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Sure! To determine how long it takes for a 10.0 g sample of [tex]\( \operatorname{Sr}-90 \)[/tex] to decompose to 0.56 g, using the concept of half-life, let's follow these steps:
1. Understand the Exponential Decay Formula:
The amount of a substance remaining after time [tex]\( t \)[/tex] considering its half-life is given by the formula:
[tex]\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \][/tex]
where:
- [tex]\( N(t) \)[/tex] is the final amount of the substance after time [tex]\( t \)[/tex],
- [tex]\( N_0 \)[/tex] is the initial amount of the substance,
- [tex]\( T_{1/2} \)[/tex] is the half-life of the substance,
- [tex]\( t \)[/tex] is the time elapsed.
2. Identify the Given Values:
- Initial amount, [tex]\( N_0 = 10.0 \)[/tex] grams,
- Final amount, [tex]\( N(t) = 0.56 \)[/tex] grams,
- Half-life, [tex]\( T_{1/2} = 28.1 \)[/tex] years.
3. Set Up the Equation:
We need to solve for [tex]\( t \)[/tex] in the exponential decay equation:
[tex]\[ 0.56 = 10 \left( \frac{1}{2} \right)^{\frac{t}{28.1}} \][/tex]
4. Simplify the Equation:
First, divide both sides by 10 to isolate the decay term:
[tex]\[ \frac{0.56}{10} = \left( \frac{1}{2} \right)^{\frac{t}{28.1}} \][/tex]
[tex]\[ 0.056 = \left( \frac{1}{2} \right)^{\frac{t}{28.1}} \][/tex]
5. Use Logarithms to Solve for [tex]\( t \)[/tex]:
Take the natural logarithm (ln) of both sides of the equation to help solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(0.056) = \ln \left( \left( \frac{1}{2} \right)^{\frac{t}{28.1}} \right) \][/tex]
Using the property of logarithms ([tex]\( \ln(a^b) = b \ln(a) \)[/tex]):
[tex]\[ \ln(0.056) = \frac{t}{28.1} \cdot \ln \left( \frac{1}{2} \right) \][/tex]
6. Solve for [tex]\( t \)[/tex]:
Isolate [tex]\( t \)[/tex] by dividing both sides by [tex]\( \ln \left( \frac{1}{2} \right) \)[/tex] and multiplying by 28.1:
[tex]\[ t = \frac{\ln(0.056)}{\ln \left( \frac{1}{2} \right)} \times 28.1 \][/tex]
7. Calculate the Time:
Using the calculated natural logarithm values:
[tex]\[ t \approx \frac{\ln(0.056)}{\ln(0.5)} \times 28.1 \][/tex]
Performing the calculation:
[tex]\[ t \approx 116.85 \text{ years} \][/tex]
Therefore, it will take approximately 116.85 years for a 10.0 g sample of [tex]\( \operatorname{Sr}-90 \)[/tex] to decompose to 0.56 g.
1. Understand the Exponential Decay Formula:
The amount of a substance remaining after time [tex]\( t \)[/tex] considering its half-life is given by the formula:
[tex]\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \][/tex]
where:
- [tex]\( N(t) \)[/tex] is the final amount of the substance after time [tex]\( t \)[/tex],
- [tex]\( N_0 \)[/tex] is the initial amount of the substance,
- [tex]\( T_{1/2} \)[/tex] is the half-life of the substance,
- [tex]\( t \)[/tex] is the time elapsed.
2. Identify the Given Values:
- Initial amount, [tex]\( N_0 = 10.0 \)[/tex] grams,
- Final amount, [tex]\( N(t) = 0.56 \)[/tex] grams,
- Half-life, [tex]\( T_{1/2} = 28.1 \)[/tex] years.
3. Set Up the Equation:
We need to solve for [tex]\( t \)[/tex] in the exponential decay equation:
[tex]\[ 0.56 = 10 \left( \frac{1}{2} \right)^{\frac{t}{28.1}} \][/tex]
4. Simplify the Equation:
First, divide both sides by 10 to isolate the decay term:
[tex]\[ \frac{0.56}{10} = \left( \frac{1}{2} \right)^{\frac{t}{28.1}} \][/tex]
[tex]\[ 0.056 = \left( \frac{1}{2} \right)^{\frac{t}{28.1}} \][/tex]
5. Use Logarithms to Solve for [tex]\( t \)[/tex]:
Take the natural logarithm (ln) of both sides of the equation to help solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(0.056) = \ln \left( \left( \frac{1}{2} \right)^{\frac{t}{28.1}} \right) \][/tex]
Using the property of logarithms ([tex]\( \ln(a^b) = b \ln(a) \)[/tex]):
[tex]\[ \ln(0.056) = \frac{t}{28.1} \cdot \ln \left( \frac{1}{2} \right) \][/tex]
6. Solve for [tex]\( t \)[/tex]:
Isolate [tex]\( t \)[/tex] by dividing both sides by [tex]\( \ln \left( \frac{1}{2} \right) \)[/tex] and multiplying by 28.1:
[tex]\[ t = \frac{\ln(0.056)}{\ln \left( \frac{1}{2} \right)} \times 28.1 \][/tex]
7. Calculate the Time:
Using the calculated natural logarithm values:
[tex]\[ t \approx \frac{\ln(0.056)}{\ln(0.5)} \times 28.1 \][/tex]
Performing the calculation:
[tex]\[ t \approx 116.85 \text{ years} \][/tex]
Therefore, it will take approximately 116.85 years for a 10.0 g sample of [tex]\( \operatorname{Sr}-90 \)[/tex] to decompose to 0.56 g.
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