A radioactive isotope, A undergoes simultaneous decay to different nuclei as : Assuming that initially neither P nor Q was present, after how many hours, amount of Q will be just double to the amount of A remaining ?

Question

A radioactive isotope, A undergoes simultaneous decay to different nuclei as : Assuming that initially neither P nor Q was present, after how many hours, amount of Q will be just double to the amount of A remaining ?, $ 6.0 \mathrm{~h} $, $ 7.0 \mathrm{~h} $, $ 8.0 \mathrm{~h} $, $ 5.0 \mathrm{~h} $

MARKS App · Accepted Answer

t 1 / 2 ( P ) = 2 t 1 / 2 ( Q ) ⇒ Q is formed at twice the rate of formation of P i.e., at any time t; [Q] = 2[P] Rate constant for decay of A(k) = k P + k Q ⇒ k = ln 2 9 + ln 2 4 . 5 = ln 2 3 h -1 Let after t-hours, [Q] = 2 [A] and [P] = x% Given, 2x = 2 (100 - 3x) ⇒ x = 2 5 ⇒ t 3 ln 2 = ln ( 1 0 0 1 0 0 - 3 x ) = ln 4 ⇒ t = 6 . 0 h

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