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

Answer

A

Explanation

`:' N/N_(0)=[1/2]^(1//T) :. 1/3 =[1/2]^(1_(1)/T)` & `2/3=[1/2]^(1/T)`
`=1/2=[1/2]^((1_(2)-1_(1))1/T)implies 1=((1_(2)-t_(1)))/Timplies T=t_(2)-t_(1)`
`implies t_(2)-t_(1)=20` min