Question

There is a stream of neutrons with a kinetic energy of $0.0327 \, eV$ . If the half-life of the neutron is $700$ second, what fraction of neutrons will decay before they travel a distance of $10 \, m$ ? 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 $v=2500\, ms ^{-1}$
With this speed, the time taken by the neutrons to travel a distance of $10 \,m$ 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}} $
$\therefore \quad \frac{\Delta N}{N}=\frac{0.693}{T_{1 / 2}}$
$=\frac{0.693}{700} \times\left(4 \times 10^{-3}\right)$
$=3.96 \times 10^{-6}$