 # 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]).

11 years ago

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)]

10 years ago

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

All the best Rajan !!!

Regards,