The half-life of a radioactive substance is 20 min. The approximate time interval (`t-(2)-t_(1)` between the time `t_(2)`, when `2/3` of it has decayed and time `t_(1)` when `1/3` of it had decayed is

Options

A. `7 min`
B. `14 min`
C. `20 min`
D. `28 min`

Answer

C

Explanation

`(2)/(3)N_(0)=N_(0)e^(-lambdat_(1))`
`(1)/(3)N_(0)=N_(0)e^(-lambdat_(2))`
`2=e^(lambda(t_(2)-t_(1)))`
`lambda(t_(2)-t_(1))= l n 2`
`(t_(2)-t_(1))= ( l n 2)/(lambda)=20 "min"`