Question

the isotope of plutonium $^{238}\text{pu}$ is used to make thermoelectric power sources for spacecraft. suppose that a space probe was launched in 2012 with 3.0 kg of $^{238}\text{pu}$. part: 0 / 2 part 1 of 2 (a) if the half - life of $^{238}\text{pu}$ is 87.7 yr, write a function of the form $q(t)=q_0e^{-kt}$ to model the quantity $q(t)$ of $^{238}\text{pu}$ left after $t$ years. round the value of $k$ to five decimal places. do not round intermediate calculations. $q(t)=square$

Explanation:

Step1: Recall the half - life formula

For the exponential decay model \(Q(t)=Q_0e^{-kt}\), when \(t = T\) (half - life), \(Q(T)=\frac{Q_0}{2}\). So, \(\frac{Q_0}{2}=Q_0e^{-kT}\).

Step2: Solve for \(k\)

Divide both sides of \(\frac{Q_0}{2}=Q_0e^{-kT}\) by \(Q_0\) (since \(Q_0
eq0\)), we get \(\frac{1}{2}=e^{-kT}\).
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=-kT\).
We know that \(T = 87.7\) (half - life of \(^{238}\text{Pu}\)), and \(\ln(\frac{1}{2})=-\ln(2)\).
So, \(k=\frac{\ln(2)}{T}\).

Step3: Calculate the value of \(k\)

Substitute \(T = 87.7\) into \(k=\frac{\ln(2)}{T}\).
\(k=\frac{\ln(2)}{87.7}\approx\frac{0.693147}{87.7}\approx0.00792\) (rounded to five decimal places).
Since \(Q_0 = 3\) (initial quantity of \(^{238}\text{Pu}\) is \(3.0\) kg), the function is \(Q(t)=3e^{-0.00792t}\).

Answer:

\(Q(t) = 3e^{-0.00792t}\)