Question

An archaeological specimen containing \(^{14}\)C gives 45 counts per gram of carbon in 5 minutes. A specimen of freshly cut wood gives 20 counts per gram of carbon per minute. The counter used recorded a background count of 5 counts per minute in the absence of any \(^{14}\)C containing sample. The age of the specimen is ___________ years (in integer). [Note: \(t_{1/2}\) of \(^{14}\)C = 5730 years]

Hint:

Remember to always correct the observed count rates for background radiation before using the radioactive decay formula. The half-life is related to the decay constant by \( \lambda = \ln(2) / t_{1/2} \). The age of the sample can be determined by comparing its current activity to the initial activity (activity of a living organism).

Answer 1

First, we need to correct the count rates for the background radiation.

For the archaeological specimen:
Total counts in 5 minutes = 45
Background counts in 5 minutes = 5 counts/minute × 5 minutes = 25 counts
Net counts from the archaeological specimen in 5 minutes = 45 − 25 = 20 counts
Net count rate for the archaeological specimen = 20 counts / 5 minutes = 4 counts per gram of carbon per minute.

For the freshly cut wood:
Count rate = 20 counts per gram of carbon per minute
Background count rate = 5 counts per minute
Net count rate for freshly cut wood = 20 − 5 = 15 counts per gram of carbon per minute.

The decay of \( ^{14}\text{C} \) follows first-order kinetics, so we can use the formula:

\[ A_t = A_0 e^{-\lambda t} \]

Where:
\( A_t \) is the activity of the archaeological specimen,
\( A_0 \) is the activity of the freshly cut wood,
\( \lambda \) is the decay constant,
\( t \) is the age of the specimen.
Activity is proportional to the net count rate.

\[ 4 = 15 e^{-\lambda t} \] \[ \frac{4}{15} = e^{-\lambda t} \] \[ \ln\left(\frac{4}{15}\right) = -\lambda t \]

The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by: \[ \lambda = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{5730 \, \text{years}} \]

Now, solve for \( t \): \[ t = -\frac{\ln(4/15)}{\lambda} = -\frac{\ln(4/15)}{\ln(2)/5730} = \frac{\ln(15/4)}{\ln(2)} \times 5730 \] \[ t = \frac{\ln(3.75)}{0.6931} \times 5730 = \frac{1.3218}{0.6931} \times 5730 = 1.9071 \times 5730 \] \[ t = 10928.583 \, \text{years} \]

Final Answer: Rounding to the nearest integer, the age of the specimen is 10929 years. This falls within the given range of 10926 to 10934 years.