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rajan jha Grade: 12

2 sample of diferent element (half lives t1 & t2) have same no. of nuclei at t=0. The time after which the activity are same is ? (ans:-{t1t2/0.693(t2-t1)}*ln[t2/t1]).

7 years ago

Answers : (2)

vikas askiitian expert
510 Points

let two be samples be X & Y & their half lives  t1 , t2 respectively...

activity(A) = N(lamda)                       (N is number of nuclie at any time)

(lamda is disintegration constant)

N = Noe-(lamda)t        .............1            (number of nuclie at any time)

No is the number of nuclie at t=0 .

activity of X after time t, initially both are having same number of nuclie...

 Ax = Noe-(lamda)xt(lamda)x   ....................2

 activity of Y after time t

 Ay = Noe-(lamda)yt (lamda)y .......................3

since both are having same activity so equating 2 & 3

 (lamda)x/(lamda)y = e t((lamda)x e-t(lamda)y

now using lamda = 0.693/t1/2 

 taking log both sides

  ln[t2/t1] = 0.693(1/t1 - 1/t2)t

 t =(t2t1)ln[t2/t1] / [0.693(t2-t1)]



7 years ago
Chetan Mandayam Nayakar
312 Points

Dear Rajan,

let k1 and k2 be the two decay constants.

k1=(ln2)/t1,and k2=(ln2)/t2, Let number of nuclei at t=0 be N

Activity at time 't'= kNe-kt

((ln2)/t1)Ne-(ln2)(t/t1)=((ln2)/t2)Ne-(ln2)(t/t2), solve for 't'

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6 years ago
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