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{\bf Question}

$^{140}B a$ decays to give $^{140}L a$, which subsequently decays
to $^{140}C e$, i.e.

$$^{140}B a \longrightarrow ^{140}L a + \beta\  \mathrm
{(half-life\ 12\ days)}$$

$$^{140}L a \longrightarrow ^{140}C e\ \mathrm {(stable)} + \beta\
\mathrm {(half-life\ 10\ hours)}$$

Calculate the decay constants $\lambda_{Ba}$ and $\lambda_{La}$.
If the initial concentration of $^{140}B a$ is 1000 units and the
initial concentration of $^{140}L a$ is zero, calculate the
concentration of $^{140}B a$ after 20 days and the concentrations
of $^{140}L a$ after both 10 hours and 20 days.

Note:
\begin{description}
\item[(i)]
Pay particular attention to units;
\item[(ii)]
If the number of an isolated radioactive element is given by
$N(t)=N(0)e^{-\lambda t}$, then $\lambda$ is the decay constant;
\item[(iii)]
The results of Question 2 may be used.
\end{description}


{\bf Answer}

Assume that time $t$ is measured in hours.  Let $B(t),\ L(t)$ be
the number of atoms of $^{140}Ba, ^{140}La$ at time $t$
respectively.

The decay from $^{140}Ba$ to $^{140}La$ is modelled by the
differential equation

$$\frac{\partial B}{\partial t} = - \lambda_{Ba} B$$

which gives $\displaystyle B(t)=B(0)e^{- \lambda_{Ba} t}$

Number of $\displaystyle ^{140}Ba$ atoms after 12 days (288 hours)
$= B(288)=B(0)e^{- \lambda_{Ba}288}$

Since half life of $^{140}Ba$ is 288 hours,

$$\frac{B(288)}{B(0)}=e^{- \lambda_{Ba} 288} = \frac{1}{2}$$

so

$$\lambda_{Ba}=-\frac{\ln \left(\frac{1}{2}\right)}{288}L^{-1}
\approx 2.407 \times 10^{-3} L^{-1}$$

Similarly, modelling decay from $^{140}La$ to $^{140}Ce$ by the
differential equation

$$\frac{\partial L}{\partial t} = - \lambda_{La} L$$

gives

$$\lambda_{La}=-\frac{\ln \left(\frac{1}{2}\right)}{10}L^{-1}
\approx 6.931 \times 10^{-2} L^{-1}$$

The decay processes are modelled by the equations found in
Question 2, where

\begin{eqnarray*}
\alpha & = & B(0) = 1000 \\ K_1 & = & \lambda_{Ba} \\ K_2 & = &
\lambda_{La}
\end{eqnarray*}

This gives

$$B(t)=B(0)e^{-\lambda_{Ba}t}=1000e^{-2.407 \times 10^{-3} \times
t}$$

\begin{eqnarray*}
L(t) & = & \frac{\lambda_{Ba}B(0)}{\lambda_{La}-\lambda_{Ba}}
\left (e^{-\lambda_{Ba} t}-e^{-\lambda_{La} t} \right) \\ &
\approx & 35.977 \left (e^{-2.407 \times 10^{-3} \times
t}-e^{-6.931 \times 10^{-2} \times t} \right)
\end{eqnarray*}

$$\begin{array} {lclcl} \rm {Concentration\ of\ } ^{140}Ba \rm {\
after\ 20\ days\ (480\ hours)\ } & = & B(480) & \approx & 315
units\\ \rm {Concentration\ of\ } ^{140}La \rm {\ after\ 10\
hours\ } & = & L(10) & \approx & 17 units
\\ \rm {Concentration\ of\ } ^{140}La \rm {\ after\ 20\ days\ (480\ hours)}
& = & L(480) & \approx & 11 units
\end{array}$$


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