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Chapter 31: Problem 46

A sample has a \(_{6}^{14} \mathrm{C}\) activity of \(0.0061 \mathrm{Bq}\) per gram of carbon. (a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 Bq. (b) Evidence suggests that the value of \(0.23 \mathrm{Bq}\) might have been as much as \(40 \%\) larger. Repeat part (a), taking into account this \(40 \%\) increase.

Short Answer

Expert verified
(a) The sample is approximately 22900 years old. (b) With a 40% increase, the sample is approximately 20100 years old.

Step by step solution

01

Understand the Given and Required

We know that in part (a), the activity of the sample is 0.0061 Bq, while the normal activity is 0.23 Bq. We need to find the age of the sample based on this information.
02

Apply the Decay Formula

The decay of carbon-14 can be expressed by the formula \( A = A_0 e^{-\lambda t} \), where \( A \) is the current activity, \( A_0 \) is the initial activity, \( \lambda \) is the decay constant, and \( t \) is the time passed. We need to use this formula to find \( t \).
03

Calculate the Decay Constant

The half-life of carbon-14 is about 5730 years. Use this to find the decay constant \( \lambda \): \[ \lambda = \frac{\ln(2)}{5730} \approx 1.2097 \times 10^{-4} \text{ per year} \].
04

Calculate the Age of the Sample (Part a)

Use \( A = 0.0061 \) and \( A_0 = 0.23 \) in the decay formula: \[ 0.0061 = 0.23 e^{-1.2097 \times 10^{-4} t} \]. Solve for \( t \):\[ t = \frac{\ln\left(\frac{0.0061}{0.23}\right)}{-1.2097 \times 10^{-4}} \approx 22900 \text{ years} \].
05

Adjust Initial Activity for Part (b)

If the initial activity was 40% greater, then \( A_0 = 0.23 + 0.092 = 0.322 \) Bq. We need to use this adjusted value for the initial activity.
06

Calculate the Age of the Sample (Part b)

Use \( A = 0.0061 \) and adjusted \( A_0 = 0.322 \) in the formula: \[ 0.0061 = 0.322 e^{-1.2097 \times 10^{-4} t} \]. Solve for \( t \):\[ t = \frac{\ln\left(\frac{0.0061}{0.322}\right)}{-1.2097 \times 10^{-4}} \approx 20100 \text{ years} \].

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Key Concepts

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

Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. Over time, this radiation causes the transformation of the parent isotope into a more stable daughter isotope. Here, carbon-14, written as \(_{6}^{14} \mathrm{C}\) for chemical notation, is a radioactive isotope of carbon.
  • It undergoes beta decay to transform into the nitrogen isotopes.
  • This process is spontaneous, and the rate at which it happens is random on a per-nucleus basis.
  • However, statistically, we can predict the behavior of large numbers of atoms very accurately.
The decay rate is crucial for understanding how we apply this to real-world dating, such as carbon dating a historical artifact.
Half-Life of Carbon-14
The half-life of a radioactive isotope is the time it takes for half of the original amount of radioactive atoms in a sample to decay. For carbon-14, the half-life is approximately 5730 years.
  • This means that after 5730 years, a given sample of carbon-14 will have decayed such that only half remains.
  • The half-life is a constant, providing a reliable measurement to estimate how long a sample has been in existence since it stopped exchanging carbon with the atmosphere.
  • This concept is essential for radiocarbon dating because it helps us calculate ages accurately over thousands of years.
In practice, the half-life allows scientists to establish a relationship between the decay constant and the understanding of decay rates in natural samples.
Exponential Decay Formula
The exponential decay formula describes how the activity of a radioactive sample changes over time. It is typically expressed as \( A = A_0 e^{-\lambda t} \), where:
  • \( A \) is the current activity of the sample.
  • \( A_0 \) is the initial activity when the sample started its decay.
  • \( \lambda \) is the decay constant, which is related to the half-life by \( \lambda = \frac{\ln(2)}{T_{1/2}} \).
  • \( t \) represents the time since the sample started to decay.
This formula allows us to calculate the age of archaeological artifacts or geological samples. Understanding how to manipulate and solve this formula with given values helps students see how theoretical physics is applied in real-world investigations.

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Most popular questions from this chapter

Find the energy that is released when a nucleus of lead \({ }^{211} \mathrm{Pb}\) (atomic mass \(=210.988735 \mathrm{u}\) ) undergoes \(\beta^{-}\) decay to become bismuth \({ }_{83}^{211} \mathrm{Bi}\) (atomic mass \(=210.987255 \mathrm{u})\)

Write the \(\beta^{+}\) decay process for each of the following nuclei, being careful to include \(Z\) and \(A\) and the proper chemical symbol for each daughter nucleus: (a) \({ }_{9}^{18} \mathrm{F}\) (b) \({ }_{8}^{15} \mathrm{O}\)

When any radioactive dating method is used, experimental error in the measurement of the sample's activity leads to error in the estimated age. In an application of the radiocarbon dating technique to certain fossils, an activity of 0.100 Bq per gram of carbon is measured to within an accuracy of \(\pm 10.0 \% .\) Find the age of the fossils and the maximum error (in years) in the value obtained. Assume that there is no error in the 5730 -year half- life of \({ }_{6}^{14} \mathrm{C}\) nor in the value of \(0.23 \mathrm{Bq}\) per gram of carbon in a living organism.

In the form \({ }_{Z}^{A} \mathrm{X}\), identify the daughter nucleus that results when (a) plutonium \({ }_{94}^{244}\) Pu undergoes \(\alpha\) decay, (b) sodium \({ }_{11}^{24}\) Na undergoes \(\beta^{-}\) decay, and (c) nitrogen \({ }_{7}^{13} \mathrm{N}\) undergoes \(\beta^{+}\) decay.

The \(\beta^{-}\) decay of phosphorus \({ }_{15}^{32} \mathrm{P}\) (atomic mass \(=31.973907 \mathrm{u}\) ) pro-duces a daughter nucleus that is sulfur \(\frac{32}{16} \mathrm{S}\) (atomic mass \(=31.972070 \mathrm{u}\) ),\(\mathrm{a} \beta^{-}\) particle, and an antineutrino. The kinetic energy of the \(\beta^{-}\) particle is \(0.90 \mathrm{MeV}\). Find the maximum possible energy (in \(\mathrm{MeV}\) ) that the antineutrino could carry away.
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