Certainly! Let's solve each part of the given problem step by step:
### Part a: Finding the Exponential Decay Model
For a substance with exponential decay, the amount of substance remaining at time [tex]\( t \)[/tex] can be modeled by the formula:
[tex]\[ A(t) = A_0 e^{kt} \][/tex]
where:
- [tex]\( A(t) \)[/tex] is the amount remaining at time [tex]\( t \)[/tex],
- [tex]\( A_0 \)[/tex] is the initial amount,
- [tex]\( k \)[/tex] is the decay constant,
- [tex]\( t \)[/tex] is time.
The decay constant [tex]\( k \)[/tex] is related to the half-life ([tex]\( t_{1/2} \)[/tex]) by the formula:
[tex]\[ k = \frac{-\ln(2)}{t_{1/2}} \][/tex]
Given the half-life [tex]\( t_{1/2} \)[/tex] of 21.8 years, we find [tex]\( k \)[/tex] as follows:
[tex]\[ k = \frac{-\ln(2)}{21.8} \][/tex]
[tex]\[ k \approx -0.032 \][/tex] (rounded to the nearest thousandth)
Therefore, the exponential decay model is:
[tex]\[ A(t) = A_0 e^{-0.032t} \][/tex]
### Part b: Time to Decay from 400 grams to 300 grams
We need to determine the time [tex]\( t \)[/tex] it takes for the substance to decay from an initial amount [tex]\( A_0 = 400 \)[/tex] grams to a target amount of 300 grams.
Using the exponential decay model:
[tex]\[ A(t) = 400 e^{-0.032t} \][/tex]
We set [tex]\( A(t) = 300 \)[/tex] and solve for [tex]\( t \)[/tex]:
[tex]\[ 300 = 400 e^{-0.032t} \][/tex]
First, divide both sides by 400:
[tex]\[ \frac{300}{400} = e^{-0.032t} \][/tex]
[tex]\[ 0.75 = e^{-0.032t} \][/tex]
Next, take the natural logarithm of both sides:
[tex]\[ \ln(0.75) = -0.032t \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.75)}{-0.032} \][/tex]
[tex]\[ t \approx 9.048 \][/tex] years (rounded to the nearest thousandth)
### Part c: Amount Remaining After 10 Years
We need to calculate the remaining amount of the substance after [tex]\( t = 10 \)[/tex] years.
Using the exponential decay model:
[tex]\[ A(t) = 400 e^{-0.032 \times 10} \][/tex]
Compute:
[tex]\[ A(10) \approx 400 \times e^{-0.32} \][/tex]
[tex]\[ A(10) \approx 400 \times 0.727 \][/tex] (using the approximate value of [tex]\( e^{-0.32} \)[/tex])
[tex]\[ A(10) \approx 291.054 \][/tex] grams (rounded to the nearest thousandth)
### Summary
a. The exponential decay model for the substance is:
[tex]\[ A(t) = A_0 e^{-0.032t} \][/tex]
b. It will take approximately [tex]\( 9.048 \)[/tex] years for the sample to decay from 400 grams to 300 grams.
c. After 10 years, approximately [tex]\( 291.054 \)[/tex] grams of the sample will remain.
