A radioactive material consists nuclides of 3 isotope which decay by $α$ -emission, $β$ -emission and deuteron emission respectively. The decay constants for these isotopes are $4 λ, 2 λ and λ$ respectively. At t=0, probability of getting $α$ , $β$ and deuteron from radio nucleides are equal. If the probability of $α$ - emission at t=1600 seconds in n/13, then find the value of n.

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Detailed Solution

$\begin{array}{l} α - e m i s s i o n β - e m i s s i o n D e u t e r o n \\ λ_{1} = 4 λ_{3} λ_{2} = 2 λ_{3} λ_{3} \end{array}$
Activity $= \frac{d N}{d t} = - λ N$
These activities same at to t=0
$\Rightarrow N_{1} : N_{2} : N_{3} = 1 : 2 : 3 w h e r e N_{1}, N_{2}, N_{3}$ are the number of initial nuclei of a, b and deuteron respectively.
$P_{α, 1600} = \frac{\frac{λ_{1} N_{1}}{16}}{\frac{λ_{1} N_{1}}{16} + \frac{λ_{2} N_{2}}{4} + \frac{λ_{3} N_{3}}{2}} = \frac{1}{13}$