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Chapter 19: Problem 76

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.

Short Answer

Expert verified
The age of the rock sample is approximately \(1.69 \times 10^{10}\) years.

Step by step solution

01

Calculate the initial amount of Rubidium-87

Initially, all the Strontium-87 present in the sample would have been Rubidium-87 as it's assumed that no Strontium-87 was present when the rock was formed. So, we can add the current amounts of Rubidium-87 and Strontium-87 in the rock to find the initial amount of Rubidium-87. Initial Rubidium-87 = Current Rubidium-87 + Current Strontium-87 Initial Rubidium-87 = \(109.7\;\mu𝑔\) + \(3.1\;\mu𝑔\) Initial Rubidium-87 = \(112.8\;\mu𝑔\)
02

Calculate the decay constant (λ)

The decay constant (λ) can be calculated using the half-life formula: Half-life (t) = \(\frac{0.693}{\lambda}\) Rearranging to solve for λ: \(\lambda = \frac{0.693}{t}\) Using the given half-life of Rubidium-87, \(4.7 \times 10^{10}\) years: \(\lambda = \frac{0.693}{4.7 \times 10^{10}}\) \(\lambda \approx 1.474\times 10^{-11}\) years\(^{-1}\)
03

Calculate the age of the rock

Now we can use the decay rate formula to calculate the age of the rock: \(N_t = N_0e^{-\lambda t}\) Rearranging to solve for t: \(t = \frac{-\ln(\frac{N_t}{N_0})}{\lambda}\) Where \(N_0\) is the initial amount of Rubidium-87, \(N_t\) is the current amount of Rubidium-87, and \(\lambda\) is the decay constant. Plugging in the values: \(t = \frac{-\ln(\frac{109.7}{112.8})}{1.474\times 10^{-11}}\) \(t \approx 1.690 \times 10^{10}\) years So the age of the rock sample is approximately \(1.69 \times 10^{10}\) years.

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Most popular questions from this chapter

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. \({ }_{1}^{3} \mathrm{H}(\beta)\) b. \({ }_{3}^{8} \mathrm{Li}(\beta\) followed by \(\alpha\) ) c. \({ }_{4}^{7}\) Be (electron capture) d. \({ }_{5}^{8}\) B (positron)

A positron and an electron can annihilate each other on colliding, producing energy as photons: $$ { }_{-1}^{0} \mathrm{e}+{ }_{+1}^{0} \mathrm{e} \longrightarrow 2{ }^{0}{ }_{0}^{0} \gamma $$ Assuming that both \(\gamma\) rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

Calculate the binding energy per nucleon for \({ }_{1}^{2} \mathrm{H}\) and \({ }_{1}^{3} \mathrm{H}\). The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410 \mathrm{u} ;\) and \({ }_{1}^{3} \mathrm{H}, 3.01605 \mathrm{u}\).

The mass defect for a lithium-6 nucleus is \(-0.03434 \mathrm{~g} / \mathrm{mol}\). Calculate the atomic mass of \({ }^{6} \mathrm{Li}\)

During the research that led to production of the two atomic bombs used against Japan in World War II, different mechanisms for obtaining a supercritical mass of fissionable material were investigated. In one type of bomb, a "gun" shot one piece of fissionable material into a cavity containing another piece of fissionable material. In the second type of bomb, the fissionable material was surrounded with a high explosive that, when detonated, compressed the fissionable material into a smaller volume. Discuss what is meant by critical mass, and explain why the ability to achieve a critical mass is essential to sustaining a nuclear reaction.
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