Question

There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half-life of neutrons is 700 s, what fraction of neutrons will decay before they travel a distance of 10 m? Given mass of neutron $=1.676\times {10}^{-27}kg$

Answer 1

The velocity v of neutron is given by:
$\frac{1}{2}m{v}^2=0.0327eV$

$\frac{1}{2}\times 1.675\times 1^{0-27}{v}^2=0.0327\times 1.6\times {10}^{-19}J$

$\therefore v^2=\frac{2\times 0.0327\times 1.6\times {10}^{-19}}{1.675\times {10}^{-27}}$

$\Rightarrow v=2.5\times {10}^{3}m{s}^{-1}$

Time taken by neutron to travel a distance of $10m$ is:

$t=\frac{distance}{velocity}=\frac{10}{2.5\times {10}^3}=4\times {10}^{-3}s$

and half-life $T=700s$.

The fraction of neutrons decayed in time $t$ is:
$\frac{N}{N_0}=e^{-\lambda t}={\left(\frac{1}{2}\right)}^{1/T}={\left(\frac{1}{2}\right)}^{4\times {10}^{-3}/700}=\left(\frac{1}{2}\right)5.7\times {10}^{-6}$$=0.999952$

Fraction of neutrons decayed is:
$1-0.999952=0.000048=4.8\times {10}^{-5}$