To solve this problem, let's break it down into the steps required for parts (a) and (b).
### Part (a): Formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex]
We need a formula that describes how the amount [tex]\( y \)[/tex] of the radioactive substance decays over time [tex]\( t \)[/tex]. Exponential decay can be modeled using the equation:
[tex]\[ y = y_0 \cdot e^{-kt} \][/tex]
Where:
- [tex]\( y_0 \)[/tex] is the initial amount of the substance,
- [tex]\( k \)[/tex] is the decay constant,
- [tex]\( t \)[/tex] is the time,
- [tex]\( y \)[/tex] is the amount remaining at time [tex]\( t \)[/tex].
Given:
- The initial amount [tex]\( y_0 = 938.9 \)[/tex] grams
- The half-life of the substance is 19 days
First, we need to find the decay constant [tex]\( k \)[/tex]. The decay constant is linked to the half-life via the formula:
[tex]\[ k = \frac{\ln(2)}{\text{half-life}} \][/tex]
Substitute the half-life into the equation:
[tex]\[ k = \frac{\ln(2)}{19} \][/tex]
Hence, the decay constant [tex]\( k \)[/tex] is:
[tex]\[ k = \frac{\ln(2)}{19} \approx 0.03648143055578659 \][/tex]
Now we can write the complete formula:
[tex]\[ y = 938.9 \cdot e^{-kt} \][/tex]
Substituting [tex]\( k \)[/tex]:
[tex]\[ y = 938.9 \cdot e^{-\left(\frac{\ln(2)}{19}\right)t} \][/tex]
So the formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 938.9 \cdot e^{-\left(\frac{\ln(2)}{19}\right)t} \][/tex]
### Part (b): Amount present in 13 days
We need to find the amount of the substance remaining after 13 days. Using the formula:
[tex]\[ y = 938.9 \cdot e^{-\left(\frac{\ln(2)}{19}\right)t} \][/tex]
Substitute [tex]\( t = 13 \)[/tex] days:
[tex]\[ y = 938.9 \cdot e^{-\left(\frac{\ln(2)}{19}\right) \cdot 13} \][/tex]
First, calculate the exponent part:
[tex]\[ -\left(\frac{\ln(2)}{19}\right) \cdot 13 \approx -0.03648143055578659 \cdot 13 \approx -0.4742585972312266 \][/tex]
Now, compute the amount [tex]\( y \)[/tex]:
[tex]\[ y = 938.9 \cdot e^{-0.4742585972312266} \][/tex]
[tex]\[ y \approx 938.9 \cdot 0.6223 \approx 584.3 \][/tex]
So, the amount remaining after 13 days is approximately 584.3 grams, rounded to the nearest tenth.
Thus, the answers are:
(a) The formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 938.9 \cdot e^{-\left(\frac{\ln(2)}{19}\right)t} \][/tex]
(b) The amount present after 13 days is:
[tex]\[ 584.3 \][/tex] grams.
