Sure, let’s address each part of the problem step-by-step.
a. Find the exponential decay model for this substance.
To develop the exponential decay model, we use the formula:
[tex]\[ A(t) = A_0 \cdot e^{kt} \][/tex]
where:
- [tex]\( A(t) \)[/tex] is the amount of the substance at time [tex]\( t \)[/tex],
- [tex]\( A_0 \)[/tex] is the initial amount of the substance,
- [tex]\( k \)[/tex] is the decay constant,
- [tex]\( t \)[/tex] is time.
To find [tex]\( k \)[/tex], we use the half-life formula:
[tex]\[ T_{\text{half}} = \frac{\ln(2)}{k} \][/tex]
Given that the half-life [tex]\( T_{\text{half}} \)[/tex] is 35.1 years, we can solve for [tex]\( k \)[/tex]:
[tex]\[ 35.1 = \frac{\ln(2)}{k} \][/tex]
This simplifies to:
[tex]\[ k = \frac{\ln(2)}{35.1} \approx 0.020 \][/tex]
So, the exponential decay model with [tex]\( k \)[/tex] rounded to the nearest thousandth is:
[tex]\[ A(t) = A_0 \cdot e^{-0.020t} \][/tex]
b. How long will it take a sample of 1000 grams to decay to 900 grams?
We start with the exponential decay model:
[tex]\[ A(t) = 1000 \cdot e^{-0.020t} \][/tex]
We need to find [tex]\( t \)[/tex] when [tex]\( A(t) = 900 \)[/tex]:
[tex]\[ 900 = 1000 \cdot e^{-0.020t} \][/tex]
This simplifies to:
[tex]\[ 0.9 = e^{-0.020t} \][/tex]
Taking the natural logarithm of both sides, we have:
[tex]\[ \ln(0.9) = -0.020t \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.9)}{-0.020} \approx 5.335 \][/tex]
So, it takes approximately 5.335 years for the sample to decay from 1000 grams to 900 grams.
c. How much of the sample of 1000 grams will remain after 20 years?
Using the same exponential decay model:
[tex]\[ A(t) = 1000 \cdot e^{-0.020t} \][/tex]
We need to find [tex]\( A \)[/tex] when [tex]\( t = 20 \)[/tex]:
[tex]\[ A(20) = 1000 \cdot e^{-0.020 \cdot 20} \][/tex]
This simplifies to:
[tex]\[ A(20) \approx 1000 \cdot e^{-0.4} \][/tex]
Evaluating [tex]\( e^{-0.4} \)[/tex]:
[tex]\[ A(20) \approx 1000 \cdot 0.670 \][/tex]
[tex]\[ A(20) \approx 673.710 \][/tex]
So, after 20 years, approximately 673.71 grams of the sample will remain.
To summarize:
a. The exponential decay model is:
[tex]\[ A(t) = A_0 \cdot e^{-0.020t} \][/tex]
b. It takes approximately 5.335 years for the sample to decay from 1000 grams to 900 grams.
c. After 20 years, approximately 673.71 grams of the sample will remain.
