The half-life of a neutron is 693 s. What fr | vedclass

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(A) Given, half-life $t_{1/2} = 693 \ s$. Kinetic energy $K = 0.084 \ eV = 0.084 	imes 1.6 	imes 10^{-19} \ J$.Using $K = \frac{1}{2}mv^2$, we find

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$0.25 	imes 10^{-5}$

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(A) Given, half-life $t_{1/2} = 693 \ s$. Kinetic energy $K = 0.084 \ eV = 0.084 	imes 1.6 	imes 10^{-19} \ J$.Using $K = \frac{1}{2}mv^2$, we find velocity $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 	imes 0.084 	imes 1.6 	imes 10^{-19}}{1.68 	imes 10^{-27}}} = \sqrt{0.16 	imes 10^8} = 0.4 	imes 10^4 \ m/s$.Time taken to travel $d = 1000 \ m$ is $t = \frac{d}{v} = \frac{1000}{0.4 	imes 10^4} = 0.25 \ s$.Using the decay law $N = N_0 e^{-\lambda t}$, the fraction decayed is $\frac{N_0 - N}{N_0} = 1 - e^{-\lambda t}$.Since $\lambda t = \frac{\ln 2}{t_{1/2}} 	imes t$ is very small, $1 - e^{-\lambda t} \approx \lambda t = \frac{\ln 2 	imes t}{t_{1/2}}$.Fraction decayed $= \frac{0.693 	imes 0.25}{693} = 0.25 	imes 10^{-3} 	imes 10^{-2} = 0.25 	imes 10^{-5}$.

The half-life of a neutron is $693 \ s$. What fraction of neutrons will decay when a beam of neutrons, having kinetic energy of $0.084 \ eV$, travels a distance of $1 \ km$? (mass of neutron $= 1.68 \times 10^{-27} \ kg$, and $\ln 2 = 0.693$)

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