If it is assumed that {}_{92}^{235}U decays o | vedclass

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(B) The emission of one $\alpha$ particle $({}_{2}^{4}He)$ decreases the mass number by $4$ and the atomic number by $2$.The emission of one $\beta$ p

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${}_{89}^{227}Ac$

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(B) The emission of one $\alpha$ particle $({}_{2}^{4}He)$ decreases the mass number by $4$ and the atomic number by $2$.The emission of one $\beta$ particle $({}_{-1}^{0}e)$ does not change the mass number but increases the atomic number by $1$.For the product ${}_{89}^{227}Ac$ from ${}_{92}^{235}U$:Change in mass number $= 235 - 227 = 8$, which corresponds to $2 \alpha$ particles $(2 	imes 4 = 8)$.Change in atomic number $= 92 - 89 = 3$.With $2 \alpha$ particles, the atomic number decreases by $2 	imes 2 = 4$. To reach $89$, we need to increase the atomic number by $1$ using one $\beta$ particle $(92 - 4 + 1 = 89)$.Thus, the decay involves $2 \alpha$ and $1 \beta$ particle, resulting in ${}_{89}^{227}Ac$.

If it is assumed that ${}_{92}^{235}U$ decays only by emitting $\alpha$ and $\beta$ particles, the possible product of the decay is:

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