Question

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.

Answer 1

(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))`