Question

A soil sample shows an average beta count of 6.8 counts per minute (cpm) per gram of organic carbon. The \( \text{C}^{14} \) count rate from organic carbon of present day vegetation is 15.26 cpm/g. The age of the sample is ________ years. \text{[round off to 1 decimal place] (Half-life of \( \text{C}^{14} \) = 5370 years)}

Hint:

For radiocarbon dating, use the formula \( t = -\frac{T_{1/2} \ln(N/N_0)}{\ln(2)} \) to calculate the age of the sample based on the ratio of the count rates.

Answer 1

Correct Answer: 6261

To find the age of the sample, we can use the formula for radiocarbon dating: \[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \] where:
- \( N \) is the current count rate of the sample (6.8 cpm),
- \( N_0 \) is the original count rate (15.26 cpm),
- \( t \) is the age of the sample (in years),
- \( T_{1/2} \) is the half-life of \( \text{C}^{14} \), which is 5370 years.
Rearranging the formula to solve for \( t \): \[ \frac{N}{N_0} = \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \] \[ \ln\left( \frac{N}{N_0} \right) = - \frac{t}{T_{1/2}} \ln(2) \] \[ t = -\frac{T_{1/2} \ln\left( \frac{N}{N_0} \right)}{\ln(2)} \] Substitute the values: \[ t = -\frac{5370 \times \ln\left( \frac{6.8}{15.26} \right)}{\ln(2)} \] \[ t = -\frac{5370 \times \ln(0.445)}{0.693} \] \[ t = -\frac{5370 \times (-0.809)}{0.693} = \frac{4331.6}{0.693} \approx 6261.0 \, \text{years}. \] Thus, the age of the sample is \( \boxed{6261.0} \) years.