A radioactive element A of decay constant `lamda_(A)` decays into another radioactive element B of decay constant `lamda_(B)`. Initially the number of

Monjur Alahi

A radioactive element A of decay constant `lamda_(A)` decays into another radioactive element B of decay constant `lamda_(B)`. Initially the number of active nuclei of A was `N_(0)` and B was absent in the sample. The maximum number of active nuclei of B is found at t=2. In `2//lamda_(A)`. The maximum number of active nuclei of B is
A. `(N_(0))/(4)`
B. `(lamda_(A))/(lamda_(B))N_(0)e^(-lamda_(B)t`
C. `(lamda_(A))/(lamda_(B)) (N_(0))/(4)`
D. None of these


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Salim Ahmed Kawsar

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Correct Answer - C
We have, `lamda_(B)N_(B)=lamda_(A)N_(A)`
`therefore" "N_(B)=(lamda_(A))/(lamda_(B)).N_(A)`
The give time is equivalent to two half-lives of A. Hence,
`N_(A)=(N_(0))/(4)" "(thereforeN=N_(0)((1)/(2))^(2)`
`N_(B)=(lamda_(A))/(lamda_(B))((N_(0))/(4))`

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