Question

In certain radioactive beta decay processes, the beta particle (an electron) leaves the atomic nucleus with a speed of99.95% the speed of light relative to the decaying nucleus. In the laboratory frame, the nucleus is moving at 75.00% the speed of light, and the electron is emitted in the same direction the nucleus is moving. Find the energy and momentum of the electron as measured in the laboratory frame and in the rest frame of the decaying nucleus.
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Consi der x is laboratory frame and x be the nucleus fram of reference If the emitted elctron travel in the same directi on of the nucleus 0.9995c 0.7500c +0.9995c 0.7500c 0.999929c (b)Ifthe emitted elctron travel in the same direction of the nucleus v+u -0.9995c 0.7500c (-0.9995c) (0.7500c) =-0.9965e c )Ifthe emitted electron in the same direction ic (0.511MeV) 0.999929 = 42.4 MeV (0.511MeV) 0.9995 15.7 MeV

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