A decay process involving three radio-active nuclei X, Y and Z is as shown. X ⟶ 3 Y + Particles + energy. Y ⟶ 2 Z + particles + energy. Initially number of 'X' was ' N 0 ' and &#

Question

A decay process involving three radio-active nuclei X, Y and Z is as shown.

$X ⟶ 3 Y + Particles + energy.$

$Y ⟶ 2 Z + particles + energy.$

Initially number of 'X' was ' $N_{0}$ ' and 'Y' and 'Z' are not present initially. If no. of nuclei of 'Y' become maximum at $t = \frac{l n ( 10 )}{λ} sec$ , find number of 'Y' at this instant. (Decay constant of 'X' is ' $2 λ$ ' and that of 'Y' is ' $λ$ ')

TestHub · Accepted Answer

dt dN x ​ ​ = − ( 2 λ ) N x ​ ⇒ N x ​ = N 0 ​ e − 2 λ t dt dN Y ​ ​ = + 3 ( 2 λ ) N x ​ − λ N Y ​ " N Y ​ " is max at t = λ l n 10 ​ : 0 = 6 λ N x ​ − λ N Y m a x ​ ⇒ N Y m a x ​ = 6 N x ​ ⇒ N Y m a x ​ = 6 N 0 ​ e − 2 λ ( λ l n 10 ​ ) = 6 N 0 ​ e − 2 l n 10 = 100 6 N 0 ​ ​ = 0.06 N 0 ​ hence (B).

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