Actinium- 227 has a half-life of approximately 22 years. Which of the following equations correctly models the amount of actinium- 227 remaining after \( t \) years? Round \( k \) to three decimal places. (1 point) \( P(t)=P_{0} e^{0.032 t} \) \( P(t)=P_{0} e^{-0.693 t} \) \( P(t)=P_{0} e^{-0.032 t} \) \( P(t)=P_{0} e^{0.693 t} \)

To find the correct model for the amount of Actinium-227 remaining after \( t \) years, we can use the formula based on exponential decay, which is given by \( P(t) = P_0 e^{-kt} \). The decay constant \( k \) can be calculated using the half-life formula: \( k = \frac{0.693}{t_{1/2}} \). Since the half-life of Actinium-227 is 22 years, we have: \[ k = \frac{0.693}{22} \approx 0.0315 \] Rounding this to three decimal places gives \( k \approx 0.032 \). Thus, the equation that correctly models the amount of Actinium-227 remaining is: \( P(t) = P_0 e^{-0.032 t} \) --- Did you know that Actinium was discovered back in 1899, making it one of the first non-primordial radioactive elements to be identified? Named after the Greek word "aktinos," meaning "ray" or "beam," it was first separated from uranium ores by the chemist Friedrich Oskar Giesel. Its discovery marked a pivotal moment in nuclear chemistry, leading to further exploration of radioactive materials and their applications. In practical scenarios, understanding the half-life of radioactive elements is vital, especially in fields like medicine for radioactive therapies or in environmental science for nuclear waste management. Miscalculating decay rates can lead to dangerous oversights, such as improper storage or disposal of nuclear materials. Always double-check your calculations and use appropriate safety measures!