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The half-life of a particle of mass $1.6 \times 10^{-26} \,kg$ is $6.9 \,s$ and a stream of such particles is travelling with the kinetic energy of a particle being $0.05 \,eV$. The fraction of particles which will decay, when they travel a distance of $1 \,m$ is
$0.1$
$0.01$
$0.001$
$0.0001$
Solution
(d)
Kinctic encrgy of particle is
$K=\frac{1}{2} m v^2$
$\Rightarrow 0.05 \times 16 \times 10^{-19}=\frac{1}{2} \times 16 \times 10^{-26} \times v^2$
$\Rightarrow \quad v^2=\frac{0.05 \times 16 \times 10^{-19} \times 2}{16 \times 10^{-26}}$
$\Rightarrow \quad v=10^3 \,ms ^{-1}$
Time taken by particle beam to travel
through $1 \,m =t=\frac{D}{v}=\frac{1}{10^3} \Rightarrow t=10^{-3} \,s$
Number of half lives occured in $10^{-3} s$
$=n=\frac{t}{T_{1 / 2}}=\frac{10^{-3} s }{6.9 s } \approx 0.0001$
So, fraction decayed $=1-\left(\frac{1}{2}\right)^{0.0001}$
$\approx 0.0001$
