-
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
-
- 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 |