The ratio of \(\frac{{^{14}\text{C}}}{{^{12}\text{C}}}\) in a piece of wood is \(\frac{1}{8}\) part that of atmosphere. If the half-life of \(^{14}\text{C}\) is 5730 years, the age of the wood sample is _______ years.
Correct Answer: 17190
Approach Solution - 1
The age can be calculated using:
t = \(\left(\ln \frac{(^{14}C/^{12}C)_{\text{initial}}}{(^{14}C/^{12}C)_{\text{sample}}}\right) \frac{t_{1/2}}{\ln 2}\)
Given \(\frac{(^{14}C/^{12}C)_{\text{sample}}}{(^{14}C/^{12}C)_{\text{initial}}} = \frac{1}{8}\),
t = \(\ln 8 \times \frac{5730}{\ln 2} = 17190\) years
Approach Solution -2
The problem is about radiocarbon dating. We need to find the age of a wooden sample given that the ratio of \(^{14}\text{C}\) to \(^{12}\text{C}\) in it is \(\frac{1}{8}\) of the atmospheric ratio, and the half-life of \(^{14}\text{C}\) is 5730 years.
Concept Used:
Radioactive decay is a first-order process. The age of a sample can be determined using the relationship between the amount of radioactive substance remaining (\(N\)), the initial amount (\(N_0\)), and the number of half-lives (\(n\)) that have passed.
The fraction of the radioactive isotope remaining is given by:
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^n \]The total age (\(t\)) of the sample is the number of half-lives (\(n\)) multiplied by the half-life period (\(t_{1/2}\)):
\[ t = n \times t_{1/2} \]For carbon dating, the ratio \(\frac{^{14}\text{C}}{^{12}\text{C}}\) in a sample is proportional to the amount of \(^{14}\text{C}\) (\(N\)), while the atmospheric ratio is proportional to the initial amount (\(N_0\)) at the time of death.
Step-by-Step Solution:
Step 1: Determine the fraction of \(^{14}\text{C}\) remaining in the wood sample.
The ratio of \(\frac{N}{N_0}\) is equal to the ratio of the \((\frac{^{14}\text{C}}{^{12}\text{C}})\) ratio in the sample to that in the atmosphere.
\[ \frac{N}{N_0} = \frac{\left(\frac{^{14}\text{C}}{^{12}\text{C}}\right)_{\text{sample}}}{\left(\frac{^{14}\text{C}}{^{12}\text{C}}\right)_{\text{atmosphere}}} \]It is given that this ratio is \(\frac{1}{8}\).
\[ \frac{N}{N_0} = \frac{1}{8} \]Step 2: Calculate the number of half-lives (\(n\)) that have passed.
We use the radioactive decay formula:
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^n \]Substitute the value of \(\frac{N}{N_0}\):
\[ \frac{1}{8} = \left(\frac{1}{2}\right)^n \]Since \(8 = 2^3\), we can rewrite the equation as:
\[ \left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n \]By comparing the exponents, we find:
\[ n = 3 \]This means that three half-lives of \(^{14}\text{C}\) have passed.
Step 3: Calculate the age of the wood sample.
The age (\(t\)) is calculated by multiplying the number of half-lives (\(n\)) by the half-life of \(^{14}\text{C}\) (\(t_{1/2}\)).
Given \( t_{1/2} = 5730 \) years.
\[ t = n \times t_{1/2} = 3 \times 5730 \, \text{years} \]Final Computation & Result:
Performing the final multiplication:
\[ t = 3 \times 5730 = 17190 \, \text{years} \]The age of the wood sample is 17190 years.
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