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Chapter 19: Problem 30

Sandals found in a cave were determined by carbon-14 dating to be \(3.9 \times 10^{2}\) years old. Assuming that carbon from living material gives \(15.3\) disintegrations/min of C-14 per gram of carbon, what is the activity of the C-14 in the sandals in disintegrations/min/g of carbon? ( \(t_{1 / 2}\) of \(\mathrm{C}-14=\) 5730 years)

Short Answer

Expert verified
Answer: The activity of the C-14 in the old sandals is approximately 12.3 disintegrations/min/g of carbon.

Step by step solution

01

Identify given information

We are given the following information: - The age of the sandals: 3.9 x 10^2 years - The activity of C-14 from living material: 15.3 disintegrations/min/g of carbon - Half-life of C-14: 5730 years
02

Consider the radioactive decay formula

The radioactive decay formula can be used to find the activity of a radioactive material at any time. The formula is given by: \(A_t = A_0 e^{-\lambda t}\) Where: - \(A_t\) is the activity at time \(t\) - \(A_0\) is the initial activity - \(\lambda\) is the decay constant - \(t\) is the time
03

Calculate the decay constant

The decay constant, \(\lambda\), can be calculated using the half-life, \(t_{1/2}\), with the following formula: \(\lambda = \frac{ln(2)}{t_{1/2}}\) Plugging in the half-life of C-14: \(\lambda = \frac{ln(2)}{5730}\)
04

Calculate the activity at 3.9 x 10^2 years

Now, we have to find the activity of C-14 in the sandals. We know the initial activity of the C-14, which is 15.3 disintegrations/min/g of carbon. We can plug in the values into the radioactive decay formula to find the activity of C-14 at 3.9 x 10^2 years: \(A_t = A_0 e^{-\lambda t}\) \(A_t = 15.3 e^{-\frac{ln(2)}{5730}(3.9 \times 10^2)}\) Calculating the activity: \(A_t \approx 12.3\) disintegrations/min/g of carbon
05

Conclusion

The activity of the C-14 in the sandals is approximately 12.3 disintegrations/min/g of carbon.

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Most popular questions from this chapter

Cobalt-60 is used extensively in medicine as a source of \(\gamma\) -rays. Its half-life is \(5.27\) years. (a) How long will it take a \(\mathrm{Co}-60\) source to decrease to \(18 \%\) of its original activity? (b) What percent of its activity remains after 29 months?

A rock from an archaeological dig was found to contain \(0.255 \mathrm{~g}\) of Pb-206 per gram of U-238. Assume that the rock did not contain any Pb-206 at the time of its formation and that U-238 decayed only to Pb-206. How old is the rock? (For \(\mathrm{U}-238, t_{1 / 2}=4.5 \times 10^{9} \mathrm{y}\).)

The principle behind the home smoke detector is described on page 516. Americium- 241 is present in such detectors. It has a decay constant of \(1.51 \times 10^{-3} \mathrm{y}^{-1}\). You are urged to check the battery in the detector at least once a year. You are, however, never encouraged to check how much Am-241 remains undecayed. Explain why.

Smoke detectors contain a small amount of americium-241. Its decay product is neptunium-237. Identify the emission from americium-241.

Krypton-87 has a rate constant of \(1.5 \times 10^{-4} \mathrm{~s}^{-1}\). What is the activity of a \(2.00-\mathrm{mg}\) sample?
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