Certainly! Let's solve the problem using the exponential decay model:
[tex]\[ A = A_0 e^{kt} \][/tex]
where:
- [tex]\(A\)[/tex] is the amount of substance remaining after 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 the time
### Step-by-Step Solution:
#### Step 1: Understand the given information
- Half-life ([tex]\(t_{1/2}\)[/tex]): The time it takes for half of the substance to decay, which is given as 30 years.
- Decay to 94% of its original amount: This means [tex]\(A = 0.94 A_0\)[/tex].
#### Step 2: Relate the half-life to the decay constant [tex]\(k\)[/tex]
The decay constant [tex]\(k\)[/tex] is related to the half-life by the formula:
[tex]\[ k = \frac{\ln(0.5)}{t_{1/2}} \][/tex]
Substitute the given half-life (30 years):
[tex]\[ k = \frac{\ln(0.5)}{30} \][/tex]
#### Step 3: Calculate the value of [tex]\(k\)[/tex]
[tex]\[ \ln(0.5) \approx -0.693 \][/tex]
Thus:
[tex]\[ k = \frac{-0.693}{30} \][/tex]
[tex]\[ k \approx -0.023104906018664842 \][/tex]
#### Step 4: Set up the exponential decay equation
We know the substance decays to 94% of its original amount:
[tex]\[ A = 0.94 A_0 \][/tex]
Substitute [tex]\(A\)[/tex] and [tex]\(k\)[/tex] into the exponential decay equation:
[tex]\[ 0.94 A_0 = A_0 e^{(-0.023104906018664842) t} \][/tex]
#### Step 5: Solve for time [tex]\(t\)[/tex]
Divide both sides by [tex]\(A_0\)[/tex]:
[tex]\[ 0.94 = e^{(-0.023104906018664842) t} \][/tex]
Take the natural logarithm of both sides to solve for [tex]\(t\)[/tex]:
[tex]\[ \ln(0.94) = -0.023104906018664842 t \][/tex]
[tex]\[
t = \frac{\ln(0.94)}{-0.023104906018664842}
\][/tex]
Calculate [tex]\(\ln(0.94)\)[/tex]:
[tex]\[
\ln(0.94) \approx -0.0618754
\][/tex]
Thus:
[tex]\[
t = \frac{-0.0618754}{-0.023104906018664842}
\][/tex]
[tex]\[
t \approx 2.6780201429126222
\][/tex]
### Answer:
It will take approximately 2.68 years for a sample of the substance to decay to 94% of its original amount.
