Petrified Wood Fossil Age Calculation

This solution outlines the method to determine the age of a petrified wood fossil by applying the principles of Carbon-14 ($^{14}C$) radioactive decay.

Given Information for Calculation

• Current amount of $^{14}C$ ($N_T$): 8 g
• Initial amount of $^{14}C$ ($N_0$): 32 g
• Decay constant ($\lambda$): $0.0001216$ year^-1
• Half-life ($t_{1/2}$) of $^{14}C$: 5700 years

Radioactive Decay Formula

The mathematical model for radioactive decay is given by:

$N_T = N_0 e^{-\lambda T}$

Where:

• $N_T$ represents the remaining quantity of the isotope at time $T$.
• $N_0$ is the initial quantity of the isotope.
• $\lambda$ is the decay constant.
• $T$ is the time elapsed, or the age of the sample.

The provided decay constant $\lambda = 0.0001216$ year^-1 is consistent with the known half-life of $^{14}C$, as $t_{1/2} = \frac{\ln(2)}{\lambda} \approx 5700$ years.

Step-by-Step Age Determination

To find the fossil's age ($T$), we follow these steps:

1. Set up the decay equation: Plug the known values into the formula:

   $8 = 32 \times e^{-0.0001216T}$

2. Isolate the exponential term: Divide both sides by the initial amount ($N_0 = 32$ g):

   $\frac{8}{32} = e^{-0.0001216T}$

   Simplify the fraction:

   $\frac{1}{4} = e^{-0.0001216T}$

3. Apply logarithms: Take the natural logarithm (ln) of both sides to solve for the exponent:

   $\ln\left(\frac{1}{4}\right) = \ln\left(e^{-0.0001216T}\right)$

   Using logarithm properties, this simplifies to:

   $-\ln(4) = -0.0001216T$

   Rearrange the equation to solve for $T$:

   $T = \frac{\ln(4)}{0.0001216}$

4. Calculate the age: Substitute the approximate value for $\ln(4) \approx 1.3863$:

   $T \approx \frac{1.3863}{0.0001216} \text{ years}$

   $T \approx 11400.5 \text{ years}$

Conclusion

The calculated age of the petrified wood fossil is approximately $11400.5$ years. This result falls within the expected range of 11300 to 11500 years.