To model the amount of actinium-227 remaining after \( t \) years, we can use the formula for exponential decay:
\[ P(t) = P_0 e^{-kt} \]
where:
- \( P(t) \) is the remaining quantity of the substance at time \( t \),
- \( P_0 \) is the initial quantity of the substance,
- \( k \) is the decay constant,
- \( t \) is time.
The decay constant \( k \) can be determined from the half-life formula. The relationship between the half-life (\( t_{1/2} \)) and the decay constant (\( k \)) is given by:
\[ k = \frac{\ln(2)}{t_{1/2}} \]
For actinium-227, the half-life \( t_{1/2} \) is approximately 22 years. Therefore:
\[ k = \frac{\ln(2)}{22} \]
Calculating this:
\[ \ln(2) \approx 0.693 \] \[ k \approx \frac{0.693}{22} \approx 0.0315 \]
Rounding \( k \) to three decimal places:
\[ k \approx 0.032 \]
Thus, the correct equation that models the amount of actinium-227 remaining after \( t \) years is:
\[ P(t) = P_0 e^{-0.032t} \]
Therefore, the correct response is:
\[ P(t) = P_0 e^{-0.032t} \]