In the given reaction,_{92}U^{235} xrightarro | vedclass

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(B) The reaction sequence is: $_{92}U^{235}$ $\xrightarrow{-\alpha } (A)$ $\xrightarrow{-\beta } (B)$ $\xrightarrow{-\beta } (C)$$1$. Alpha decay of $

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$_{92}U^{235}$ and $C$

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(B) The reaction sequence is: $_{92}U^{235}$ $\xrightarrow{-\alpha } (A)$ $\xrightarrow{-\beta } (B)$ $\xrightarrow{-\beta } (C)$$1$. Alpha decay of $_{92}U^{235}$ results in $(A)$:$_{92}U^{235} \xrightarrow{-\alpha } _{90}A^{231}$$2$. Beta decay of $(A)$ results in $(B)$:$_{90}A^{231} \xrightarrow{-\beta } _{91}B^{231}$$3$. Beta decay of $(B)$ results in $(C)$:$_{91}B^{231} \xrightarrow{-\beta } _{92}C^{231}$Isotopes are atoms of the same element having the same atomic number but different mass numbers. Here,$_{92}U^{235}$ and $_{92}C^{231}$ both have an atomic number of $92$. Therefore,the isotopes are $_{92}U^{235}$ and $C$.

In the given reaction,$_{92}U^{235}$ $\xrightarrow{-\alpha } (A)$ $\xrightarrow{-\beta } (B)$ $\xrightarrow{-\beta } (C)$,the isotopes are:

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