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

Fraction remains after n half-lives
`(N)/(N_(0))=((1)/(2))^(n)=((1)/(2))^((t)/(T))`
Given , `N=(N_(0))/(e )rArr(N_(0))/(eN_(0))=((1)/(2))^((5)/(T))`
`rArr(1)/(e )=((1)/(2))^((5)/(T))`
Taking log on both sides, we get
`log1-loge=(5)/(T)"log"(1)/(2)`
`rArr-1=(5)/(T)(-log2)`
`rArrT=5log_(e )2`
Now , let t. be the time after which activity reduces to half
`((1)/(2))=((1)/(2))^(t//5log_(e)2)` ,br> `rArrt.5log_(e)2`