InterviewSolution
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InterviewSolution
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A radio nuclide `A_(1)` with decay constant `lambda_(1)` transforms into a radio nuclide `A_2` with decay constant `lambda_(2)`. Assuming that at the initial moment, the preparation contained only the radio nuclide `A_(1)` (a) Find the equation decribing accumulation of radio nuclide `A_(2)` with time. (b) Find the time interval after which the activity of radio nuclide `A_(2)` reaches its maximum value. |
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Answer» (a) This part has already been discussed in section on succesive disintegrations with the result. `N_(2) (t)=(lambda_(1) N_(0))/(lambda_(2) -lambda_(1))(e^(-lambda_(1)t -e^(-lambda_(2)t)))` (b) The activity of nuclide `A_(2)` is `lambda_(2) N_(2)`. This is mximum when `N_(2)` is maximum, i.e., when `dN_(2)//dt=0`, activity maximum. `(d[N_(2) (t)])/(dt)=(d)/(dt)[(lambda_(1) (N_(0)))/(lambda_(2) -lambda_(1))(e^(-lambda_(1)t -e^(-lambda_(2)t)))]` `(lambda_(1)N_(0))/(lamda_(2)-lambda_(1))(lambda_(1)e^(-lambda_(1)t)+lambda_(2)e^(-lambda_(2)t))=0` `e_((lambda_(1) -lambda_(2))) t=(lambda_(2))/(lambda_(1))` `t=(1n (lambda_(2))//lambda_(1))/(lambda_(2)-lambda_(1))` . |
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