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Nucleus A decays into B with decay const...

Nucleus A decays into B with decay constant `lambda_(1)` and B decays into C with decay constant `lambda_(2)` Initially at t=0 number of nuclei of A and B are `2N_(0)` and `N_(0)` respectively . At `t=t_(0)`, number of nuclei of B become constant If at this instant number of buyclei of B are `(3N_(0))/(2)`

A

`(1)/(lamda_(1))ln[(4lamda_(1))/(3lamda_(2))]`

B

`(1)/(lamda_(2))ln[(4lamda_(1))/(3lamda_(2))]`

C

`(lamda_(1)+lamda_(2))ln[(4lamda_(1))/(3lamda_(2))]`

D

`(1)/(lamda_(1))ln[(4lamda_(1))/(3lamda_(2))]`

Text Solution

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The correct Answer is:
A
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