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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]).
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)]
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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CHETAN MANDAYAM NAYAKAR
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