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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 activity of a radioactive sample is decreased by 25% of the initia...
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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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? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about 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? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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