KCET 2017 Physics Solved Paper

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−