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 $1km?$
(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, $t_{1/2}=693sec$
Kinetic energy of neutron $=0.084eV$
$\frac{1}{2}m{v}^2=0.084\times 1.6\times 10^{-19}$
$v^2=\frac{2\times 0.084\times 1.6\times 10^{-19}}{m}$
$=\frac{2\times 0.084\times 1.6\times 10^{-19}}{1.68\times 10^{-27}}=0.16\times 10^8$
$v=0.4\times 10^{4}m/s$
Time taken by neutron to travelled in $1km$,
$t=\frac{1000}{0.4\times 10^4}\Rightarrow t=0.25s$
By Radioactive Decay's law,
$N=N_0{\left(\frac{1}{2}\right)}^n\Rightarrow \frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$
$\frac{N_0}{N}=2^n$
$\ln \frac{N_0}{N}=\ln 2^n=n\ln 2=\frac{t}{t_{1/2}}\ln 2=\frac{0.25\times 0.693}{693}$
$=0.25\times 10^{-5}[\because \ln 2=0.693]$