Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 43

A living plant contains approximately the same fraction of carbon-14 4 as in atmospheric carbon dioxide. Assuming that the observed rate of decay of carbon-14 4 from a living plant is 13.6 counts per minute per gram of carbon, how many counts per minute per gram of carbon will be measured from a \(15,000\) -year-old sample? Will radiocarbon dating work well for small samples of 10 \(\mathrm{mg}\) or less? (For \(^{14} \mathrm{C}, t_{1 / 2}=5730\) years.)

Short Answer

Expert verified
In a 15,000-year-old sample, the activity measured will be approximately 7.58 counts per minute per gram of carbon. Radiocarbon dating may not work well for small samples of 10 mg or less, as the number of carbon-14 atoms will be much fewer and difficult to accurately measure.

Step by step solution

01

Calculate the decay constant

Using the half-life formula, we can determine the decay constant (λ) for carbon-14: \[t_{1/2} = \frac{ln(2)}{\lambda}\] Rearrange the equation to solve for λ: \[\lambda = \frac{ln(2)}{t_{1/2}}\] Now, plug in the given half-life, \(t_{1/2}=5730\) years, to get the decay constant: \[\lambda = \frac{ln(2)}{5730} = 1.21 \times 10^{-4} \ \text{year}^{-1}\]
02

Determine the remaining carbon-14 atoms

Using the decay equation, we can find the number of remaining carbon-14 atoms in the 15,000-year-old sample: \[N(t) = N_0 e^{-\lambda t}\] Where \(N(t)\) is the number of remaining carbon-14 atoms at time t, \(N_0\) is the initial number of carbon-14 atoms, and t is the time in years. We want to find \(N(t)\) when \(t = 15,000\) years: \[N(15,000) = N_0 e^{- (1.21 \times 10^{-4})(15,000)}\]
03

Calculate the original number of carbon-14 atoms

We know the initial activity of the carbon-14 in the sample, so we can calculate the initial number of carbon-14 atoms using the given activity of 13.6 counts per minute per gram of carbon. The activity (A) is related to the decay constant and the number of atoms by: \[A = \lambda N\] Therefore, we can find the initial number of carbon-14 atoms: \[N_0 = \frac{A}{\lambda} = \frac{13.6}{1.21 \times 10^{-4}} = 1.13 \times 10^5\ \text{atoms}\]
04

Determine the counts per minute per gram of carbon

Now we can plug in the initial number of carbon-14 atoms into our decay equation and find the remaining number of carbon-14 atoms at 15,000 years: \[N(15,000) = (1.13 \times 10^5) e^{- (1.21 \times 10^{-4})(15,000)} = 6.26 \times 10^4\ \text{atoms}\] We can now find the activity in counts per minute per gram of carbon for the 15,000-year-old sample using the activity formula: \[A(t) = \lambda N(t) = (1.21 \times 10^{-4})(6.26 \times 10^4) = 7.58 \ \text{counts per minute per gram of carbon}\]
05

Discuss the viability of radiocarbon dating for small samples

Since the activity of carbon-14 in the 15,000-year-old sample is 7.58 counts per minute per gram of carbon, the activity is significantly lower than the initial activity of 13.6 counts per minute per gram of carbon. For small samples of 10 mg or less, the number of carbon-14 atoms will be much fewer, making it difficult to accurately measure the activity and radiocarbon date the sample. Therefore, radiocarbon dating may not work well for small samples of 10 mg or less.

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.

Start your free trial

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

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

The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July \(16,1945 .\) What percentage of the strontium- 90\(\left(t_{1 / 2}=28.9 \text { years) originally produced }\right.\) by that explosion still remains as of July \(16,2017 ?\)

Photosynthesis in plants can be represented by the following overall equation: $$ 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\mathrm{} Light \mathrm{}}{\longrightarrow} C_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ Algae grown in water containing some $^{18} \mathrm{O}\left(\text { in } \mathrm{H}_{2}^{18} \mathrm{O}\right)$ evolve oxygen gas with the same isotopic composition as the oxygen bin the water. When algae growing in water containing only \(^{18} \mathrm{O}\) were furnished carbon dioxide containing \(^{18} \mathrm{O}\) no \(^{18} \mathrm{O}\) was found to be evolved from the oxygen gas produced. What conclusions about photosynthesis can be drawn from these experiments?

The most significant source of natural radiation is radon-222. $^{222} \mathrm{Rn},\( a decay product of \)^{238} \mathrm{U},$ is continuously generated in the earth's crust, allowing gaseous Rn to seep into the basements of buildings. Because \(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of 3.82 days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when $^{238} \mathrm{U}\( decays to \)^{222} \mathrm{Rn}$ ? What nuclei are produced when \(^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \(^{222} \mathrm{Rn}\) ? c. Another problem associated with \(^{222} \mathrm{Rn}\) is that the decay of \(^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer $\left(t_{1 / 2}=3.11 \mathrm{min} \text { ) that is a solid. What is the identity of the }\right.$ solid? Give the balanced equation of this species decaying by \(\alpha\) -particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that 222 Rn levels not exceed 4 \(\mathrm{pCi}\) per liter of air $\left(1 \mathrm{Ci}=1 \text { curie }=3.7 \times 10^{10} \text { decay events per second; }\right.$ \(1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\) . Convert 4.0 \(\mathrm{pCi}\) per liter of air into concentrations units of \(^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(^{2222} \mathrm{Rn}\) per liter of air.

A positron and an electron can annihilate each other on colliding, producing energy as photons: $$ -_{-1}^{0} e+_{+1}^{0} e \longrightarrow 2_{0}^{0} \gamma $$ Assuming that both g rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

Uranium-2355 undergoes a series of \(\alpha\) -particle and \(\beta\) -particle productions to end up as lead-207. How many \(\alpha\) particles and \(\beta\) particles are produced in the complete decay series?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation
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