A radioactive nucleus is initially at rest. I | vedclass

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Question

(C) Let the momentum of the electron be $\vec{P_e}$ and the momentum of the neutrino be $\vec{P_n}$. Since they are emitted at right angles,the result

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$\pi - 	an^{-1}(2)$

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(C) Let the momentum of the electron be $\vec{P_e}$ and the momentum of the neutrino be $\vec{P_n}$. Since they are emitted at right angles,the resultant momentum of the electron and neutrino is $\vec{P_{en}} = \vec{P_e} + \vec{P_n}$.By the law of conservation of linear momentum,the initial momentum of the nucleus is zero,so the recoil momentum of the nucleus $\vec{P_R}$ must be equal and opposite to the resultant momentum of the electron and neutrino: $\vec{P_R} = -(\vec{P_e} + \vec{P_n})$.The magnitude of the resultant momentum is $P_{en} = \sqrt{P_e^2 + P_n^2}$.The angle $	heta$ that the resultant momentum $\vec{P_{en}}$ makes with the electron's momentum $\vec{P_e}$ is given by $	an 	heta = \frac{P_n}{P_e} = \frac{6.4 	imes 10^{-23}}{3.2 	imes 10^{-23}} = 2$,so $	heta = 	an^{-1}(2)$.Since the recoil momentum $\vec{P_R}$ is in the opposite direction to $\vec{P_{en}}$,the angle $\phi$ that the recoil momentum makes with the electron's momentum is $\phi = \pi - 	heta = \pi - 	an^{-1}(2)$.

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