Question

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 describing 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.

Answer 1

(a) Suppose $N_1$ and $N_2$ are the number of two radionuclides $A_1,A_2$ at time t.

Then

$\frac{d{N}_1}{dt}=-{\lambda }_1{N}_1$ $\left(1\right)$

$\frac{d{N}_2}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$ $\left(2\right)$

From $\left(1\right)$

$N_1=N_{10}{e}^{-{\lambda }_{1}t}$

where $N_{10}$ is the initial number of nuclides $A_1$ at time $t=0$

From (2)

$(\frac{d{N}_2}{dt}+{\lambda }_2{N}_2)e^{{\lambda }_{2}t}={\lambda }_1{N}_{10}{e}^{-({\lambda }_1-{\lambda }_2)t}$

or $\left(N_2{e}^{{\lambda }_{2}t}\right)=$ const $\frac{{\lambda }_1{N}_{10}}{{\lambda }_1-{\lambda }_2}{e}^{-({\lambda }_1-{\lambda }_2)t}$

since $N_2=0$ at $t=0$

constant $N_2=\frac{{\lambda }_1{N}_{10}}{{\lambda }_1-{\lambda }_2}$

Thus = $\frac{{\lambda }_1{N}_{10}}{{\lambda }_1-{\lambda }_2}\left(e^{-{\lambda }_{2}t-e^{-{\lambda }_{1}t}}\right)$

(b) The activity of nuclide $A_2$ is ${\lambda }_2{N}_2$. This is maximum when $N_2$ is maximum.

That happens when

$\frac{d{N}_2}{dt}=0$

This require ${\lambda }_2{e}^{-{\lambda }_2{t}_m={\lambda }_1{e}^{-{\lambda }_2{t}_m}}$

or $t_m=\frac{ln\left({\lambda }_1/{\lambda }_2\right)}{{\lambda }_1-{\lambda }_2}$