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. `5 log _(10) 2`
D. `5 log_(e) 2`

Answer

D

Explanation

According to activity law
`R = R _(0) e ^(- lamda t) -" "(i)`
Where,
`R _(0)` initial activity at `t =0`
`R =` activity at time t
`lamda =` decay constnat
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 _(0) e ^(- 5 lamda )`
`e ^(-1) =e ^(-5 lamda)`
` e^(-1) = e ^(-5 lamda)`
` 5 lamda =1 or lamda = (1)/(5) ` per minute
At `t =T _(1//2),` the activity R reduces to `(R _(0))/(2)`
wheree `T _(1//2)=` half life of a radioactive sample
From equation (i), we get
` (R _(0))/(2) = R _(0) e^(- lamda T _(1//2))`
`e ^( lamda T _(1//2)) =2`
Taking natural logarithms on both sides of above equation, we get
`lamda T _(1//2) = log _(e) 2`
or ` T _(1//2) = (log _(e) 2 )/( lamda ) = (log _(2) 2 )/( ((1)/(5)))=5 log _(2) 2` minutes