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`rd of its has decayed and time `t_1` when `1/3`rd of it had decayed is -

(A) 14 min
(B) 20 min
(C) 28 min
(D) 7 min

Options

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

Answer

B

Explanation

Given , half -life , `T="In2"/lambda`=20min
We know that , `N=N_0e^(-lambdat)`
`therefore (1-2/3)N_0=N_0e^(-lambdat_2)`
or,`1/3N_0=N_0e^(-lambdat_2)`….(1)
Again, `(1-1/3)N_0=N_0e^(-lambdat_1)`
or , `2/3N_0=N_0e^(-lambdat_1)`…(2)
Diviiding equation (1) by equation (2) , we get
`1/2=e^(-lambda(t_2-t_1))`
`therefore t_2-t_1="In2"/lambda`=20 min