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

Answer

B

Explanation

`tau_(1//2)=20` min `rArr lambda=(ln(2))/20`
`N=N_(0)e^(-lambdat_(1))`
`N_(0)/3 =N_(0)e^(-lambdat_(2))`
Again, `2/3N_(0)=N_(0)e^(-lambdat_(2))`
`ln(2/3) =-lambdat_(2) rArr lambdat_(2)=ln(3//2)`
`|t_(2)-t_(1)|=1/lambda[ln3/2-ln(3)]`
`=20/(ln(2))[ln(3)-ln(2)-ln(3)]`
`=20/(ln(2))[-ln(2)]=20`min