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. `14 mi n`
B. `20 mi n`
C. `28 mi n`
D. `7mi n`

Answer

B

Explanation

Here, `T=20 "minutes"`
In time `t_(1)`, the radioactive sample left behind
`=1-1/3=2/3`.
In time `t_(2)`, the radioactive sample left behind
`=1-1/3=2/3`.
As `n/(N_(0))=(1/2)^(t//T) :. 2/3(1/2)^(t_(1)//T).........(i)`
and `1/3=(1/2)^(t_(2)//T)`
Dividing (ii) by (i), we get
`1/2=(1/2)^((t_(2)-t_(1))//T)` or `t_(2)-t_(1)=T=20min`.