratio of \(^{176}\mathrm{Lu}\) to \(^{176}\mathrm{Hf}\) in the rock, the age of the rock can be estimated. The formula to calculate the age is: \[ t = \frac{1}{\lambda} \ln \left(1 + \frac{N_\mathrm{Hf}}{N_\mathrm{Lu}}\right) \] where \(t\) is the age of the rock, \(\lambda\) is the decay constant (which is related to the half-life by the formula \(\lambda = \frac{\ln 2}{T_{1/2}}\), where \(T_{1/2}\) is the half-life), \(N_\mathrm{Hf}\) is the amount of \(^{176}\mathrm{Hf}\) in the sample, and \(N_\mathrm{Lu}\) is the amount of \(^{176}\mathrm{Lu}\) in the sample.
04

Estimate the age of the Earth's crust

In this exercise, scientists have used this method to estimate that the Earth's crust was formed 4.3 billion years ago. By analyzing the ratios of \(^{176}\mathrm{Lu}\) to \(^{176}\mathrm{Hf}\) in very old rocks, they were able to determine the age of formation through radioactive decay processes. This method is critical in understanding the geological history and development of our planet.

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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 natural process where unstable atomic nuclei transform into more stable ones by emitting energy. This transformation can release particles like alpha, beta, or gamma rays. For the element \(^{176} ext{Lu}\), radioactive decay results in its conversion to \(^{176} ext{Hf}\). It's a slow process, requiring billions of years.
  • This decay helps estimate the age of geological formations.
  • It's like a natural clock ticking away over millions of years.
Understanding radioactive decay is key to determining the lifetime of materials on Earth.
Half-life
The concept of half-life is crucial in radiometric dating. It's the time it takes for half of a radioactive substance to decay. For \(^{176} ext{Lu}\), this period is a whopping 37 billion years.
  • Half-life provides a measure of stability for an isotope.
  • Long half-lives mean the substance changes very slowly.
By knowing the half-life, scientists can use the remaining radioactive material to calculate how old a rock or fossil is. This straightforward concept is a cornerstone of geochronology.
Isotopic Ratios
Isotopic ratios involve comparing the amounts of different isotopes within a sample. For dating rocks, the ratio of \(^{176} ext{Lu}\) to \(^{176} ext{Hf}\) is used.
  • This ratio indicates how much decaying has occurred.
  • A greater amount of \(^{176} ext{Hf}\) means more time has passed since the rock formed.
By calculating these ratios, scientists can determine the age of rocks, which is vital for understanding Earth's history. It's like having a snapshot of the past stored in stone.
Geochronology
Geochronology is the science of determining the age of rocks and fossils. It's like piecing together a massive chronological puzzle of Earth's history.
  • Techniques like radiometric dating provide accurate time estimates.
  • This field helps us explore the timeline of Earth's development.
Studying isotopic ratios and half-lives contributes directly to geochronology. It allows researchers to map out significant events like the formation of Earth's crust, offering insights into the dynamics of our planet over billions of years.

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Most popular questions from this chapter

Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. The fuel rods in most nuclear reactors therefore are often made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction $$ \mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad 8^{\circ}=-2.36 \mathrm{V} $$ a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium. c. Calculate \(\mathscr{G} \circ, \Delta G^{\circ},\) and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accidents at Three Mile Island in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred? If \(1.00 \times 10^{3} \mathrm{kg}\) Zreacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at 1.0 \(\mathrm{atm}\) and \(1000 .^{\circ} \mathrm{C}\) is produced? e. At Chernobyl in \(1986,\) hydrogen was produced by the reaction of superheated steam with the graphite reactor core: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ It was not possible to prevent a chemical explosion at Chernobyl. In light of this, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.

The most stable nucleus in terms of binding energy per nucleon is \(^{56}\mathrm{Fe}\) . If the atomic mass of \(^{56}\mathrm{Fe}\) is \(55.9349 \mathrm{u},\) calculate the binding energy per nucleon for \(^{56} \mathrm{Fe} .\)

Many transuranium elements, such as plutonium-232 , have very short half- lives. (For \(^{232} \mathrm{Pu}\) , the half-life is 36 minutes.) However, some, like protactinium- 231 (half-life \(=3.34 \times 10^{4}\) years), have relatively long half-lives. Use the masses given in the following table to calculate the change in energy when 1 mole of \(^{232} \mathrm{Pu}\) nuclei and 1 mole of \(^{231} \mathrm{Pa}\) nuclei are each formed from their respective number of protons and neutrons. (Since the masses of \(^{232} \mathrm{Pu}\) and \(^{231} \mathrm{Pa}\) are atomic masses, they each include the mass of the electrons present. The mass of the nucleus will be the atomic mass minus the mass of the electrons.)

The stable isotopes of boron are boron-10 and boron-11. Four radioactive isotopes with mass numbers 8, 9, 12, and 13 are also known. Predict possible modes of radioactive decay for the four radioactive isotopes of boron.

Define third-life in a similar way to half-life, and determine the "third- life" for a nuclide that has a half-life of 31.4 years.

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