In the nuclear reaction _{85}X^{297} to Y + 4 | vedclass

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Question

(C) The given nuclear reaction is $_{85}X^{297} 	o Y + 4\alpha$.An alpha particle is represented as $_2He^4$.Therefore,the reaction can be written as

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$_{77}Y^{281}$

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(C) The given nuclear reaction is $_{85}X^{297} 	o Y + 4\alpha$.An alpha particle is represented as $_2He^4$.Therefore,the reaction can be written as $_{85}X^{297} 	o Y + 4(_2He^4)$.To find the atomic number $(Z)$ and mass number $(A)$ of $Y$,we apply the conservation of charge and mass number.For the mass number $(A)$: $297 = A + 4 	imes 4 \implies 297 = A + 16 \implies A = 297 - 16 = 281$.For the atomic number $(Z)$: $85 = Z + 4 	imes 2 \implies 85 = Z + 8 \implies Z = 85 - 8 = 77$.Thus,the resulting nucleus $Y$ is $_{77}Y^{281}$.

In the nuclear reaction $_{85}X^{297} \to Y + 4\alpha$,$Y$ is:

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