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(B) In a nuclear reaction,both the total atomic number $(Z)$ and the total mass number $(A)$ are conserved.For reaction $(I): _{92}^{235}U + _{0}^{1}n

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$(I) \, 57, 148; (II) \, 2, 4$

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(B) In a nuclear reaction,both the total atomic number $(Z)$ and the total mass number $(A)$ are conserved.For reaction $(I): _{92}^{235}U + _{0}^{1}n \rightarrow X + _{35}^{85}Br + 3 _{0}^{1}n$Mass number of $X$: $A_X = 235 + 1 - 85 - 3(1) = 236 - 88 = 148$.Atomic number of $X$: $Z_X = 92 + 0 - 35 - 3(0) = 57$.So,$X$ is $_{57}^{148}X$.For reaction $(II): _{3}^{6}Li + _{1}^{2}H \rightarrow Y + _{2}^{4}He$Mass number of $Y$: $A_Y = 6 + 2 - 4 = 4$.Atomic number of $Y$: $Z_Y = 3 + 1 - 2 = 2$.So,$Y$ is $_{2}^{4}Y$.Thus,for $X$,$Z=57, A=148$ and for $Y$,$Z=2, A=4$.

For the following two nuclear reactions,the atomic number and mass number of nuclei $X$ and $Y$ are respectively:
$(I) \, _{92}^{235}U + _{0}^{1}n \rightarrow X + _{35}^{85}Br + 3 _{0}^{1}n$
$(II) \, _{3}^{6}Li + _{1}^{2}H \rightarrow Y + _{2}^{4}He$

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For the following two nuclear reactions,the atomic number and mass number of nuclei $X$ and $Y$ are respectively:$(I) \, _{92}^{235}U + _{0}^{1}n \rightarrow X + _{35}^{85}Br + 3 _{0}^{1}n$$(II) \, _{3}^{6}Li + _{1}^{2}H \rightarrow Y + _{2}^{4}He$