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Chapter 19: Problem 6

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 ratio of \({ }^{176}\mathrm{Lu}\) to \({ }^{176}\mathrm{Hf}\) is utilized to date very old rocks by examining the changes in the radioactive isotope \(^{176}\mathrm{Lu}\) over time due to radioactive decay. Since \({ }^{176}\mathrm{Hf}\) does not change (it's not radioactive), it serves as a reference point for measuring the changes in the amount of \({ }^{176}\mathrm{Lu}\). The half-life equation, \( N_t = N_0(1/2)^{t/T} \), is used to calculate the remaining amount of \({ }^{176}\mathrm{Lu}\) after a given time period, and by comparing this ratio with the initial ratio, scientists can estimate the time that has passed, thus determining the age of the rock.

Step by step solution

01

Understanding Radioactive Decay and Half-Life

Radioactive decay is a process by which unstable atomic nuclei lose energy by emitting radiation. This decay eventually leads to the formation of stable isotopes. A radioactive substance's half-life is the time it takes for half of the radioactive atoms to decay. In this exercise, we are given the radioactive nuclide \({ }^{176}\mathrm{Lu}\) that decays into \({ }^{176}\mathrm{Hf}\). The half-life of Lu-176 is 37 billion years, meaning that after 37 billion years, half of the initial amount of Lu-176 will decay into Hf-176.
02

The Initial Ratio of \({ }^{176}\mathrm{Lu}\) and \({ }^{176}\mathrm{Hf}\)

Initially, during the Earth's formation approximately 4.3 billion years ago, we can assume that the rocks had a certain amount of Lu-176 and Hf-176. As time progresses, the amount of Lu-176 decreases (due to radioactive decay), while the amount of Hf-176 increases. Since the amount of Hf-176 does not change (it's not radioactive), it serves as a reference point for measuring the changes in the amount of Lu-176. The ratio between these two isotopes can be used to determine the age of the rock.
03

Calculating the Remaining Amount of \({ }^{176}\mathrm{Lu}\)

To determine the age of a rock using the isotopes, we need to find the remaining fraction of the radioactive isotope (Lu-176) after a given time period has passed. Using the half-life equation, we can write: \( N_t = N_0(1/2)^{t/T} \) Where: \(N_t\) = remaining amount of Lu-176 after time t, \(N_0\) = initial amount of Lu-176, t = time (in years), and T = half-life of Lu-176 (37 billion years).
04

Using the Ratio to Determine the Age of the Rock

The ratio of \({ }^{176}\mathrm{Lu}\) to \({ }^{176}\mathrm{Hf}\) can then be used to estimate the age of the rock. As the radioactive decay takes place, the ratio will change, and by comparing this ratio with the initial ratio, we can estimate the time that has passed. R = \(\frac{N_t}{N_{Hf}}\), where R is the ratio of Lu-176 to Hf-176. By measuring the current ratio of these isotopes, scientists can compare it to the initial ratio and use this information to estimate the age of the rock. Using the known half-life, it becomes possible to determine the age of rocks dating back billions of years, including the Earth's crust itself.

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Most popular questions from this chapter

In each of the following radioactive decay processes, supply the missing particle. a. \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+\) ? b. \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+?\) c. \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+?\) d. \({ }^{241} \mathrm{Cm}+? \rightarrow{ }^{241} \mathrm{Am}\)

Which do you think would be the greater health hazard: the release of a radioactive nuclide of 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 \(\mathrm{Sr}\) and Xe. Why are the chemical properties of a radioactive substance important in assessing its potential health hazards?

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \({ }^{68} \mathrm{Ga}\) (electron capture) c. \({ }^{212} \mathrm{Fr}(\alpha)\) b. \(^{62} \mathrm{Cu}\) (positron) d. \({ }^{129} \mathrm{Sb}(\beta)\)

A \(0.20-\mathrm{mL}\) sample of a solution containing \({ }_{1}^{3} \mathrm{H}\) that produces \(3.7 \times\) \(10^{3}\) cps is injected into the bloodstream of an animal. After allowing circulatory equilibrium to be established, a \(0.20-\mathrm{mL}\) sample of blood is found to have an activity of 20 . cps. Calculate the blood volume of the animal.

The curie (Ci) is a commonly used unit for measuring nuclear radioactivity: 1 curie of radiation is equal to \(3.7 \times 10^{10}\) decay events per second (the number of decay events from \(1 \mathrm{~g}\) radium in \(1 \mathrm{~s}\) ). a. What mass of \(\mathrm{Na}_{2}{ }^{38} \mathrm{SO}_{4}\) has an activity of \(10.0 \mathrm{mCi}\) ? Sulfur38 has an atomic mass of \(38.0\) and a half-life of \(2.87 \mathrm{~h}\). b. How long does it take for \(99.99 \%\) of a sample of sulfur- 38 to decay?
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