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`        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

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'
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Rajan !!!

Regards,
CHETAN MANDAYAM NAYAKAR
```
6 years ago
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