In a hypothetical fission reaction{ }_{92} X^ | vedclass

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(C) In a nuclear reaction,both the total atomic number $(Z)$ and the total mass number $(A)$ must be conserved.For the given reaction: ${ }_{92} X^{23

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(C) In a nuclear reaction,both the total atomic number $(Z)$ and the total mass number $(A)$ must be conserved.For the given reaction: ${ }_{92} X^{236} \rightarrow{ }_{56} Y^{141}+{ }_{36} Z^{92}+3 R$Checking the atomic number $(Z)$:$LHS$: $Z = 92$$RHS$: $Z = 56 + 36 = 92$Since $92 = 92$,the atomic number is conserved.Checking the mass number $(A)$:$LHS$: $A = 236$$RHS$: $A = 141 + 92 + 3(A_R) = 233 + 3(A_R)$For conservation of mass number: $236 = 233 + 3(A_R)$$3(A_R) = 3$$A_R = 1$Since the particle $R$ has a mass number of $1$ and an atomic number of $0$ (as $Z$ is already balanced),the particle $R$ is a neutron $({ }_{0} n^{1})$.

In a hypothetical fission reaction
${ }_{92} X^{236} \rightarrow{ }_{56} Y^{141}+{ }_{36} Z^{92}+3 R$
The identity of emitted particles $(R)$ is :

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In a hypothetical fission reaction${ }_{92} X^{236} \rightarrow{ }_{56} Y^{141}+{ }_{36} Z^{92}+3 R$The identity of emitted particles $(R)$ is :