Question

There is a stream of neutrons with a kinetic energy of $0.32\text{eV}$. If the half-life of neutrons is $700\text{seconds}$, what fraction (in ${10}^{–3})$ of neutrons will decay before they travel a distance of $8\text{km}$? Given, mass of a neutron $=1.6\times {10}^{–27}\text{kg}$ (Round off to nearest integer)

Answer 1

Given, $E=\frac{1}{2}\times m{v}^2=0.32\text{eV}$
$\Rightarrow \frac{1}{2}\times 1.6\times {10}^{-27}{v}^2=3.2\times 16\times {10}^{-20}$
$\Rightarrow v^2=6.4\times {10}^7=64\times {10}^6$
$\therefore v=8\times {10}^3\text{m/s}=8\text{km/s}$

Time of travel, $t=\frac{8\text{km}}{8\text{km/s}}=1\text{sec}$
Given, $T_{1/2}=700\text{s}$

If fraction of neutrons left in time $t=\frac{N}{N_0}$,
$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}={\left(\frac{1}{2}\right)}^{\frac{1}{700}}=0.999$

Therefore, fraction of neutrons decayed in time $t$
$=1-\frac{N}{N_0}=1-0.999$
$=0.001$ or $1\times {10}^{-3}$