In the transformation of {}_{92}^{238}U to {} | vedclass

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(C) The nuclear transformation is ${}_{92}^{238}U ightarrow {}_{92}^{234}U + 	ext{emissions}$.First,the emission of one $\alpha$-particle $({}_{2}^

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two $\beta^-$

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(C) The nuclear transformation is ${}_{92}^{238}U ightarrow {}_{92}^{234}U + 	ext{emissions}$.First,the emission of one $\alpha$-particle $({}_{2}^{4}He)$ results in: ${}_{92}^{238}U ightarrow {}_{90}^{234}Th + {}_{2}^{4}He$.To reach the final product ${}_{92}^{234}U$,the ${}_{90}^{234}Th$ must undergo beta decay: ${}_{90}^{234}Th ightarrow {}_{91}^{234}Pa + {}_{-1}^{0}e$ and ${}_{91}^{234}Pa ightarrow {}_{92}^{234}U + {}_{-1}^{0}e$.Thus,the total emissions are one $\alpha$-particle and two $\beta^-$-particles.

In the transformation of ${}_{92}^{238}U$ to ${}_{92}^{234}U$,if one emission is an $\alpha$-particle,what should be the other emission$(s)$?

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