Question

There is a stream of neutrons with a kinetic energy of $0.0327eV$ . If the half-life of the neutron is $700$ second, what fraction of neutrons will decay before they travel a distance of $10m$ ? Given the mass of neutron = $1.675\times 10^{-27}kg$ .

• A. $3.96\times 10^{-6}$
• B. $3.90\times 10^{-6}$
• C. $3.85\times 10^{-6}$
• D. $4.86\times 10^{-6}$

Answer 1

Answer: (A)

From the given kinetic energy of the neutrons we first calculate their velocity. Thus
$\frac{1}{2}m{v}^2=0.0327\times 1.6\times 10^{-19}$
$\therefore v^2=\frac{2\times 0.0327\times 1.6\times 10^{-19}}{1.675\times 10^{-27}}=625\times 10^4$
$\frac{1}{2}m{v}^2=0.0327\times 1.6\times 10^{-19}$
$\therefore v^2=\frac{2\times 0.0327\times 1.6\times 10^{-19}}{1.675\times 10^{-27}}$
$=625\times 10^4$ or $v=2500m{s}^{-1}$
With this speed, the time taken by the neutrons to travel a distance of $10m$ is,
$\Delta t=\frac{10}{2500}=4\times 10^{-3}s$
The fraction of neutrons decayed in time $\Delta t$ second is,
$\frac{\Delta N}{N}=\lambda \Delta t$
Also, $\lambda =\frac{0.693}{T_{1/2}}$
$\therefore \frac{\Delta N}{N}=\frac{0.693}{T_{1/2}}$
$=\frac{0.693}{700}\times \left(4\times 10^{-3}\right)$
$=3.96\times 10^{-6}$