The particles emitted in the decay of { }_{9 | vedclass

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Question

(A) The given decay reaction is $ { }_{92}^{238} U \rightarrow { }_{92}^{234} U $. An $\alpha$-decay is represented as $ { }_{Z}^{A} X \rightarrow { }

Vedclass · Accepted Answer

$1 \alpha$ and $2 \beta$

Vedclass · Answer

(A) The given decay reaction is $ { }_{92}^{238} U ightarrow { }_{92}^{234} U $. An $\alpha$-decay is represented as $ { }_{Z}^{A} X ightarrow { }_{Z-2}^{A-4} Y + { }_{2}^{4} 	ext{He} $. $A$ $\beta$-decay is represented as $ { }_{Z}^{A} X ightarrow { }_{Z+1}^{A} Y + { }_{-1}^{0} e $. In the decay of $ { }_{92}^{238} U $ to $ { }_{92}^{234} U $,the change in mass number $\Delta A = 238 - 234 = 4$. Since each $\alpha$-particle carries a mass number of $4$,the number of $\alpha$-particles emitted is $4/4 = 1$. Emission of $1$ $\alpha$-particle reduces the atomic number $Z$ by $2$,so the atomic number becomes $92 - 2 = 90$. However,the final atomic number is $92$. To increase the atomic number from $90$ to $92$,we need $2$ $\beta$-decays,as each $\beta$-decay increases $Z$ by $1$. Therefore,$1$ $\alpha$ and $2$ $\beta$ particles are emitted.

The particles emitted in the decay of $ { }_{92}^{238} U $ to $ { }_{92}^{234} U $ are:

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