The activity of a radioactive sample is measures as `N_0` counts per minute at `t = 0` and `N_0//e` counts per minute at `t = 5 min`. The time (in minute) at which the activity reduces to half its value is.

Options

A. `log_(e)(2)/(5)`
B. `(5)/(log_(e)2)`
C. `5log_(10)2`
D. `5log_(e)2`

Answer

D

Explanation

Applying law of radioactivity
`R=R_(0)e^(-lamdat)` . . . (i)
where `R_(0)=`initial activity at t=0
R=activity at time t
`lamda`=decay constant.
According to given problem,
`R_(0)=N_(0)` counts per minute
`R=(N_(0))/(e)` counts per minute
t=5 min
Putting these values is Eq. (i) we get
`(N_(0))/(e)=N_(0)e^(-5lamda)`
`impliese^(-1)=e^(-5lamda)`
or, `5lamda=1` or `lamda=(1)/(5)` per minute.
At `t=T_(1//2)`, the activity
R reduces to `(R_(0))/(2)`.
where `T_(1//2)=` half-life of a radioactive sample
from eq. (i) we get
`(R_(0))/(2)=R_(0)e^(-lamdaT_(1//2))`
`e^(lamdaT_(1//2))=2`
Taking natural logarithms on both sides, we get
`lamdaT_(1//2)=log_(e)2`
or `T_(1//2)=(log_(e)2)/(lamda)=(log_(e)2)/(((1)/(5)))`
`=5log_(e)2`min.