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Chapter 29: Problem 77

A space rock contains \(3.00 \mathrm{~g}\) of \({ }_{62}^{147} \mathrm{Sm}\) and \(0.150 \mathrm{~g}\) of ${ }_{60}^{143} \mathrm{Nd} .{ }_{62}^{147} \mathrm{Sm} \alpha\( decays to \){ }_{60}^{143} \mathrm{Nd}\( with a half-life of \)1.06 \times 10^{11} \mathrm{yr}\(. If the rock originally contained no \){ }_{60}^{143} \mathrm{Nd}$ how old is it?

Short Answer

Expert verified
Answer: The approximate age of the space rock is \(2.25 \times 10^{10} \mathrm{~yr}\).

Step by step solution

01

Calculate the decay constant

To find the decay constant, we can use the half-life value in the formula: \(\lambda = \frac{\ln 2}{t_{1/2}}\) Using the given half-life of \(1.06 \times 10^{11} \mathrm{yr}\), we can calculate: \(\lambda = \frac{\ln 2}{1.06 \times 10^{11} \mathrm{~yr}} = 6.54 \times 10^{-12} \mathrm{~yr^{-1}}\)
02

Formulate the decay equation

We have the decay equation: \(N_t = N_0 \cdot e^{-\lambda t}\) Initially, we had only \({}_{62}^{147}\mathrm{Sm}\) in the rock, so \(N_0 = 3.00 \mathrm{~g}\). After kinetic decay, the remaining amount of \({}_{62}^{147}\mathrm{Sm}\) is \(3.00 - 0.150 \mathrm{~g} = 2.85 \mathrm{~g}\) Now we want to find the time \(t\), with \(N_t = 2.85 \mathrm{~g}\) and \(\lambda = 6.54 \times 10^{-12} \mathrm{~yr^{-1}}\). \(2.85 = 3.00 \cdot e^{-6.54 \times 10^{-12} \mathrm{~yr^{-1}} \cdot t}\)
03

Solve for the age of the rock

To find the age of the rock, we need to solve for \(t\). First, divide both sides by \(3.00\): \(\frac{2.85}{3.00} = e^{-6.54 \times 10^{-12} \mathrm{~yr^{-1}} \cdot t}\) Now take the natural logarithm of both sides: \(\ln\left(\frac{2.85}{3.00}\right) = -6.54 \times 10^{-12} \mathrm{~yr^{-1}} \cdot t\) Then, solve for \(t\) by dividing both sides by \(-6.54 \times 10^{-12} \mathrm{~yr^{-1}}\): \(t = \frac{\ln\left(\frac{2.85}{3.00}\right)}{-6.54 \times 10^{-12} \mathrm{~yr^{-1}}} = 2.25 \times 10^{10}\mathrm{~yr}\) So, the age of the space rock is approximately \(2.25 \times 10^{10} \mathrm{~yr}\).

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