Definition of decay constant
Calculation of half life
Calculation of initial number of nuclei at $\mathit{t}=0$
The decay constant $\left(\mathit{\lambda }\right)$ of a radioactive nucleus equals the ratio of the instantaneous rate of decay $\left(\frac{∆\mathit{N}}{∆\mathit{t}}\right)$ to the corresponding instantaneous number of radioactive nuclei.
Alternatively,
The decay constant $\left(\mathit{\lambda }\right)$ of a radioactive nucleus is the constant of proportionality in the relation between its rate of decay and number of its nuclei at any given instant.
Alternatively,
$\left(\frac{∆\mathit{N}}{∆\mathit{t}}\right)\propto \mathit{N}$
$\left(\frac{∆\mathit{N}}{∆\mathit{t}}\right)=\mathit{\lambda }\mathit{N}$
The constant $\left(\mathit{\lambda }\right)$ is known as the decay constant
Alternatively,
The decay constant equals the reciprocal of the mean life of a given radioactive nucleus.
$\mathit{\lambda }=\frac{1}{\mathit{\tau }}$
where,
Alternatively,
The decay constant equal the ratio of $\ln 2$ to the half life of the given radioactive element.
$\mathit{\lambda }=\frac{\ln 2}{{\mathit{T}}_{1∕2}}$
where ${\mathit{T}}_{1∕2}=$ Half life
Alternatively,
The decay constant of a radioactive element, is the reciprocal of the time in which the number of its nuclei reduces to $\frac{1}{\mathit{e}}$ of its original number.
We have
$\mathit{R}=\mathit{\lambda }\mathit{N}$
$\mathit{R}\left(20\mathrm{hrs}\right)=10000=\mathit{\lambda }{\mathit{N}}_{20}$
$\mathit{R}\left(30\mathrm{hrs}\right)=5000=\mathit{\lambda }{\mathit{N}}_{30}$
$\therefore \frac{{\mathit{N}}_{20}}{{\mathit{N}}_{30}}=2$
This means that the number of nuclei, of the given radioactive nucleus, gets halved in a time of $\left(30-20\right)$ hours $=10$ hours
$\therefore$ Half life $=10$ hours
This means that in 20 hours ( $=2$ half lives), the original number of nuclei must have gone down by a factor of 4 .
Hence Rate of decay at $\mathit{t}=0$
$\mathit{\lambda }{\mathit{N}}_0=4\mathit{\lambda }{\mathit{N}}_{20}$
$=4\times 10000=40,000$ disintegrations per second
[Note : Award full marks of the last part of this question even if student does not calculate initial number of nuclei and calculates correctly rate of disintegration at $\mathit{t}=0$ ]
i.e., ${\mathit{R}}_0=40,000$ disintegrations per second
${\mathit{N}}_0=\frac{40000}{\mathit{\lambda }}=\frac{40000}{\ln 2}$ $\times 10\times 60\times 60$
${\mathit{N}}_0=\frac{144\times {10}^7}{0.693}$