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Sagot :
Certainly! Let's walk through the given problem step by step.
1. Understanding the Problem:
- We have a radioactive substance, Fermium-257, with an initial amount of [tex]\( A \)[/tex].
- The half-life of Fermium-257 is given as 100 days.
- We need to determine the age of the sample when it has decayed to 60% of its original amount.
2. Using the Decay Formula:
- The formula for the remaining amount of a substance after [tex]\( t \)[/tex] days is given by:
[tex]\[ P(t) = A \left( \frac{1}{2} \right)^{\frac{t}{h}} \][/tex]
where [tex]\( P(t) \)[/tex] is the amount remaining after [tex]\( t \)[/tex] days, [tex]\( A \)[/tex] is the initial amount, and [tex]\( h \)[/tex] is the half-life.
3. Given Information:
- [tex]\( P(t) = 0.60A \)[/tex] (since 60% of the original amount remains)
- [tex]\( h = 100 \)[/tex] days
4. Setting Up the Equation:
- We set the decay formula to match the given condition:
[tex]\[ 0.60A = A \left( \frac{1}{2} \right)^{\frac{t}{100}} \][/tex]
5. Simplifying the Equation:
- We can divide both sides of the equation by [tex]\( A \)[/tex] (assuming [tex]\( A \neq 0 \)[/tex]):
[tex]\[ 0.60 = \left( \frac{1}{2} \right)^{\frac{t}{100}} \][/tex]
6. Solving for [tex]\( t \)[/tex]:
- First, we take the natural logarithm of both sides to help isolate [tex]\( t \)[/tex]:
[tex]\[ \ln(0.60) = \ln\left( \left( \frac{1}{2} \right)^{\frac{t}{100}} \right) \][/tex]
- By using the logarithm power rule, [tex]\( \ln\left( a^b \right) = b \ln(a) \)[/tex], this simplifies to:
[tex]\[ \ln(0.60) = \frac{t}{100} \ln\left( \frac{1}{2} \right) \][/tex]
[tex]\[ \ln(0.60) = \frac{t}{100} \ln(0.5) \][/tex]
7. Isolating [tex]\( t \)[/tex]:
- We then solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.60)}{\ln(0.5)} \times 100 \][/tex]
8. Calculating [tex]\( t \)[/tex]:
- Using the values [tex]\( \ln(0.60) \approx -0.5108 \)[/tex] and [tex]\( \ln(0.5) \approx -0.6931 \)[/tex]:
[tex]\[ t = \frac{-0.5108}{-0.6931} \times 100 \][/tex]
[tex]\[ t \approx 0.737 \times 100 \][/tex]
[tex]\[ t \approx 73.7 \text{ days} \][/tex]
9. Finding the Nearest Given Option:
- The closest given option to 73.7 days is 74 days.
Therefore, the sample is approximately 74 days old.
1. Understanding the Problem:
- We have a radioactive substance, Fermium-257, with an initial amount of [tex]\( A \)[/tex].
- The half-life of Fermium-257 is given as 100 days.
- We need to determine the age of the sample when it has decayed to 60% of its original amount.
2. Using the Decay Formula:
- The formula for the remaining amount of a substance after [tex]\( t \)[/tex] days is given by:
[tex]\[ P(t) = A \left( \frac{1}{2} \right)^{\frac{t}{h}} \][/tex]
where [tex]\( P(t) \)[/tex] is the amount remaining after [tex]\( t \)[/tex] days, [tex]\( A \)[/tex] is the initial amount, and [tex]\( h \)[/tex] is the half-life.
3. Given Information:
- [tex]\( P(t) = 0.60A \)[/tex] (since 60% of the original amount remains)
- [tex]\( h = 100 \)[/tex] days
4. Setting Up the Equation:
- We set the decay formula to match the given condition:
[tex]\[ 0.60A = A \left( \frac{1}{2} \right)^{\frac{t}{100}} \][/tex]
5. Simplifying the Equation:
- We can divide both sides of the equation by [tex]\( A \)[/tex] (assuming [tex]\( A \neq 0 \)[/tex]):
[tex]\[ 0.60 = \left( \frac{1}{2} \right)^{\frac{t}{100}} \][/tex]
6. Solving for [tex]\( t \)[/tex]:
- First, we take the natural logarithm of both sides to help isolate [tex]\( t \)[/tex]:
[tex]\[ \ln(0.60) = \ln\left( \left( \frac{1}{2} \right)^{\frac{t}{100}} \right) \][/tex]
- By using the logarithm power rule, [tex]\( \ln\left( a^b \right) = b \ln(a) \)[/tex], this simplifies to:
[tex]\[ \ln(0.60) = \frac{t}{100} \ln\left( \frac{1}{2} \right) \][/tex]
[tex]\[ \ln(0.60) = \frac{t}{100} \ln(0.5) \][/tex]
7. Isolating [tex]\( t \)[/tex]:
- We then solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.60)}{\ln(0.5)} \times 100 \][/tex]
8. Calculating [tex]\( t \)[/tex]:
- Using the values [tex]\( \ln(0.60) \approx -0.5108 \)[/tex] and [tex]\( \ln(0.5) \approx -0.6931 \)[/tex]:
[tex]\[ t = \frac{-0.5108}{-0.6931} \times 100 \][/tex]
[tex]\[ t \approx 0.737 \times 100 \][/tex]
[tex]\[ t \approx 73.7 \text{ days} \][/tex]
9. Finding the Nearest Given Option:
- The closest given option to 73.7 days is 74 days.
Therefore, the sample is approximately 74 days old.
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