A radionuclide `A_(1)` goes through the transformation chain `A_(1) rarr A_(2) rarr A_(3)` (stable) with respective decay constants `lambda_(1)` and `lambda(2)`. Assuming that at the initial moment the preparation contained only the radionuclide `A_(1)` equal in quantity to `N_(10)` nuclei, find the equation describing accumulation of the stable isotope `A_(3)`.

Explanation

Here we hav the equaitons
`(dN_(1))/(dt)= -lambda_(1)N_(1)`
`(dN_(2))/(dt)= lambda_(1)N_(1)-lambda_(2)N_(2) and (dN_(3))/(dt)= lambda_(2)N_(2)`
From problem `229`
`N_(1)=N_(10)e^(-lambda_(1)t)`
`N_(2)=(lambda_(1)N_(10))/(lambda_(1)-lambda_(2))(e^(-lambda_(2)t)-e^(-lambda_(1)t))`
Then `(dN_(3))/(dt)=(lambda_(1)lambda_(2))/(lambda_(1)-lambda_(2))N_(10)(e^(lambda_(2)t)-e^(-lambda_(1)t))`
or `N_(3)= Const-(lambda_(1)lambda_(2))/(lambda_(1)-lambda_(2))((e^(-lambda_(2)t))/(lambda_(2))-(e^(-lambda_(1)t))/(lambda_(1)))N_(10)`
Since `N_(3)=0` initially
`Const =(lambda_(1)lambda_(2))/(lambda_(1)-lambda_(2))N_(10)((1)/(lambda_(2))-(1)/(lambda_(2)))`
So `N_(3)=(lambda_(1)lambda_(2)N_(10))/(lambda_(1)-lambda_(2))[(1)/(lambda_(2))(1-e^(-lambda_(2)t))-(1)/(lambda_(1))(1-e^(-lambda_(1)t))]`
`=N_(10)[1+(lambda_(1)e^(-lambda_(2)t)-lambda_(2)e^(-lambda_(1)t))/(lambda_(2)-lambda_(1))]`