Define the term ‘decay constant’ of a radioactive sample. The rate of disintegration of a given radioactive nucleus is 10000 disintegrations and 5,000 disintegrations after 20 hr. and 30 hr. respectively from start. Calculate the half life and initial number of nuclei at t = 0.
 

Decay constant’ of a radioactive sample is defined as the ratio of its instantaneous rate of disintegration to the number of atoms present at that time.
i.e., $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{\lambda n},$
$\mathrm{\lambda }=\frac{-\mathrm{dN}/\mathrm{dt}}{\mathrm{N}}$
Let, the initial number of nuclei be N_o at t=0
$\begin{array}{l}\frac{\mathrm{dN}}{\mathrm{dt}}=10,000\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ at }\mathrm{t}=20\mathrm{hr}\\ \mathrm{and} \frac{{\mathrm{dN}}^{\mathrm{\prime }}}{\mathrm{dt}}=5,000\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ at }\mathrm{t}=30\mathrm{hr}\\ \mathrm{we} \mathrm{have}, \frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{\lambda N},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\end{array}$
For first case,  
$\begin{array}{r}10,000=\mathrm{\lambda N},\Rightarrow \mathrm{N}=\frac{10,000}{\mathrm{\lambda }}\\ \mathrm{and} \mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda }.\mathrm{t}}\\ \frac{10,000}{\mathrm{\lambda }}={\mathrm{N}}_0{\mathrm{e}}^{-20\mathrm{\lambda }} ...\left(\mathrm{i}\right)\end{array}$
For second case, 
$\begin{array}{l}5,000={\mathrm{\lambda N}}^{\mathrm{\prime }}\\ {\mathrm{N}}^{\mathrm{\prime }}=\frac{5,000}{\mathrm{\lambda }}\\ \mathrm{and} {\mathrm{N}}^{\mathrm{\prime }}={\mathrm{N}}_0{\mathrm{e}}^{-30\mathrm{\lambda }}\\ \frac{5,000}{\mathrm{\lambda }}={\mathrm{N}}_0{\mathrm{e}}^{-30\mathrm{\lambda }} ....\left(\mathrm{ii}\right)\end{array}$
on dividing equation (i) by equation (ii), we get 
$\Rightarrow \frac{{\mathrm{e}}^{-20\mathrm{\lambda }}}{{\mathrm{e}}^{-30\mathrm{\lambda }}}=2$
Taking log on both sides,
$\begin{array}{l}\Rightarrow -20\mathrm{\lambda }+30\mathrm{\lambda }={\log }_{\mathrm{e}}2\\ \Rightarrow 10\mathrm{\lambda }=2.302\times 0.3010\\ \Rightarrow \mathrm{\lambda }=0.0693\\ \therefore \text{ Half life, }\mathrm{r}=\frac{0.693}{0.0693}=10\mathrm{hr}\\ \end{array}$