There is a steam of neutrons with a kinetic energy of 0.335 eV. If the half-life of neutron is 700 s — Nuclear Chemistry and Radioactivity Chemistry Question
Question
There is a steam of neutrons with a kinetic energy of 0.335 eV. If the half-life of neutron is 700 s, what per cent will decay before they travel a distance of 80 km. Mass of neutron is $1.675 \times 10^{-27} \text{ kg}$. ($\ln 1.01 = 0.01$, $\ln 2 = 0.7$)
💡 Solution & Explanation
Kinetic Energy $E = 0.335 \text{ eV} = 0.335 \times 1.6 \times 10^{-19} \text{ J} = 5.36 \times 10^{-20} \text{ J}$. Velocity $v = \sqrt{2E/m} = \sqrt{\frac{2 \times 5.36 \times 10^{-20}}{1.675 \times 10^{-27}}} = 8 \times 10^3 \text{ m/s} = 8 \text{ km/s}$. Travel time $t = \frac{d}{v} = \frac{80 \text{ km}}{8 \text{ km/s}} = 10 \text{ s}$. Decay constant $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.7}{700} = 10^{-3} \text{ s}^{-1}$. Fraction decayed is $1 - e^{-\lambda t} = 1 - e^{-0.01}$. Since $\ln 1.01 = 0.01$, $e^{0.01} = 1.01 \implies e^{-0.01} = \frac{1}{1.01} \approx 0.99$. The fraction decayed is $1 - 0.99 = 0.01$, which is 1%. Therefore, correct answer is 1.