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Chapter 19: Problem 32

A sample of a wooden artifact gives \(5.0\) disintegrations \(/ \mathrm{min} / \mathrm{g}\) carbon. The half-life of carbon-14 is 5730 years, and the activity of C-14 in wood just cut down from a tree is \(15.3\) disintegrations \(/ \mathrm{min} / \mathrm{g}\) carbon. How old is the wooden artifact?

Short Answer

Expert verified
Answer: The wooden artifact is approximately 4016 years old.

Step by step solution

01

Identify the decay formula

We will use the decay formula for radioactive materials to model the decay of carbon-14 in the wooden artifact. This formula is given by: \(N(t) = N_0 \cdot (\frac{1}{2})^{\frac{t}{t_{1/2}}}\) where: - \(N(t)\) = activity (disintegrations \(/ \mathrm{min} / \mathrm{g}\) carbon) at time \(t\) - \(N_0\) = initial activity (disintegrations \(/ \mathrm{min} / \mathrm{g}\) carbon) - \(t\) = time (in years) - \(t_{1/2}\) = half-life (in years)
02

Write the equation for the decay of the artifact

We can set up the equation as follows, substituting the given values for \(N(t)\), \(N_0\), and \(t_{1/2}\): \(5.0 = 15.3 \cdot (\frac{1}{2})^{\frac{t}{5730}}\)
03

Solve for time \(t\)

To solve for \(t\), we first need to isolate the exponential term. For that, divide both sides by 15.3: \(\frac{5.0}{15.3} = (\frac{1}{2})^{\frac{t}{5730}}\) Now, to remove the exponent, take the logarithm base-2 (log2) of both sides: \(\log_2{(\frac{5.0}{15.3})} = \frac{t}{5730}\) Finally, multiply both sides by the half-life (5730) to solve for \(t\): \(t = 5730 \cdot \log_2{(\frac{5.0}{15.3})}\)
04

Calculate the age of the artifact

Using a calculator, find the age of the artifact: \(t = 5730 \cdot \log_2{(\frac{5.0}{15.3})} \approx 4016 \ \mathrm{years}\) So, the wooden artifact is approximately 4016 years old.

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Most popular questions from this chapter

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