Answer :
To determine how much silicon-32 will be present after 300 years, given an initial amount of 30 grams with a half-life of 710 years, we use the exponential decay formula:
[tex]\[ A = A_0 e^{kt} \][/tex]
Here:
- [tex]\( A \)[/tex] is the amount remaining after time [tex]\( t \)[/tex].
- [tex]\( A_0 \)[/tex] is the initial amount, which is 30 grams.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time elapsed, which is 300 years.
- [tex]\( e \)[/tex] is the base of the natural logarithm.
### Step 1: Calculate the decay constant [tex]\( k \)[/tex]
The decay constant [tex]\( k \)[/tex] is related to the half-life [tex]\( T_{1/2} \)[/tex] by the formula:
[tex]\[ k = \frac{\ln(0.5)}{T_{1/2}} \][/tex]
Using the given half-life of 710 years:
[tex]\[ k = \frac{\ln(0.5)}{710} \][/tex]
### Step 2: Substitute [tex]\( k \)[/tex] and [tex]\( t \)[/tex]
Let’s substitute [tex]\( k \)[/tex] and [tex]\( t = 300 \)[/tex] years into our decay formula:
[tex]\[ A = 30 \times e^{k \times 300} \][/tex]
### Step 3: Calculate the exponent
[tex]\[ k \times 300 \][/tex]
Using the calculated [tex]\( k \approx -0.0009762636345914723 \)[/tex]:
### Step 4: Calculate the amount remaining
Plugging the values into the decay formula, we get:
[tex]\[ A \approx 30 \times e^{-0.0009762636345914723 \times 300} \][/tex]
[tex]\[ A \approx 30 \times e^{-0.2928789903774417} \][/tex]
[tex]\[ A \approx 30 \times 0.7461123474059929 \][/tex]
[tex]\[ A \approx 22.383370422179787 \][/tex]
### Step 5: Round the answer
After rounding to three decimal places:
[tex]\[ A \approx 22.383 \][/tex]
So, after 300 years, approximately 22.383 grams of silicon-32 will be present. The correct answer is:
[tex]\[ \boxed{22.383} \][/tex]
[tex]\[ A = A_0 e^{kt} \][/tex]
Here:
- [tex]\( A \)[/tex] is the amount remaining after time [tex]\( t \)[/tex].
- [tex]\( A_0 \)[/tex] is the initial amount, which is 30 grams.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time elapsed, which is 300 years.
- [tex]\( e \)[/tex] is the base of the natural logarithm.
### Step 1: Calculate the decay constant [tex]\( k \)[/tex]
The decay constant [tex]\( k \)[/tex] is related to the half-life [tex]\( T_{1/2} \)[/tex] by the formula:
[tex]\[ k = \frac{\ln(0.5)}{T_{1/2}} \][/tex]
Using the given half-life of 710 years:
[tex]\[ k = \frac{\ln(0.5)}{710} \][/tex]
### Step 2: Substitute [tex]\( k \)[/tex] and [tex]\( t \)[/tex]
Let’s substitute [tex]\( k \)[/tex] and [tex]\( t = 300 \)[/tex] years into our decay formula:
[tex]\[ A = 30 \times e^{k \times 300} \][/tex]
### Step 3: Calculate the exponent
[tex]\[ k \times 300 \][/tex]
Using the calculated [tex]\( k \approx -0.0009762636345914723 \)[/tex]:
### Step 4: Calculate the amount remaining
Plugging the values into the decay formula, we get:
[tex]\[ A \approx 30 \times e^{-0.0009762636345914723 \times 300} \][/tex]
[tex]\[ A \approx 30 \times e^{-0.2928789903774417} \][/tex]
[tex]\[ A \approx 30 \times 0.7461123474059929 \][/tex]
[tex]\[ A \approx 22.383370422179787 \][/tex]
### Step 5: Round the answer
After rounding to three decimal places:
[tex]\[ A \approx 22.383 \][/tex]
So, after 300 years, approximately 22.383 grams of silicon-32 will be present. The correct answer is:
[tex]\[ \boxed{22.383} \][/tex]
