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Chapter 19: Problem 84

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 \mug of \(^{87} \mathrm{Rb}\) and 3.1\(\mu \mathrm{g}\) of $^{87} \mathrm{Sr} ?\( Assume that no \)^{87}$ 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 \(3.03 \times 10^9\) years.

Step by step solution

01

Identify the given information

We are given: 1. Half-life of Rubidium-87 (\(T_{1/2}\)) = \(4.7 \times 10^{10}\) years 2. Final amount of Rubidium-87 (\(N_t\)) = 109.7 µg 3. Final amount of Strontium-87 (\(N_{Sr}\)) = 3.1 µg 4. Initial amount of Rubidium-87 (\(N_0\)) = \(N_t + N_{Sr}\) (since no \(^{87} \mathrm{Sr}\) was present initially)
02

Calculate the initial amount of Rubidium-87

To find the initial amount of Rubidium-87, we add the final amounts of Rubidium-87 and Strontium-87. \(N_0 = N_t + N_{Sr} = 109.7 \, \mathrm{µg} + 3.1 \, \mathrm{µg} = 112.8 \, \mathrm{µg}\)
03

Find the decay constant

We are given half-life of Rubidium-87, so we can find the decay constant λ using: \(T_{1/2} = \frac{\ln{2}}{\lambda}\) Now, solve for \(\lambda\): \(\lambda = \frac{\ln{2}}{T_{1/2}} = \frac{\ln{2}}{4.7 \times 10^{10} \, \mathrm{years}}\)
04

Apply the decay formula to find the age of the rock sample

The decay formula is given by: \(N_t = N_0 \cdot e^{-\lambda t}\) We need to solve for t (time in years). We can rearrange the formula for t: \(t = \frac{\ln{\frac{N_0}{N_t}}}{\lambda}\) Plug in the values: \(t = \frac{\ln{\frac{112.8 \, \mathrm{µg}}{109.7 \, \mathrm{µg}}}}{\frac{\ln{2}}{4.7 \times 10^{10} \, \mathrm{years}}} = \frac{\ln{1.028}}{\frac{\ln{2}}{4.7 \times 10^{10} \, \mathrm{years}}}\) Now, calculate t: \(t \approx 3.03 \times 10^9 \, \mathrm{years}\) So, the age of the rock sample is approximately \(3.03 \times 10^9\) years.

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Most popular questions from this chapter

The number of radioactive nuclides in a sample decays from $1.00 \times 10^{20}\( to \)2.50 \times 10^{19}$ in 10.0 minutes. What is the half-life of this radioactive species?

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Calculate the binding energy in J/nucleon for carbon-12 (atomic mass \(=12.0000\) u) and uranium-235 (atomic mass \(=\) 235.0439 u). The atomic mass of \(_{1}^{1} \mathrm{H}\) is 1.00782 \(\mathrm{u}\) and the mass of a neutron is 1.00866 u. The most stable nucleus known is \(^{56}\) Fe $(\text { see Exercise } 50)\( . Would the binding energy per nucleon for \)^{56} \mathrm{Fe}$ be larger or smaller than that of \(^{12} \mathrm{C}\) or \(^{235} \mathrm{U}\) ? Explain.

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