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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

Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

Each of the following nuclei undergoes either beta decay or positron emission. Predict the type of emission for each: \((\mathbf{a})\) tritium, \(_{1}^{3} \mathrm{H},(\mathbf{b})_{38}^{89} \mathrm{Sr},(\mathbf{c})\) iodine-120, \((\mathbf{d})\) (d) silver-102.

Based on the following atomic mass values \(-^{1} \mathrm{H}, 1.00782\) \(\mathrm{amu} ;^{2} \mathrm{H}, 2.01410 \mathrm{amu}\); \(^{3} \mathrm{H}, 3.01605 \mathrm{amu} ;^{3} \mathrm{He}, 3.01603\) \(\mathrm{amu} ;^{4} \mathrm{He}, 4.00260 \mathrm{amu}-\) amu—and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: \begin{equation}(\mathbf{a})\quad_{1}^{2} \mathrm{H}+_{1}^{3} \mathrm{H} \longrightarrow _{4}^{2} \mathrm{He}+_{1}^{0} \mathrm{n}\end{equation} \begin{equation}(\mathbf{b})\quad_{1}^{2} \mathrm{H}+_{1}^{2} \mathrm{H} \longrightarrow_{2}^{3} \mathrm{He}+_{0}^{1} \mathrm{n}\end{equation} \begin{equation}(\mathbf{c})\quad_{1}^{2} \mathrm{H}+_{2}^{3} \mathrm{He} \longrightarrow_{2}^{4} \mathrm{He}+_{1}^{1} \mathrm{H}\end{equation}

How much energy must be supplied to break a single \(^{21} \mathrm{Ne}\) nucleus into separated protons and neutrons if the nucleus has a mass of 20.98846 amu? What is the nuclear binding energy for 1 mol of \(^{21} \mathrm{Ne} ?\)

Give the symbol for \((\mathbf{a})\) a neutron, \((\mathbf{b})\) an alpha particle, \((\mathbf{c})\) gamma radiation.

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