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.)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The only stable isotope of fluorine is fluorine-19. Predict possible modes of decay for fluorine-21, fluorine-18, and fluorine- \(17 .\)

Rubidium- 87 decays by \(\beta\) -particle production to strontium- 87 with a half-life of \(4.7 \times 10^{10}\) years. What is the age of a rock sample that contains \(109.7 \mu \mathrm{g}\) of \({ }^{87} \mathrm{Rb}\) and \(3.1 \mu \mathrm{g}\) of \({ }^{87} \mathrm{Sr}\) ? Assume that no \({ }^{87} \mathrm{Sr}\) was present when the rock was formed. The atomic masses for \({ }^{87} \mathrm{Rb}\) and \({ }^{87} \mathrm{Sr}\) are \(86.90919 \mathrm{u}\) and \(86.90888\) u, respectively.

Which do you think would be the greater health hazard: the release of a radioactive nuclide of \(\mathrm{Sr}\) or a radioactive nuclide of Xe into the environment? Assume the amount of radioactivity is the same in each case. Explain your answer on the basis of the chemical properties of Sr and Xe. Why are the chemical properties of a radioactive substance important in assessing its potential health hazards?

Radioactive copper-64 decays with a half-life of \(12.8\) days. a. What is the value of \(k\) in \(\mathrm{s}^{-1}\) ? b. A sample contains \(28.0 \mathrm{mg}^{64} \mathrm{Cu}\). How many decay events will be produced in the first second? Assume the atomic mass of \({ }^{64} \mathrm{Cu}\) is \(64.0 \mathrm{u}\). c. A chemist obtains a fresh sample of \({ }^{64} \mathrm{Cu}\) and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below \(25 \%\) of the initial measured value. How long does she have to do the experiment?

Iodine-131 is used in the diagnosis and treatment of thyroid disease and has a half-life of \(8.0\) days. If a patient with thyroid disease consumes a sample of \(\mathrm{Na}^{131} \mathrm{I}\) containing \(10 . \mu \mathrm{g}{ }^{131} \mathrm{I}\), how long will it take for the amount of \({ }^{131} \mathrm{I}\) to decrease to \(1 / 100\) of the original amount?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free