Complete the following nuclear reaction:{ }_{ | vedclass

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(B) In a nuclear fission reaction,both the total mass number $(A)$ and the total atomic number $(Z)$ must be conserved.Given reaction: ${ }_{0}^{1} n+

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${ }_{51}^{133} Sb, 4{ }_{0}^{1} n$

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(B) In a nuclear fission reaction,both the total mass number $(A)$ and the total atomic number $(Z)$ must be conserved.Given reaction: ${ }_{0}^{1} n+{ }_{92}^{235} U \rightarrow{ }_{92}^{236} U \rightarrow X + { }_{41}^{99} Nb + Y{ }_{0}^{1} n$.Conservation of atomic number $(Z)$:$92 = Z_X + 41 \Rightarrow Z_X = 92 - 41 = 51$.Conservation of mass number $(A)$:$236 = A_X + 99 + Y(1)$.For option $(B)$,$X = { }_{51}^{133} Sb$ and $Y = 4$.$A_X = 133$.$133 + 99 + 4 = 236$.Since both $A$ and $Z$ are conserved,option $(B)$ is correct.

Complete the following nuclear reaction:
${ }_{0}^{1} n+{ }_{92}^{235} U \rightarrow{ }_{92}^{236} U \rightarrow$ $\qquad$ $+{ }_{41}^{99} Nb+$ $\qquad$.

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Complete the following nuclear reaction:${ }_{0}^{1} n+{ }_{92}^{235} U \rightarrow{ }_{92}^{236} U \rightarrow$ $\qquad$ $+{ }_{41}^{99} Nb+$ $\qquad$.