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Nuclei of radioactive element A are prod...

Nuclei of radioactive element `A` are produced at rate `'t^(2')` (where `t` is time) at any time `t`. The element `A` has decay constant `lambda`. Let `N` be the number of nuclei of element `A` at any time `t`. At time `t=t_(0), dN//dt` is minimum. The number of nuclei of element `A` at time `t=t_(0)` is

A

`(-2t_(0)+lambdat_(0)^(2))/(lambda^(2))`

B

`(t_(0)-lambdat_(0)^(2))/(lambda^(2))`

C

`(2t_(0)+lambdat_(0)^(2))/(lambda)`

D

`(2t_(0)-lambdat_(0)^(2))/(lambda)`

Text Solution

Verified by Experts

The correct Answer is:
A

`(dN)/(dt)=t^(2)-lambdaN rArr(d^(2)N)/(dt^(2))=2t-lambda(dN)/(dt)`
For `(dN)/(dt)` to be min, `(d^(2)N)/(dt^(2))=0`
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