Question

A radioactive element decays by $\alpha -emission$. A detector records n beta particles in 2 s and in next 2 s, it records 0.75 n beta particles. Find mean life correct to nearest whole number. Given $ln2=0.6031,ln3=1.0986$

Answer 1

Let $N_0$ be initial number of nuclei at time $t=0$.
The number of undecayed nuclei in time $t$ is $N=N_0{e}^{\lambda t}$
The number of nuclei decayed in time $t$ is:
$n=N_0-N0-N_0{e}^{-\lambda t}$ (i)

The number of undecayed nuclei in next time $t$,
$N^{\prime }=N{e}^{-\lambda t}=\left(N_0{e}^{-\lambda t}\right)e^{-\lambda t}=N_0{e}^{-2\lambda t}$ (ii)

Number of decayed nuclei in next time $t$,
$0.75n=N-N^{\prime }=N_0{e}^{-\lambda t}-N_0{e}^{-2\lambda t}$
$=N_0{e}^{-\lambda t}(1-e^{-\lambda t})$ (iii)

Dividing Eq. (ii) by Eq. (i), we get:
$0.75=e^{-\lambda t}\Rightarrow \frac{4}{3}=e^{\lambda t}$

Taking natural logarithm:
$ln\frac{4}{3}=\lambda t\Rightarrow \lambda =\frac{ln\frac{4}{3}}{t}$

Given $t=2s$.
$\therefore \lambda =\frac{ln4-ln3}{2}=\frac{2ln2-ln3}{2}=\frac{2\times 0.6931-1.0986}{2}$$=0.1438s^{-1}$

Mean life, $\tau =\frac{1}{\lambda }=\frac{1}{0.1438}=7s$ (whole number)