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 = N_oe^-(lamda)t        .............1            (number of nuclie at any time)

N_o is the number of nuclie at t=0 .

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

 A_x = N_oe^-(lamda)_xt(lamda)_x   ....................2

 activity of Y after time t

 A_y = N_oe^-(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/t_1/2 

 taking log both sides

  ln[t₂/t_1] = 0.693(1/t1 - 1/t₂)t

 t =(t₂t₁)ln[t₂/t_1] / [0.693(t₂-t₁)]

Approved

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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