PhysicsEasy183×since 2002Q6070When a U238{U^{238}}U238 nucleus originally at rest, decays by emitting an alpha particle having a speed ′u′,'u',′u′, the recoil speed of the residual nucleus isA4μ238{{4\mu } \over {238}}2384μB−4μ234- {{4\mu } \over {234}}−2344μC4μ234{{4\mu } \over {234}}2344μD−4μ238- {{4\mu } \over {238}}−2384μCheck answerSkip