Question

A radioactive material decays by $\beta -particle$ emission. During the first $2$ seconds of a measurement, $n\beta -particles$ are emitted and in the next $2$ seconds $0.75n\beta -particles$ are emitted. If the mean-life of this material in seconds is given by $69.5\times {10}^{-n}$. Then $n$ will be given by
$(ln3=1.0986$ and $ln2=0.6931)$.

Answer 1

$\begin{array}{c}\text{Given:}\\ N=n\\ N^{\prime }=n+0.75n=1.75n\end{array}$

$\begin{array}{c}{N}_0=\text{initial}\\ N=\text{final}\end{array}$

$\lambda =\ln \left[\frac{N_0}{N}\right]$

$\begin{array}{c}\text{For t}=2\\ \begin{array}{c}2\lambda =\ln \left[\frac{N_0}{N_{0}n}\right]⟶\left(i\right)\\ \text{For}t=4\\ 4\lambda =\ln \left[\frac{N_0}{N_0-1.75n}\right]⟶\left(ii\right)\end{array}\\ \begin{array}{cc}\text{Mean life}& =\tau =\frac{1}{\lambda }=69.5\times {10}^{-n}\\ \Rightarrow \lambda & =\frac{{10}^n}{69.5}\end{array}\end{array}$

$\begin{array}{c}\text{Dividing}\left(i\right)/\left(ii\right)\\ \text{We get}{N}_0=4n\\ \text{pulting}{N}_0=4n\text{and}\lambda \text{in}\left(1\right)\\ \text{We get}n=1⟶\text{Answer}\end{array}$