Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life Calculation
Half-life is the time it takes for half of a sample of a radioactive substance to decay. It is a vital concept in the study of radioactive materials, including Carbon-14. Half-life calculations are crucial for determining the age of objects using radioactive decay. To calculate the age of a sample, we need to know:
  • The half-life of the radioactive substance, which is 5715 years for Carbon-14.
  • The ratio of the substance's activity in a sample to that in a living organism.
This information helps us estimate how many half-lives have occurred, thus giving insight into the age of the sample.
Radioactive Decay
Radioactive decay is a process where unstable nucleus emits radiation to become more stable. In Carbon-14 dating, we examine how this decay alters the amount of Carbon-14 in an object over time. For example, when a plant is alive, it continuously absorbs Carbon-14. But once it dies, no more Carbon-14 is absorbed, and the Carbon-14 already present starts decaying. This decay can be measured by counting the disintegrations over a given period. Understanding the decay rate of Carbon-14 helps us to find out the age of archaeological samples, such as fossils. The decreased radioactivity in a fossil compared to a living organism indicates how long it has stopped absorbing Carbon-14.
Archaeological Dating
Archaeological dating using Carbon-14, or radiocarbon dating, is a method to determine the age of an object containing organic material. It measures the remaining Carbon-14 levels compared to living organisms. When archaeologists find a fossil, they measure the Carbon-14 activity and compare it with a living equivalent to find its age.
  • The more the Carbon-14 has decayed, the older the sample.
  • This method helps date objects up to 50,000 years with reasonable accuracy.
Radiocarbon dating has revolutionized archaeology by providing a more precise timeline of human historical events and ancient life.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) What is the function of the moderator in a nuclear reactor? (b) What substance acts as the moderator in a pressurized water generator? (c) What other substances are used as a moderator in nuclear reactor designs?

Naturally found uranium consists of \(99.274 \%{ }^{238} \mathrm{U},\) \(0.720 \%^{235} \mathrm{U},\) and \(0.006 \%^{233} \mathrm{U}\). As we have seen, \({ }^{235} \mathrm{U}\) is the isotope that can undergo a nuclear chain reaction. Most of the \({ }^{235} \mathrm{U}\) used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, \(\mathrm{UF}_{6}(g) .(\mathbf{a})\) What is the mass of \(\mathrm{UF}_{6}\) in a 30.0 -L vessel of \(\mathrm{UF}_{6}\) at a pressure of 695 torr at \(350 \mathrm{~K} ?\) (b) What is the mass of \({ }^{235} \mathrm{U}\) in the sample described in part (a)? (c) Now suppose that the UF \(_{6}\) is diffused through a porous barrier and that the change in the ratio of \({ }^{238} \mathrm{U}\) and \({ }^{235} \mathrm{U}\) in the diffused gas can be described by Equation \(10.23 .\) What is the mass of \({ }^{235} \mathrm{U}\) in a sample of

A \(26.00-g\) sample of water containing tritium, \({ }_{1}^{3} \mathrm{H},\) emits \(1.50 \times 10^{3}\) beta particles per second. Tritium is a weak beta emitter with a half-life of \(12.3 \mathrm{yr}\). What fraction of all the hydrogen in the water sample is tritium?

Which type or types of nuclear reactors have these characteristics? (a) Does not use a secondary coolant (b) Creates more fissionable material than it consumes (c) Uses a gas, such as \(\mathrm{He}\) or \(\mathrm{CO}_{2}\), as the primary coolant

A radioactive decay series that begins with 90 Th ends with formation of the stable nuclide \({ }^{208} \mathrm{~Pb} .\) How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free