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  1. 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. loge 2/5
    2. 5/loge 2
    3. 5 log 102
    4. 5 loge 2
Correct Option: D

N = N0.e- λt
Here, t = 5 minutes

N0
= N0.e- 5λ
e

⇒ 5λ = 1,
λ =
1
5

Now, T1/2 =
ln2
= 5 ln2
λ



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