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The activity of a radioavtive sample is ...

The activity of a radioavtive sample is measured as `N_(0)` counts per minute at `t=0and (N_(0))/(e)` counts per minute at t=5 min. The time (in minutes) at which the activity reduces to half its value is

A

`log _(e) 2//5`

B

`(5)/(log_(e)2)`

C

`5 log _(10)2`

D

`5 log _(e) 2`

Text Solution

Verified by Experts

The correct Answer is:
D

`N=N_(0)e^(-lamdat) implies (N_(0))/(e )=N_(0)e^(-lamda(5))implies lamda =(1)/(5)`
`Now (N_(0))/(2)=N_(0)e^(-lamda(t)) implies t=(1)/(lamda)ln2=5ln2`
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