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(C) To find $X$ and $Y$,we apply the law of conservation of atomic number $(Z)$ and mass number $(A)$ to both reactions.For reaction $I$: ${ }_{7}^{14

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$X$ is a proton and $Y$ is a neutron

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(C) To find $X$ and $Y$,we apply the law of conservation of atomic number $(Z)$ and mass number $(A)$ to both reactions.For reaction $I$: ${ }_{7}^{14} N +{ }_{2}^{4} He \longrightarrow{ }_{8}^{17} O + X$Sum of $Z$: $7 + 2 = 8 + Z_X \implies Z_X = 1$Sum of $A$: $14 + 4 = 17 + A_X \implies A_X = 1$Thus,$X$ is ${ }_{1}^{1} H$ (a proton).For reaction $II$: ${ }_{4}^{9} Be +{ }_{2}^{4} He \longrightarrow{ }_{6}^{12} C + Y$Sum of $Z$: $4 + 2 = 6 + Z_Y \implies Z_Y = 0$Sum of $A$: $9 + 4 = 12 + A_Y \implies A_Y = 1$Thus,$Y$ is ${ }_{0}^{1} n$ (a neutron).Therefore,$X$ is a proton and $Y$ is a neutron.

Consider the following nuclear reactions:
$I$. ${ }_{7}^{14} N +{ }_{2}^{4} He \longrightarrow{ }_{8}^{17} O + X$
$II$. ${ }_{4}^{9} Be +{ }_{2}^{4} He \longrightarrow{ }_{6}^{12} C + Y$
Then,

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Consider the following nuclear reactions:$I$. ${ }_{7}^{14} N +{ }_{2}^{4} He \longrightarrow{ }_{8}^{17} O + X$$II$. ${ }_{4}^{9} Be +{ }_{2}^{4} He \longrightarrow{ }_{6}^{12} C + Y$Then,