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_e2/5`
B. `5/(log_e2)`
C. `5log_10 2`
D. `5log_e 2`

Answer

D

Explanation

According to activity law
`R=R_0e^(-lambdat)` …(i)
where,
`R_0`=initial activity at t=0
R=activity at time t
`lambda`=decay constant
According to given problem,
`R_0=N_0` counts per minute
`R=N_0/e` counts per minute
t=5 minutes
Substituting these values in equation (i), we get `N_0/e=N_0e^(-5lambda)`
`e^(-1)=e^(-5lambda)`
`5lambda=1` or `lambda=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 equation (i), we get
`R_0/2=R_0 e^(-lambdaT_(1//2))`
`e^(lambdaT_(1//2))=2`
Taking natural logarithms on both sides of above equation, we get
`lambdaT_(1//2) = log_e2`
or `T_(1//2)=(log_e 2)/lambda=(log_e 2) /((1/5))=5 log_e 2` minutes