Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.