The isotope of plutonium ${ }^{238} \mathrm{Pu}$ is used to make thermoelectric power sources for spacecraft. Suppose that a space probe was launched in 2012 with $3.5 \mathrm{~kg}$ of ${ }^{238} \mathrm{Pu}$. Part: $0 / 2$ Part 1 of 2 (a) If the half-life of ${ }^{238} \mathrm{Pu}$ is $87.7 \mathrm{yr}$, write a function of the form $Q(t)=Q_{0} e^{-k t}$ to model the quantity $Q(t)$ of ${ }^{238}$ Pu left after $t$ years. Round the value of $k$ to five decimal places. Do not round intermediate calculations. \[ Q(t)=\square\left[\begin{array}{ccc} \square^{\square} & e & \\ \times & 5 \end{array}\right. \]

Step 1 ：Calculate the decay constant \( k \) using the formula \( k = \frac{\ln(2)}{T_{1/2}} \) where \( T_{1/2} \) is the half-life of the substance.

Step 2 ：Given the half-life \( T_{1/2} \) of \( ^{238} \mathrm{Pu} \) is 87.7 years, substitute this value into the formula to find \( k \): \( k = \frac{\ln(2)}{87.7} \approx 0.00790 \) (rounded to five decimal places).

Step 3 ：Write the exponential decay function \( Q(t) \) using the initial quantity \( Q_0 = 3.5 \mathrm{~kg} \) and the decay constant \( k \) calculated in the previous step: \( Q(t) = 3.5 \mathrm{~kg} \cdot e^{-0.00790 \cdot t} \).

Step 4 ：The final function that models the quantity of \( ^{238} \mathrm{Pu} \) left after \( t \) years is: \(\boxed{Q(t) = 3.5 e^{-0.00790 t}}\)