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

Options

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

Answer

C

Explanation

`N = N_(0)e^(-lambdat)`
`(2)/(3)N_(0) = N_(0)e^(-lambdat_(1)) rArr e^(-lambdat_(1)) = (2)/(3) ….(i)`
`(1)/(3)N_(0) = N_(0)e^(-lambdat_(2)) rArr e^(-lambdat_(2)) = (1)/(3) …(ii)`
`(i)//(ii) rArr e^(lambda(t_(2)-t_(1)) = 2`
Taking log: `lambda(t_(2)-t_(1)) = ln2`
`t_(2)-t_(1) = (In2)/(lambda) = T_(1//2) = 20min`