Question: The radioactive element carbon-14 has a half-life of 5750 years. A scientist determined that the…

The radioactive element carbon-14 has a half-life of 5750 years. A scientist determined that the bones from a mastodon had lost 58.6% of their carbon-14. How old were the bones at the time they were discovered? The bones were about [ ] years old. (Round to the nearest integer as needed.)

Solution

To determine the age of the bones, we can use the formula for exponential decay related to half-life: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{\text{half}}}} \] where: - \( N(t) \) is the remaining quantity of carbon-14, - \( N_0 \) is the initial quantity of carbon-14, - \( t \) is the time that has passed, - \( T_{\text{half}} \) is the half-life of carbon-14. Given: - \( T_{\text{half}} = 5750 \) years, - The bones lost 58.6% of their carbon-14, meaning 41.4% remains. Thus, \( N(t) = 0.414 N_0 \). Set up the equation: \[ 0.414 = \left(\frac{1}{2}\right)^{\frac{t}{5750}} \] Take the natural logarithm of both sides to solve for \( t \): \[ \ln(0.414) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{5750}}\right) \] Using the power rule for logarithms: \[ \ln(0.414) = \frac{t}{5750} \ln\left(\frac{1}{2}\right) \] Solve for \( t \): \[ t = \frac{\ln(0.414)}{\ln\left(\frac{1}{2}\right)} \times 5750 \] Calculate: \[ t \approx \frac{-0.882}{-0.693} \times 5750 \approx 9539.57 \] Rounding to the nearest integer, the bones were approximately 9540 years old.