Questions: Part A An archeological artifact has a carbon-14 decay rate of 2.75 dis/mingC. If the rate of decay of a living organism is 15.3 dis/mingC, how old is this artifact? Assume that t1 / 2 for carbon-14 is 5730 yr. 1.42 x 10^4 yr 6.16 x 10^3 yr 1.03 x 10^3 yr 1.209 x 10^-4 yr

Solution Steps

We need to determine the age of an archeological artifact based on its carbon-14 decay rate compared to that of a living organism. The half-life of carbon-14 is given as 5730 years.

The decay of carbon-14 follows an exponential decay model. The formula for the remaining quantity of a radioactive substance is: \[ N(t) = N_0 e^{-\lambda t} \] where:

• \( N(t) \) is the remaining quantity at time \( t \),
• \( N_0 \) is the initial quantity,
• \( \lambda \) is the decay constant,
• \( t \) is the time elapsed.

The decay rate is proportional to the quantity of carbon-14. Therefore, we can write: \[ \frac{N(t)}{N_0} = \frac{2.75}{15.3} \]

The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by: \[ \lambda = \frac{\ln(2)}{t_{1/2}} \] Substituting the given half-life: \[ \lambda = \frac{\ln(2)}{5730} \approx 0.00012097 \, \text{yr}^{-1} \]

Using the relationship between the decay rates: \[ \frac{2.75}{15.3} = e^{-\lambda t} \] Taking the natural logarithm on both sides: \[ \ln\left(\frac{2.75}{15.3}\right) = -\lambda t \] Solving for \( t \): \[ t = \frac{\ln\left(\frac{15.3}{2.75}\right)}{\lambda} \] \[ t = \frac{\ln(5.5636)}{0.00012097} \] \[ t \approx \frac{1.7160}{0.00012097} \] \[ t \approx 14188.5 \, \text{yr} \]

Final Answer