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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
The ratios of \({ }^{176} \mathrm{Lu}\) to \({ }^{176} \mathrm{Hf}\) are utilized to date very old rocks by first calculating the number of half-lives that have elapsed since the Earth's crust was formed. Then, the fraction of remaining \({ }^{176} \mathrm{Lu}\) is calculated through the formula, Fraction remaining = \((\frac{1}{2})^{\text{number of half-lives}}\). This ratio is then compared to the measured ratio in the rock samples, allowing scientists to deduce the time when the rock was formed and thus date the formation of the Earth's crust accurately. The formula for the \({ }^{176} \mathrm{Lu}\) to \({ }^{176} \mathrm{Hf}\) ratio is: Ratio = \(\frac{(\frac{1}{2})^{\frac{4.3}{37}}}{1 - (\frac{1}{2})^{\frac{4.3}{37}}}\)

Step by step solution

01

(Step 1: Understand Radioactive Decay)

(Radioactive decay is a process in which an unstable atomic nucleus loses energy through releasing radiation. In this case, \({ }^{176} \mathrm{Lu}\) decays into \({ }^{176} \mathrm{Hf}\). The half-life of an element is the time required for half of the initial quantity of the element to decay into its daughter product. As time passes, the amount of \({ }^{176} \mathrm{Lu}\) will decrease while the amount of \({ }^{176} \mathrm{Hf}\) will increase.)
02

(Step 2: Calculate the Number of Half-Lives)

(We need to determine the number of half-lives that have elapsed since the formation of the Earth's crust. Since the Earth's crust formed 4.3 billion years ago and the half-life of \({ }^{176} \mathrm{Lu}\) is 37 billion years, we can find the number of half-lives by dividing the crust's age by the half-life: Number of half-lives = \(\frac{\text{crust's age}}{\text{half-life}}\) Number of half-lives = \(\frac{4.3 \, \text{billion years}}{37 \, \text{billion years}}\) )
03

(Step 3: Calculate the Remaining \({ }^{176} \mathrm{Lu}\))

(Now we calculate the fraction of the radioactive nuclide remaining after a certain number of half-lives. The formula for this is: Fraction remaining = \((\frac{1}{2})^{\text{number of half-lives}}\) Fraction remaining = \((\frac{1}{2})^{\frac{4.3}{37}}\) )
04

(Step 4: Calculate the \({ }^{176} \mathrm{Lu}\) to \({ }^{176} \mathrm{Hf}\) Ratio)

(The amount of \({ }^{176} \mathrm{Hf}\) is equal to the initial amount of \({ }^{176} \mathrm{Lu}\) minus the current amount of \({ }^{176} \mathrm{Lu}\). Therefore, the \({ }^{176} \mathrm{Lu}\) to \({ }^{176} \mathrm{Hf}\) ratio is given by: Ratio = \(\frac{\text{current amount of } { }^{176} \mathrm{Lu}}{\text{initial amount of } { }^{176} \mathrm{Lu} - \text{current amount of } { }^{176} \mathrm{Lu}}\) Ratio = \(\frac{(\frac{1}{2})^{\frac{4.3}{37}}}{1 - (\frac{1}{2})^{\frac{4.3}{37}}}\) )
05

(Step 5: Date Formation)

(By measuring the \({ }^{176} \mathrm{Lu}\) to \({ }^{176} \mathrm{Hf}\) ratio in old rocks, scientists can compare this ratio with the ratio calculated in step 4. They can then deduce the time when the rock was formed by calculating the number of half-lives elapsed during that time. This allows them to date very old rocks and the formation of the Earth's crust accurately.)

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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 the process where an unstable atomic nucleus loses its energy by emitting radiation. During this process, a parent isotope transforms into a daughter product. In the example from our exercise, the radioactive nuclide \(^{176}\text{Lu}\)decays into \(^{176}\text{Hf}\). This occurs over time as the unstable \(^{176}\text{Lu}\) nucleus releases particles and energy, leading to the formation of the more stable \(^{176}\text{Hf}\).

Decay processes are generally slow and continuous, allowing them to be used as a reliable "clock." As time passes, the amount of the parent nuclide (\(^{176}\text{Lu}\)) diminishes, while the amount of its decay product (\(^{176}\text{Hf}\)) increases. Understanding this principle is essential for using radioactive decay in geological dating, as it provides insights into the time frame over which changes occurred.
Half-Life
The concept of half-life relates to the time required for half of a sample of a radioactive isotope to decay into its daughter product. In simpler terms, it's the time taken for half of the original radioactive material to transform. For instance, \(^{176}\text{Lu}\) has a half-life of 37 billion years. This means every 37 billion years, half of an initial amount of \(^{176}\text{Lu}\) will have decayed into \(^{176}\text{Hf}\).

When using half-lives to measure geological time, scientists can calculate how many half-lives have passed since the formation of a rock. This is done by dividing the age of the rock (e.g., 4.3 billion years) by the half-life of the isotope (37 billion years).
  • Number of half-lives = \(\frac{4.3 \text{ billion years}}{37 \text{ billion years}}\)
Through this calculation, scientists can estimate how much of the isotope has decayed and, consequently, how long the process has been ongoing since its initiation.
Isotope Ratio
The isotope ratio in radiometric dating compares the amount of the parent nuclide left to the amount of the daughter product formed. To date very old rocks, scientists measure the \(^{176}\text{Lu}\) to \(^{176}\text{Hf}\) ratio. As \(^{176}\text{Lu}\) decays over time,

this ratio changes. Calculating this ratio helps scientists track the decay process.
  • The ratio = \(\frac{\text{current amount of }^{176}\text{Lu}}{\text{initial amount of }^{176}\text{Lu} - \text{current amount of }^{176}\text{Lu}}\)
  • Using the remaining fraction of \(^{176}\text{Lu}\) and the decayed amount, scientists form precise time measurements.
Understanding the isotope ratio is crucial because it allows scientists to deduce the timing of geological events. By comparing measured ratios with theoretical calculations, they can determine the age of rock formations and, by extension, the Earth's crust.

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Most popular questions from this chapter

Breeder reactors are used to convert the nonfissionable nuclide \({ }_{92}^{238} \mathrm{U}\) to a fissionable product. Neutron capture of the \({ }_{92}^{238} \mathrm{U}\) is followed by two successive beta decays. What is the final fissionable product?

One type of commercial smoke detector contains a minute amount of radioactive americium- \(241\left({ }^{241} \mathrm{Am}\right)\), which decays by \(\alpha\) -particle production. The \(\alpha\) particles ionize molecules in the air, allowing it to conduct an electric current. When smoke particles enter, the conductivity of the air is changed and the alarm buzzes. a. Write the equation for the decay of \({ }^{241} \mathrm{Am}\) by \(\alpha\) -particle production. b. The complete decay of \({ }^{241}\) Am involves successively \(\alpha, \alpha\), \(\beta, \alpha, \alpha, \beta, \alpha, \alpha, \alpha, \beta, \alpha\), and \(\beta\) production. What is the final stable nucleus produced in this decay series? c. Identify the 11 intermediate nuclides.

Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \({ }^{241} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a \(5.00-\mathrm{g}\) sample of \({ }^{241} \mathrm{Am} ?\)

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.

Phosphorus-32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is \(14.3\) days. What mass of phosphorus-32 is left from an original sample of \(175 \mathrm{mg}\) \(\mathrm{Na}_{3}{ }^{32} \mathrm{PO}_{4}\) after \(35.0\) days? Assume the atomic mass of \({ }^{32} \mathrm{P}\) is \(32.0 \mathrm{u}\).

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