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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
 
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- loge 2/5
 - 5/loge 2
 - 5 log 102
 - 5 loge 2
 
 
Correct Option: D
N = N0.e- λt
Here, t = 5 minutes
| = N0.e- 5λ | ||
| e | 
⇒ 5λ = 1,
| λ = | ||
| 5 | 
| Now, T1/2 = | = 5 ln2 | |
| λ |