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The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minuteat t = 5 minutes. The time (in minutes) at which the activity reduces to half its value: a.5 loge2 b.2log5e c.5log2e d.5 log10 5?
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The activity of a radioactive sample is measured as N0 counts per minu...
Explanation:
- Given that the activity reduces to half its value at t = 5 minutes.
- The activity at t = 0 is N0 counts per minute.
- The activity at t = 5 minutes is N0/e counts per minute.
Calculating the Half-Life:
- The decay of a radioactive sample follows an exponential decay model: A(t) = A0 * e^(-kt), where A(t) is the activity at time t, A0 is the initial activity, k is the decay constant, and e is the base of the natural logarithm.
- At t = 0, the activity is N0 counts per minute, so A(0) = N0 = A0.
- At t = 5 minutes, the activity is N0/e counts per minute, so A(5) = N0/e = N0 * e^(-5k).
- Dividing the two equations gives: N0/e = N0 * e^(-5k) / N0.
- Simplifying further, we get: 1/e = e^(-5k).
- Taking the natural logarithm of both sides gives: ln(1/e) = ln(e^(-5k)).
- ln(1/e) = -1, and ln(e^(-5k)) = -5k.
- Therefore, -1 = -5k, and k = 1/5 = loge(2).
- The half-life of the radioactive sample is given by the formula T1/2 = ln(2) / k = ln(2) / (1/5) = 5ln(2) = 5loge(2).
Answer: The time at which the activity reduces to half its value is 5 loge(2) minutes.
- Given that the activity reduces to half its value at t = 5 minutes.
- The activity at t = 0 is N0 counts per minute.
- The activity at t = 5 minutes is N0/e counts per minute.
Calculating the Half-Life:
- The decay of a radioactive sample follows an exponential decay model: A(t) = A0 * e^(-kt), where A(t) is the activity at time t, A0 is the initial activity, k is the decay constant, and e is the base of the natural logarithm.
- At t = 0, the activity is N0 counts per minute, so A(0) = N0 = A0.
- At t = 5 minutes, the activity is N0/e counts per minute, so A(5) = N0/e = N0 * e^(-5k).
- Dividing the two equations gives: N0/e = N0 * e^(-5k) / N0.
- Simplifying further, we get: 1/e = e^(-5k).
- Taking the natural logarithm of both sides gives: ln(1/e) = ln(e^(-5k)).
- ln(1/e) = -1, and ln(e^(-5k)) = -5k.
- Therefore, -1 = -5k, and k = 1/5 = loge(2).
- The half-life of the radioactive sample is given by the formula T1/2 = ln(2) / k = ln(2) / (1/5) = 5ln(2) = 5loge(2).
Answer: The time at which the activity reduces to half its value is 5 loge(2) minutes.
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The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minuteat t = 5 minutes. The time (in minutes) at which the activity reduces to half its value: a.5 loge2 b.2log5e c.5log2e d.5 log10 5?
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The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minuteat t = 5 minutes. The time (in minutes) at which the activity reduces to half its value: a.5 loge2 b.2log5e c.5log2e d.5 log10 5? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minuteat t = 5 minutes. The time (in minutes) at which the activity reduces to half its value: a.5 loge2 b.2log5e c.5log2e d.5 log10 5? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minuteat t = 5 minutes. The time (in minutes) at which the activity reduces to half its value: a.5 loge2 b.2log5e c.5log2e d.5 log10 5?.
The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minuteat t = 5 minutes. The time (in minutes) at which the activity reduces to half its value: a.5 loge2 b.2log5e c.5log2e d.5 log10 5? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minuteat t = 5 minutes. The time (in minutes) at which the activity reduces to half its value: a.5 loge2 b.2log5e c.5log2e d.5 log10 5? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minuteat t = 5 minutes. The time (in minutes) at which the activity reduces to half its value: a.5 loge2 b.2log5e c.5log2e d.5 log10 5?.
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