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Chapter 21: Problem 42

A wooden artifact from a Chinese temple has a \(^{14} \mathrm{C}\) cactivity of 38.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. From the half-life for \(^{14} \mathrm{C}\) decay, 5715 yr, determine the age of the artifact.

Short Answer

Expert verified
The age of the artifact is approximately 3616.4 years.

Step by step solution

01

Identify given values

We are given the following values: - \(N_t\): the remaining activity count of the artifact, which is 38.0 counts per minute. - \(N_0\): the initial activity count of the standard, which is 58.2 counts per minute. - \(T\): the half-life of the \(\mathrm{^{14}C}\) decay, which is 5715 years. We need to find the decay time \(t\) to determine the age of the artifact.
02

Setup the decay formula

Now we set up the decay formula: \(N_t = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}\)
03

Plug in the given values

Plug in the given values to the decay formula: \(38.0 = 58.2 \cdot \left(\frac{1}{2}\right)^{\frac{t}{5715}}\)
04

Solve for t

We need to solve the equation for \(t\): 1. Divide both sides by 58.2: \(\frac{38.0}{58.2} = \left(\frac{1}{2}\right)^{\frac{t}{5715}}\) 2. Take the natural logarithm of both sides: \(ln\left(\frac{38.0}{58.2}\right) = ln\left(\left(\frac{1}{2}\right)^{\frac{t}{5715}}\right)\) 3. Use the property of logarithms that states \(ln(a^b) = b \cdot ln(a)\): \(ln\left(\frac{38.0}{58.2}\right) = \frac{t}{5715} \cdot ln\left(\frac{1}{2}\right)\) 4. Multiply both sides by the reciprocal of \(ln\left(\frac{1}{2}\right)\) to isolate \(t\): \(t = 5715\frac{ln\left(\frac{38.0}{58.2}\right)}{ln\left(\frac{1}{2}\right)}\) 5. Calculate the value of \(t\): \(t \approx 3616.4\) Therefore, the age of the artifact is approximately 3616.4 years.

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Key Concepts

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

Carbon-14 Decay
Carbon-14 decay is a key concept in radiocarbon dating, which is a method used to determine the age of an artifact. Carbon-14, or \(^{14} \text{C}\), is a radioactive isotope of carbon that is naturally found in the atmosphere. It is absorbed by all living organisms at a constant rate during their lifetime.
  • After the organism dies, it stops absorbing Carbon-14, and the isotope begins to decay.

  • This decay follows an exponential pattern, meaning that it decreases rapidly at first and then more slowly over time.

  • The rate of decay is measured in what's called 'half-life', which is the time it takes for half of the initial amount of Carbon-14 to decay.
Understanding this decay is crucial because it allows scientists to date the artifact by measuring the remaining radioactivity in it.
Half-Life Calculation
The half-life of Carbon-14 is 5715 years, meaning that every 5715 years, half of the Carbon-14 in a sample will have decayed. This stable rate of decay is what makes Carbon-14 such a reliable tool for dating artifacts.
When calculating the age of an artifact using half-life, you use the equation:\[N_t = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}\]where:
  • \(N_t\) is the remaining count of Carbon-14 activity.

  • \(N_0\) is the original count at the time of the organism's death.

  • \(t\) is the time that has passed since the organism's death.

  • \(T\) is the half-life of Carbon-14.
This formula helps us to determine exactly how much time has passed since the sample stopped absorbing Carbon-14, thereby indicating its age.
Natural Logarithm in Chemistry
Natural logarithms, denoted by \(ln\), play a crucial role in calculating the decay time in radiocarbon dating. In chemistry, natural logarithms are used because they are particularly suited to dealing with exponential decay, like that of radioisotopes.
  • Converting a decay equation into a logarithmic form simplifies solving for the time \(t\).

  • The equation \(ln\left(\frac{N_t}{N_0}\right) = \frac{t}{T} \cdot ln\left(\frac{1}{2}\right)\) is derived from the earlier exponential decay formula.

  • The property \(ln(a^b) = b \cdot ln(a)\) is especially helpful to extract the time variable.
By using the natural logarithm, solving for the age of an artifact becomes more manageable, providing an accurate estimate of its age. This makes natural logarithms an indispensable tool in the field of radiocarbon dating.

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Most popular questions from this chapter

Tests on human subjects in Boston in 1965 and 1966, following the era of atomic bomb testing, revealed average quantities of about 2 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 kg, calculate the number of rads and rems of radiation in 1 yr from such a level of plutonium.

Which of the following statements about the uranium used in nuclear reactors is or are true? (i) Natural uranium has too little \(^{235} U\) to be used as a fuel. (ii) \(^{238} U\) cannot be used as a fuel because it forms a supercritical mass too easily. (iii) To be used as fuel, uranium must be enriched so that it is more than 50\(\%^{235} \mathrm{U}\) in composition. (iv) The neutron-induced fission of \(^{235} \mathrm{U}\) releases more neutrons per nucleus than fission of \(^{238} \mathrm{U}\).

Give the symbol for \((\mathbf{a})\) a proton, \((\mathbf{b})\) a beta particle, \((\mathbf{c})\) a positron.

Potassium-40 decays to argon-40 with a half-life of \(1.27 \times 10^{9}\) yr. What is the age of a rock in which the mass ratio of \(^{40} \mathrm{Ar}\) to \(^{40} \mathrm{K}\) is 4.2?

Which statement best explains why nuclear transmutations involving neutrons are generally easier to accomplish than those involving protons or alpha particles? \begin{equation} \begin{array}{l}{\text { (a) Neutrons are not a magic number particle. }} \\\ {\text { (b) Neutrons do not have an electrical charge. }} \\ {\text { (c) Neutrons are smaller than protons or alpha particles. }} \\ {\text { (d) Neutrons are attracted to the nucleus even at long distances,}} \\ \quad {\text { whereas protons and alpha particles are repelled. }}\end{array} \end{equation}
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