The particles emitted in the decay of \( { }_{92}^{238} U \) to \( { }_{92}^{234} U \):

Given decay reaction is
\({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{92}{ }^{234} \mathrm{U}\)
We know that, a-decay is
\(z^{A} X \rightarrow z-2^{A-4} Y+\alpha-\) particle
\beta-decay is
\(z^{A} X \rightarrow z-1^{A} X+\beta-\) 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 a particles emitted is \(1 .\)
Now, emission of an a-particle implies atomic number reduces by 2 but there is no change in atomic number
Therefore, number of \(\beta\) particles emitted is 2 . Thus, the given reaction
\({ }_{92}^{238} U \rightarrow{ }_{92}^{234} U+1 \alpha+2 \beta^{-}\)