Question

There is a stream of neutrons with a kinetic energy of 0.0327eV. If the half-life of neutrons is 700 second, the percentage fraction of neutron will decay before they travel a distance of 10 km is (multiply answer by 20).

$(massofneutron=1.675\times {10}^{-27}kg)$

Answer 1

The K.E is $1/2m{V}^2=0.0327eV=0.0327\times 1.602\times {10}^{-19}J=5.2385\times {10}^{-21}J$
$V^2=\frac{2\times 5.2385\times {10}^{-21}}{1.675\times {10}^{-27}}=6254973$

The speed, $V=2501m/s=2.5km/s$
The time required to travel distance of 10 km is $t=\frac{D}{V}=\frac{10km}{2.5km/s}=4s$
The decay constant is $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{700}$
The decay expression is $\frac{N}{N_0}=e^{-\lambda t}=e^{-\frac{0.693}{700}\times 4}=e^{-3.96\times {10}^{-3}}=0.996$
The percentage fraction of neutron will decay before they travel a distance of 10 km is $\frac{1-0.996}{1}\times 100=0.40$