Respuesta :
Answer:
[tex]A=50e^{-0.035t}[/tex]There will be 17.5 grams of carbon-11 left after 30 minutes.
The exponential decay equation is noted by:
[tex]A=A_0e^{kt}[/tex]Where:
A = amount present over time
A₀ = initial amount
k = rate of decay
t = time in minutes
Let us first find the value of k using the given that we have
From the problem, we have:
A = 1/2 gram
A₀ = 1 gram
k = ?
t = 20 minutes
Substituting these values and we will get:
[tex]\begin{gathered} A=A_0e^{kt} \\ \frac{1}{2}=1e^{k(20)} \\ \frac{1}{2}=1e^{k(20)} \\ k=-0.035 \end{gathered}[/tex]Therefore, our model is going to be:
[tex]A=50e^{-0.035t}[/tex]To find how much sample is still radioactive after 30 minutes, we just have to substitute t = 30 to the equation:
[tex]\begin{gathered} A=50e^{-0.035t} \\ A=50e^{-0.035(30)} \\ A=17.49688\approx17.5 \end{gathered}[/tex]Therefore, there are 17.5 grams of carbon-11 left after 30 minutes.
