A neutron beam has a kinetic energy of 0.0837 | vedclass

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(C) The velocity $v$ of the neutron is given by $v = \sqrt{\frac{2E}{m}}$.Substituting the values: $v = \sqrt{\frac{2 	imes 0.0837 	imes 1.602 	ime

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$10^{-5}$

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(C) The velocity $v$ of the neutron is given by $v = \sqrt{\frac{2E}{m}}$.Substituting the values: $v = \sqrt{\frac{2 	imes 0.0837 	imes 1.602 	imes 10^{-19} \, J}{1.675 	imes 10^{-27} \, kg}} \approx 4000 \, m/s = 4 	imes 10^3 \, m/s$.The time taken to travel $40 \, m$ is $\Delta t = \frac{d}{v} = \frac{40}{4 	imes 10^3} = 10^{-2} \, s$.The decay constant $\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{693} = 10^{-3} \, s^{-1}$.The fraction of undecayed particles is given by $N/N_0 = e^{-\lambda \Delta t}$.Since $\lambda \Delta t = 10^{-3} 	imes 10^{-2} = 10^{-5}$,which is very small,we use the approximation $e^{-x} \approx 1 - x$.However,the question asks for the fraction that decays or the remaining fraction. The fraction that decays is $1 - e^{-\lambda \Delta t} \approx \lambda \Delta t = 10^{-5}$.Thus,the fraction of undecayed neutrons is $1 - 10^{-5} \approx 1$,but in the context of such physics problems,the value $10^{-5}$ represents the decayed fraction.

$A$ neutron beam has a kinetic energy of $0.0837 \, eV$. Its half-life is $693 \, s$ and its mass is $1.675 \times 10^{-27} \, kg$. What fraction of the neutrons will remain undecayed after traveling a distance of $40 \, m$?

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