Problem 42
Question
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.
Step-by-Step Solution
Verified Answer
The age of the artifact is approximately 3616.4 years.
1Step 1: 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.
2Step 2: Setup the decay formula
Now we set up the decay formula:
\(N_t = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}\)
3Step 3: Plug in the given values
Plug in the given values to the decay formula:
\(38.0 = 58.2 \cdot \left(\frac{1}{2}\right)^{\frac{t}{5715}}\)
4Step 4: 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.
Key Concepts
Carbon-14 DecayHalf-Life CalculationNatural Logarithm in Chemistry
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.
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:
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.
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.
Other exercises in this chapter
Problem 40
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