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The activity of a radioactive sample is decreased by 25% of the initial value after 30 days. The half - life (in days) of the sample is approximately
- a)38
- b)45
- c)59
- d)69
Correct answer is option 'D'. Can you explain this answer?
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The Problem Statement
A radioactive sample decreases its activity by 25% after 30 days. We need to find the approximate half-life of the sample.
Understanding Activity Decrease
- If the initial activity is A, after 30 days, the remaining activity is:
- Remaining Activity = A - 0.25A = 0.75A
Exponential Decay Formula
- The activity of a radioactive sample decays exponentially:
- N(t) = N0 * e^(-kt), where:
- N(t) = remaining activity after time t
- N0 = initial activity
- k = decay constant
- t = time
Setting Up the Equation
- For our case, after 30 days:
- 0.75A = A * e^(-30k)
- Simplifying gives:
- 0.75 = e^(-30k)
Taking the Natural Log
- Taking the natural logarithm of both sides:
- ln(0.75) = -30k
- Therefore, the decay constant (k) can be expressed as:
- k = -ln(0.75) / 30
Finding Half-Life
- The half-life (T1/2) is given by:
- T1/2 = ln(2) / k
- Substituting the expression for k:
- T1/2 = ln(2) * 30 / -ln(0.75)
Calculating Values
- Using approximate values:
- ln(2) ≈ 0.693
- ln(0.75) ≈ -0.2877
- Thus:
- T1/2 ≈ 30 * 0.693 / 0.2877 ≈ 69 days
Conclusion
- The approximate half-life of the radioactive sample is therefore around 69 days, which aligns with option 'D'.
A radioactive sample decreases its activity by 25% after 30 days. We need to find the approximate half-life of the sample.
Understanding Activity Decrease
- If the initial activity is A, after 30 days, the remaining activity is:
- Remaining Activity = A - 0.25A = 0.75A
Exponential Decay Formula
- The activity of a radioactive sample decays exponentially:
- N(t) = N0 * e^(-kt), where:
- N(t) = remaining activity after time t
- N0 = initial activity
- k = decay constant
- t = time
Setting Up the Equation
- For our case, after 30 days:
- 0.75A = A * e^(-30k)
- Simplifying gives:
- 0.75 = e^(-30k)
Taking the Natural Log
- Taking the natural logarithm of both sides:
- ln(0.75) = -30k
- Therefore, the decay constant (k) can be expressed as:
- k = -ln(0.75) / 30
Finding Half-Life
- The half-life (T1/2) is given by:
- T1/2 = ln(2) / k
- Substituting the expression for k:
- T1/2 = ln(2) * 30 / -ln(0.75)
Calculating Values
- Using approximate values:
- ln(2) ≈ 0.693
- ln(0.75) ≈ -0.2877
- Thus:
- T1/2 ≈ 30 * 0.693 / 0.2877 ≈ 69 days
Conclusion
- The approximate half-life of the radioactive sample is therefore around 69 days, which aligns with option 'D'.
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The activity of a radioactive sample is decreased by 25% of the initial value after 30 days. The half - life (in days) of the sample is approximatelya)38b)45c)59d)69Correct answer is option 'D'. Can you explain this answer?
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