Question

A nucleus 220X at rest decays emitting an alpha particle. If the energy of the daughter nucleus is 0.2 MeV, what is the Q value of the reaction?

Answer 1

zX220 → z-2Y216 + 2He4 From energy and momentum conservation we can derive the equation for K.E. of alpha particle as K.E. of alpha particle = [(A−4)/4] x K.E. of daughter nucleus = [(220−4)/4] x 0.2 MeV = 10.8 MeV Q value of the reaction is given by Q = K.E of alpha particle + K.E of daughter nucleus = 10.8 MeV + 0.2 MeV = 11 MeV

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