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\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

The rate at which a radioactive substance splits up is given by

$$\ds\frac{dN}{dt}=-\lambda N,$$

where $N$ is the number of atoms present after $t$ seconds and
$lambda$ is a constant. Show that $N=N_0 e^{-\lambda t}$ where
$N_0$ is the number of atoms present initially. Find the times in
\emph{years} for half the number of atoms of a given mass of
radium to disintegrate if $\lambda=1.37 \times 10^{-11}$ per
second.

\medskip

{\bf Answer}

${}$

$\begin{array} {crcl} &amp; \ds\frac{dN}{dt} &amp; = &amp; -\lambda N\\
\Rightarrow &amp; \ds\int\ds\frac{dN}{N} &amp; = &amp; -\lambda \ds\int dt\\
\Rightarrow &amp; \ln N &amp; = &amp; -\lambda t=c\\ \rm{or} &amp; N &amp; = &amp;
e^{-\lambda t+c}\\ &amp; N &amp; = &amp; e^c e^{-\lambda t} \end{array}$

so let $e^c=N_0$

$\un{N=N_0e^{-\lambda t}}$

What is $N_0$?

Set $t=0$ to get $N=N_0e^{-\lambda \cdot 0}=N_0$ i.e., $N=N_0$
when $t=0$.

We want the time, $T$, when $N=\ds\frac{N_0}{2}$

i.e.,

$\begin{array} {crcl} &amp; \ds\frac{N_0}{2} &amp; = &amp; N_0e^{-\lambda T}\\
\Rightarrow &amp; \ds\frac{1}{2} &amp; = &amp; e^{-\lambda T}\\ &amp;
\ln\left(\ds\frac{1}{2}\right) &amp; = &amp; -\lambda T\\ \rm{or} &amp; T &amp; =
&amp; -\ds\frac{1}{\lambda}\ln\left(\ds\frac{1}{2}\right)\\ &amp; &amp; = &amp;
+\ds\frac{\ln(2)}{1.37\times10^{-11} (per sec)}\\ &amp; &amp; = &amp;
50,594,684,712 secs\\&amp; &amp; = &amp;
\ds\frac{50,594,684,712}{60\times60\times24\times365}
years\end{array}$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\un{=1604.3\ years}$
\end{document}
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