Given decay reaction is 92238U⟶92234U We know that, α-decay is ZAX→Z−2A−4Y+α − particle β-decay is ZAX→Z−1AX+β − particle In above decay of uranium, change in mass number A is 4 and change in atomic number Z is 0. Change in mass number A implies number of α particles emitted is 1. Now, emission of an α-particle implies atomic number reduces by 2 but there is no change in atomic number Therefore, number of β particles emitted is 2. Thus, the given reaction 92238U⟶92234U+1α+2β−