A radioactive nucleus A decays into B with a decay constnat `lambda_(1)`. B is alos radioactive and it decays into C with a decay constant of `lambda_(2)`. At time t=0 activity of A was recorded as `A_(1)^(0)` and that of B was `A_(2)^(0)=0` , in a sample.
(a) If `N_(1)` and `N_(2)` are population of nuclei of A and B respectively at time t, set up a differential equation in `N_(2)`.
(b) The solution to the differential equation obtained in part(s) is given by
(b) The solution to the differential equation obtained in part (a) is given by
`N_(2)=(lambda_(1))/(lambda_(2)-lambda_(1))N_(1)^(0)(e^(-lambda_(1)l)-e^(-lambda_(2)l))`
Where `N_(1)^(0)` is initial population of A. Based on this answer following questions-
(i) If half life of A is quite large compared to B (for example hafl life of A is 8 hr and that of B is 1hr). show that ratio of activity of B to that of A approaches a constant value after a sufficiently long time. Find this ratio, in terms of `lambda_(1)` and `lambda_(2)`.
(ii) If half of A is inifinitely large compared to that of B (for example, A has half life `-10^(9)` yr and for B half life `-10^(5)` yr), show that both A and B will have equal activity after a very long time.
(iii) If `lambda_(1)gtlambda_(2)` find the time at which activity of B reaches a maximum.

Answer

(a)`(dN_(2))/(dt)=-lambda_(2)N_(2)+lambda_(1)N_(1)^(0)e^(-lambda_(1)t)`
(b) `(A_(2))/(A_(1))=(lambda_(2))/(lambda_(2)-lambda_(1))`
(c) `((1)/(lambda_(2)-lambda_(1)))ln((lambda_(2))/(lambda_(1)))`