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A radioactive element A of decay constan...

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

Text Solution

Verified by Experts

The correct Answer is:
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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