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Chapter 21: Problem 40

A wooden artifact from a Chinese temple has a \({ }^{14} \mathrm{C}\) activity of 38.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. From the halflife for \({ }^{14} \mathrm{C}\) decay, \(5715 \mathrm{yr}\), determine the age of the artifact.

Short Answer

Expert verified
The age of the artifact is approximately \(2797\) years.

Step by step solution

01

Calculate the decay constant (\(\lambda\))

We know the halflife of \({}^{14}\mathrm{C}\), which is \(5715 \mathrm{yr}\). We can use the formula: \(\lambda = \frac{ln(2)}{T_{1/2}}\) Where: - \(\lambda\) is the decay constant - \(T_{1/2}\) is the halflife Plugging in the values, \(\lambda = \frac{ln(2)}{5715 \mathrm{yr}} = 1.21 \times 10^{-4} \mathrm{yr}^{-1}\)
02

Apply the decay equation to find the age of the artifact

We need to find the time \(t\), using the decay equation: \(N = N_0 \times e^{-\lambda t}\) Rearrange the equation to solve for \(t\): \(t = \frac{ln(\frac{N}{N_0})}{-\lambda}\) Plugging in the values, where \(N = 38.0\) counts/min and \(N_0 = 58.2\) counts/min, \(t = \frac{ln(\frac{38.0}{58.2})}{-1.21 \times 10^{-4} \mathrm{yr}^{-1}} = 2797.49 \mathrm{yr}\)
03

Final Answer

The age of the artifact is approximately \(2797\) years.

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