The half life of a stream of radioactive particles moving along a straight path with a constant kinetic…
The half life of a stream of radioactive particles moving along a straight path with a constant kinetic energy of $4 \mathrm{eV}$ is 1 minute. The percentage of particles which decay before travelling a distance of $3.6 \mathrm{~km}$ is (Mass of the radioactive particles $=3.2 \times 10^{-21} \mathrm{~kg}$ and charge of the electron $=1.6 \times 10^{-19} \mathrm{C}$ ).
87.5
175
37.5
75
Solution
$\begin{aligned}
& \text {As, } K . E .=\frac{1}{2} m v^2 \Rightarrow \mathrm{V}=\sqrt{\frac{2 \times K . E .}{m}} \\
& \Rightarrow \mathrm{V}=\sqrt{\frac{2 \times 4 \times 16 \times 10^{-19}}{32 \times 10^{-21}}}=\sqrt{\frac{2 \times 4}{2 \times 10^{-2}}} \\
& \Rightarrow \mathrm{V}=20 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Time to travel a distance of $3.6 \mathrm{~km}$ is
$\begin{aligned}
& t=\frac{D}{v}=\frac{3.6 \times 1000}{20} s=\frac{3.6 \times 1000}{20 \times 60} \mathrm{~min} \\
& =3 \mathrm{~min}=3 \text { half-time }
\end{aligned}$
In three half life times, the number of particles decayed
$\begin{aligned}
& =\mathrm{N}_0-\frac{\mathrm{N}_0}{2^3}=\frac{7}{8} \mathrm{~N}_0 \\
& \% \text { req. }=\frac{\frac{7}{8}}{\mathrm{~N}_0} \mathrm{~N}_{\mathrm{o}} \times 100=87.5 \%
\end{aligned}$