A radioactive sample is observed for 5 days. After 5 days the amount of the sample is found to be 0.25% of the initial amount. Estimate the half-life of the given sample.
A radioactive sample is observed for 5 days. After 5 days the amount of the sample is found to be 0.25% of the initial amount. Estimate the half-life of the given sample.
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![**Problem Statement:**
A radioactive sample is observed for 5 days. After 5 days, the amount of the sample is found to be 0.25% of the initial amount. Estimate the half-life of the given sample.
**Explanation:**
To solve this problem, we will use the formula for exponential decay:
\[ N(t) = N_0 \times (0.5)^{t/T} \]
where:
- \( N(t) \) is the amount of substance remaining after time \( t \),
- \( N_0 \) is the initial amount of substance,
- \( T \) is the half-life of the substance,
- \( t \) is the time elapsed.
Given:
- \( N(t) = 0.0025 \times N_0 \) (since 0.25% is 0.0025 of the initial),
- \( t = 5 \) days.
We need to solve for \( T \) (the half-life):
\[ 0.0025 = (0.5)^{5/T} \]
Taking the natural logarithm of both sides:
\[ \ln(0.0025) = \frac{5}{T} \ln(0.5) \]
Solving for \( T \):
\[ T = \frac{5 \ln(0.5)}{\ln(0.0025)} \]
Plug the values into a calculator to find \( T \):
\[ T \approx \frac{5 \times (-0.6931)}{-5.9915} \]
\[ T \approx 0.578 \text{ days} \]
Therefore, the estimated half-life of the given sample is approximately 0.578 days.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18a542a1-0899-4ebe-acd6-685d74e53e0e%2F43f46830-79c2-474b-9cb6-fd22442f33ac%2Fri0g71d_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A radioactive sample is observed for 5 days. After 5 days, the amount of the sample is found to be 0.25% of the initial amount. Estimate the half-life of the given sample.
**Explanation:**
To solve this problem, we will use the formula for exponential decay:
\[ N(t) = N_0 \times (0.5)^{t/T} \]
where:
- \( N(t) \) is the amount of substance remaining after time \( t \),
- \( N_0 \) is the initial amount of substance,
- \( T \) is the half-life of the substance,
- \( t \) is the time elapsed.
Given:
- \( N(t) = 0.0025 \times N_0 \) (since 0.25% is 0.0025 of the initial),
- \( t = 5 \) days.
We need to solve for \( T \) (the half-life):
\[ 0.0025 = (0.5)^{5/T} \]
Taking the natural logarithm of both sides:
\[ \ln(0.0025) = \frac{5}{T} \ln(0.5) \]
Solving for \( T \):
\[ T = \frac{5 \ln(0.5)}{\ln(0.0025)} \]
Plug the values into a calculator to find \( T \):
\[ T \approx \frac{5 \times (-0.6931)}{-5.9915} \]
\[ T \approx 0.578 \text{ days} \]
Therefore, the estimated half-life of the given sample is approximately 0.578 days.
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