The isotope of plutonium $^{238} Pu$ is used to make thermoelectric power sources for spacecraft. Suppose that a space probe was launched in 2012 with $3.5 kg$ of $^{238} Pu$ .
Part: $0 / 2$
Part 1 of 2
(a) If the half-life of $^{238} Pu$ is $87.7 yr$ , 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.
$Q (t) = ◻ [\begin{array}{ccc} ◻^{◻} & e \\ \times & 5 \end{array}$

Answer

Expert–verified

Hide Steps

Answer

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

Steps

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 $^{238} Pu$ 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 $Q_{0} = 3.5 kg$ and the decay constant $k$ calculated in the previous step: $Q (t) = 3.5 kg \cdot e^{- 0.00790 \cdot t}$ .

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

Answer

Popular Problems