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A radionuclide A_(1) with decay constant lambda_(1) transforms into a radionuclide A_(2) with decay constant lambda_(2). Assuming that at the initial moment the preparation contained only the radionuclide A_(1), find: (a) the equation describing accumulation of the radionuclide A_(2) with time, (b) the time interval after which the activity value. |
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Answer» Solution :(a)Suppose `N_(1)` and `N_(2)` are the number of TWO radionuclides `A_(1),A_(2)` at time `t`. Then `(dN_(1))/(dt)= 0lambda_(1)N_(1)` (1) `(dN_(2))/(dt)= lambda_(1)N_(1)-lambda_(2)N_(2)` (2) From (1) `N_(1)= N_(10)E^(-lambda_(1 t)` where `N_(10)` is the inital number of nuclides `A_(1)` at time `t=0` From (2) `((dN_(2))/(dt)+lambda_(2)N_(2))e^(lambda_(2)t)=lambda_(1)N_(10)e^(-(lambda_(1)-lambda_(2))t` or `(N_(2)e^(lambda_(2t)))= CONST(lambda_(1)N_(0))/(lambda_(1)-lambda_(2))e^(-(lambda_(1)-lambda_(2))t)` since `N_(2)=0 at t=0` Constant `N_(2)=(lambda_(1)N_(10))/(lambda_(1)-lambda_(2))` Thus `=(lambdaN_(10))/(lambda_(1)-lambda_(2))(e^(-lambda_(2)t)-e^(-lambda_(1)t))` (b) The activity of nuclie `A_(2)` is `lambda_(2)N_(2)`. This is maximum when `N_(2)` is maximum. That hapens when `(dN_(2))/(dt)=0` This requires `lambda_(2)e^(-lambda_(2)t+_(m))` or `t_(m)=(In(lambda_(1)//lambda_(2)))/(lambda_(1)-lambda_(2))` |
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