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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 minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

1. log₂ 2/5
2. log₂ 5
3. 5 log₂ 2
4. 5 log₂ e

Correct answer: 5 log₂ 2

Solution

The activity of a radioactive sample decreases exponentially with time. Given that the activity reduces to N₀/e in 5 minutes, the decay constant λ can be calculated as λ = 1/5. The half-life is related to λ by the formula T₁/₂ = ln(2)/λ. Substituting λ = 1/5, we get T₁/₂ = 5 ln(2), which is equivalent to 5 log₂ 2.

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