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.


Solution:

(i) 

  • 41

by solving that we get t=57.95


  • 8

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

  • -12

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 ,    eln2/30 *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..

  • -5

T1/2  = 0.693 / k  ( k is decay  constant)

30 =  0.693 / k

therefore, k = 30 / 0.693 = 43 .29

2) After  second half life ,  sample undecayed = 1/4  

or,  sample decayed  = 3/4

hence , time taken by the sample  to  decay  3/4  of its  initial value = 30* 2 = 60 seconds. 

  • 2
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