91Ó°ÊÓ

A radioactive decay problem where the Radium decays to Radon which decays to Polonium. The decay constant of polonium is given to be${\mathrm{λ}}_3$.

For a linear differential equation$y\text{'}+Py=Q$,

$I=∫Pdx$ …… (3.4)

And the general solution is,

$\begin{array}{r}y{e}^I=∫Q{e}^{I}dx+c_{,}or\\ y=e^{-I}∫Q{e}^{I}dx+c{e}^{-I},\end{array}\}\mathbf{\text{where}}\mathit{I}\mathbf{=}\mathbf{∫}\mathit{P}\mathit{d}\mathit{x}$ …â¶Ä¦(3.9)

Consider the decay of Polonium.

The rate of change is the difference,

That is,$\frac{d{N}_3}{dt}+{\mathrm{λ}}_3{N}_3=\frac{{\mathrm{λ}}_1{\mathrm{λ}}_2{N}_0}{{\mathrm{λ}}_2-{\mathrm{λ}}_1}\left(e^{-{\mathrm{λ}}_{1}t}-e^{-{\mathrm{λ}}_{2}t}\right)$

Hence, the equation will be in the form, $y\text{'}+Py=Q$.

By equation (3.4),

$$\begin{gathered} I=∫{\mathrm{λ}}_{3}dt \\ ={\mathrm{λ}}_{3}t \end{gathered}$$

That is,$e^I=e^{{\mathrm{λ}}_{3}t}$

By equation (3.9), we can write,

$$\begin{gathered} N_3{e}^I=∫\left[\frac{{\mathrm{λ}}_1{\mathrm{λ}}_2{N}_0}{{\mathrm{λ}}_2-{\mathrm{λ}}_1}\left(e^{-{\mathrm{λ}}_{1}t}-e^{-{\mathrm{λ}}_{2}t}\right)\right]e^{{\mathrm{λ}}_{3}t}dt \\ =\frac{{\mathrm{λ}}_1{\mathrm{λ}}_2{N}_0}{{\mathrm{λ}}_2-{\mathrm{λ}}_1}∫\left[e^{\left({\mathrm{λ}}_3-{\mathrm{λ}}_1\right)t}-e^{\left({\mathrm{λ}}_3-{\mathrm{λ}}_2\right)t}\right]dt \\ =\frac{{\mathrm{λ}}_1{\mathrm{λ}}_2{N}_0}{{\mathrm{λ}}_2-{\mathrm{λ}}_1}\left[\frac{e^{\left({\mathrm{λ}}_3-{\mathrm{λ}}_1\right)t}}{{\mathrm{λ}}_3-{\mathrm{λ}}_1}-\frac{e^{\left({\mathrm{λ}}_3-{\mathrm{λ}}_2\right)t}}{{\mathrm{λ}}_3-{\mathrm{λ}}_2}\right]+C \end{gathered}$$

Hence,

$N_3=\frac{{\mathrm{λ}}_1{\mathrm{λ}}_2{N}_0}{{\mathrm{λ}}_2-{\mathrm{λ}}_1}\left[\frac{e^{-{\mathrm{λ}}_{1}t}}{{\mathrm{λ}}_3-{\mathrm{λ}}_1}-\frac{e^{-{\mathrm{λ}}_{2}t}}{{\mathrm{λ}}_3-{\mathrm{λ}}_2}\right]+C{e}^{-{\mathrm{λ}}_{3}t}$ …… (1)

Find the value of C, using the initial condition.

That is, take $t=0,N_3=0$, to get,

$$\begin{gathered} 0=\frac{{\mathrm{λ}}_1{\mathrm{λ}}_2{N}_0}{{\mathrm{λ}}_2-{\mathrm{λ}}_1}\left[\frac{1}{{\mathrm{λ}}_3-{\mathrm{λ}}_1}-\frac{1}{{\mathrm{λ}}_3-{\mathrm{λ}}_2}\right]+C \\ C=-\frac{{\mathrm{λ}}_1{\mathrm{λ}}_2{N}_0}{{\mathrm{λ}}_2-{\mathrm{λ}}_1}\left[\frac{1}{{\mathrm{λ}}_3-{\mathrm{λ}}_2}-\frac{1}{{\mathrm{λ}}_3-{\mathrm{λ}}_1}\right] \\ =\frac{{\mathrm{λ}}_1{\mathrm{λ}}_2{N}_0}{\left({\mathrm{λ}}_3-{\mathrm{λ}}_1\right)\left({\mathrm{λ}}_3-{\mathrm{λ}}_2\right)} \end{gathered}$$

Substitute back in equation (1), to get,

$N_3={\mathrm{λ}}_1{\mathrm{λ}}_2{N}_0\left[\frac{e^{-{\mathrm{λ}}_{1}t}}{\left({\mathrm{λ}}_2-{\mathrm{λ}}_1\right)\left({\mathrm{λ}}_3-{\mathrm{λ}}_1\right)}+\frac{e^{-{\mathrm{λ}}_{2}t}}{\left({\mathrm{λ}}_1-{\mathrm{λ}}_2\right)\left({\mathrm{λ}}_3-{\mathrm{λ}}_2\right)}+\frac{e^{-{\mathrm{λ}}_{3}t}}{\left({\mathrm{λ}}_1-{\mathrm{λ}}_3\right)\left({\mathrm{λ}}_2-{\mathrm{λ}}_3\right)}\right]$