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Chapter 21: Problem 41

A 10.00 -g plant fossil from an archaeological site is found to have a ${ }^{14} \mathrm{C}$ activity of 3094 disintegrations over a period of ten hours. A living plant is found to have a \({ }^{14} \mathrm{C}\) activity of 9207 disintegrations over the same period of time for an equivalent amount of sample with respect to the total contents of carbon. Given that the half-life of \({ }^{14} \mathrm{C}\) is 5715 years, how old is the plant fossil?

Short Answer

Expert verified
The plant fossil is approximately 4489 years old, calculated by determining the activity ratio of \({ }^{14} \mathrm{C}\) between the fossil and a living plant, and then using the half-life formula for \({ }^{14} \mathrm{C}\) which has a half-life of 5715 years.

Step by step solution

01

Calculate the activity ratio

First, calculate the activity ratio of the fossil's \({ }^{14} \mathrm{C}\) activity to the living plant's \({ }^{14} \mathrm{C}\) activity: Activity ratio = \(\dfrac{\text{Activity in fossil}}{\text{Activity in living plant}}\) Activity ratio = \(\dfrac{3094}{9207}\) Now, calculate the exact value.
02

Find the exact value of the activity ratio

Activity ratio = \(\dfrac{3094}{9207} \approx\) 0.3361
03

Use the half-life formula

Now, we can use the half-life formula to find the age of the fossil. The formula for half-life is: \(N_t = N_0 \times (\dfrac{1}{2})^\dfrac{t}{t_{1/2}}\) Where: - \(N_t\) is the amount of radioactive substance at time \(t\) - \(N_0\) is the initial amount of radioactive substance - \(t_{1/2}\) is the half-life - \(t\) is the time that has passed In this case, we have the activity ratio (\(\dfrac{N_t}{N_0}\)), which is approximately 0.3361. The half-life of \({ }^{14} \mathrm{C}\) is 5715 years. We can rearrange the formula to solve for \(t\): \(t = t_{1/2} \times \dfrac{\log(\dfrac{N_t}{N_0})}{\log(\dfrac{1}{2})}\) Now, plug in the values.
04

Calculate the age of the plant fossil

By plugging in the values, we get: \(t = 5715 \times \dfrac{\log(0.3361)}{\log(0.5)}\) Now, calculate the exact age.
05

Find the exact age of the plant fossil

By finding the exact age, we get: \(t \approx 4489.3 \, \text{years}\) Thus, the plant fossil is approximately 4489 years old.

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Most popular questions from this chapter

Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of 8.02 days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of NaI, in which only a small fraction of the iodide is radioactive. (a) Why is NaI a good choice for the source of iodine? (b) If a Geiger counter is placed near the person's thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thyroid will take up about \(12 \%\) of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to \(0.01 \%\) of the original amount?

When two protons fuse in a star, the product is \({ }^{2} \mathrm{H}\) plus a positron. Write the nuclear equation for this process.

According to current regulations, the maximum permissible dose of strontium-90 in the body of an adult is $1 \mu \mathrm{Ci}\left(1 \times 10^{-6} \mathrm{Ci}\right)\(. Using the relationship rate \)=k N$, calculate the number of atoms of strontium-90 to which this dose corresponds. To what mass of strontium-90 does this correspond? The half-life for strontium-90 is $28.8 \mathrm{yr}$

Tests on human subjects in Boston in 1965 and \(1966,\) following the era of atomic bomb testing, revealed average quantities of about \(2 \mathrm{pCi}\) of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits $8 \times 10^{-13} \mathrm{~J}\( of energy and if the average person weighs \)75 \mathrm{~kg},$ calculate the number of grays and sieverts of radiation in 1 yr from such a level of plutonium.

Why is it important that radioisotopes used as diagnostic tools in nuclear medicine produce gamma radiation when they decay? Why are alpha emitters not used as diagnostic tools?
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