91Ó°ÊÓ

Iodine 131 is a radioactive material that decays according to the function$A\left(t\right)=A_0{e}^{-0.087t}$

The given function is $A\left(t\right)=A_0{e}^{-0.087t}$

The amount A of radioactive material present at time t is represented by equation $A\left(t\right)=A_0{e}^{kt}$

Compare the given function with the equation.

$$\begin{gathered} A\left(t\right)=A_0{e}^{-0.087t} \\ A\left(t\right)=A_0{e}^{kt} \\ k=-0.087 \\ k=-8.7\% \end{gathered}$$

The decay rate of iodine 131 is -8.7% per year.

The graph of the function $A\left(t\right)=A_0{e}^{-0.087t}$is

Where A is the fraction of iodine 131 present and x is the time.

Substitute $A=100,t=9$, in the given function role="math" localid="1647272236622" $A\left(t\right)=A_0{e}^{-0.087t}$

$$\begin{gathered} A\left(t\right)=A_0{e}^{-0.087t} \\ A\left(9\right)=A_0{e}^{-0.087×9} \\ A\left(9\right)=45.70grams \end{gathered}$$

After 9 days, there will be about 45.7 grams of Iodine 131 remaining.

Substitute $A\left(t\right)=70,A_0=100$in the given function role="math" localid="1647272778218" $A\left(t\right)=A_0{e}^{-0.087t}$

role="math" localid="1647272624312" $$\begin{gathered} A\left(t\right)=A_0{e}^{-0.087t} \\ 70=100e^{-0.087t} \\ 0.7=e^{-0.087t} \\ \ln \left(0.7\right)=\ln \left(e^{-0.087t}\right) \\ \ln \left(0.7\right)=-0.087t \\ t=\frac{\ln \left(0.7\right)}{-0.087} \\ t≈4.10 \end{gathered}$$

After 4.10 days, there will be 70 grams of Iodine 131 left.

The half-life of iodine 131 will be the half-life of the given function.

Half-life of function $A=\frac{1}{2}{A}_0$

Substitute $A=\frac{1}{2}{A}_0$in the given function

$$\begin{gathered} A\left(t\right)=A_0{e}^{-0.087t} \\ \frac{1}{2}{A}_0=A_0{e}^{-0.087t} \\ \ln \left(0.5\right)=\ln \left(e^{-0.087t}\right) \\ \ln \left(0.5\right)=-0.087t \\ t=\frac{\ln \left(0.5\right)}{-0.087} \\ t≈7.97 \end{gathered}$$

The half-life of Iodine 131 is about 7.97 days.