Question

The half life of neutron is $693$ seconds. What fraction of neutrons will decay when a beam of neutrons, having kinetic energy of $0.084$ $eV$, travells a distance of $1 km ?$
(mass of neutron $=1.68 \times 10^{-27} kg$, and $\ln 2=0.693)$

• A. $60 \times 10^{-5}$
• B. $15 \times 10^{-5}$
• C. $0.25 \times 10^{-5}$
• D. $50 \times 10^{-5}$

Answer 1

Answer: (C)

Given, the half-life of neutron, $t_{1 / 2}=693\,sec$
Kinetic energy of neutron $=0.084 \,eV$
$\frac{1}{2} \,m v^{2} =0.084 \times 1.6 \times 10^{-19} $
$v^{2} =\frac{2 \times 0.084 \times 1.6 \times 10^{-19}}{m}$
$=\frac{2 \times 0.084 \times 1.6 \times 10^{-19}}{1.68 \times 10^{-27}}=0.16 \times 10^{8} $
$ v =0.4 \times 10^{4} \,m / s$
Time taken by neutron to travelled in $1 \,km$,
$t=\frac{1000}{0.4 \times 10^{4}} \Rightarrow t=0.25 s$
By Radioactive Decay's law,
$N=N_{0}\left(\frac{1}{2}\right)^{n} \Rightarrow \frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n} $
$\frac{N_{0}}{N}=2^{n}$
$\ln \frac{N_{0}}{N}=\ln 2^{n} =n \ln 2=\frac{t}{t_{1 / 2}} \ln 2=\frac{0.25 \times 0.693}{693} $
$=0.25 \times 10^{-5} [\because \ln 2=0.693] $