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A radioactive substance (A) is not produced at a constant rate which decays with a decay constant `lambda` to form stable substance (B).If production of A starts at t=0 . Find (i)The number of nuclei of A and (ii)Number of nuclei at any time t. |
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Answer» (i)Let N be the number of nuclei of A at any time t `"dN"/"dt"=alpha-lambdaN` `int_0^N"dN"/(alpha-lambdaN)=int_0^t"dt"` On solving , we will get `N=alpha/lambda(1-e^(-lambdat))` (ii) Number of nuclei of B at any time t `N_B=alphat-N_A` `=alphat-alpha/lambda(1-e^(-lambdat))` `=alpha/lambda(lambdat-1+e^(-lambdat))` |
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