The half –life of a radioactive substance is 30 s. Calculate:

(i ) the decay constant, and

(ii) Time taken by the sample to decay 3/4 of its initial value.

by solving that we get t=57.95

t=.063/kfrom this k=.063/t by substituing this value we get k=.023

2)3/4v=ve^-kt cancelling bothsides we get 3/4=e^-kt by solving this weget t=57.95

1. t1/2 =ln2/lamda , 

30 = ln2/lamda , so lamda=ln2/30

2. N=N0e^{-lamda * t}

3N0/4 =N0e-ln2/30 * t    ,  3/4=e^{-ln2/30 *t} ,    e^{ln2/30 *t} = 4/3

take log both sides , ln2/30*t =ln(4/3) , so t=30(ln4-ln3) /ln2

now u can put values of ln2 & ln3..