The two possible nuclei to complete the reactions given below are, respectively

(1) ${^{}_7N}^{14}+ {^{}_2He}^4 = {^{}_8O}^{17} +----$
(2) ${^{}_8O}^{16} + {^{}_1H}^2 = {^{}_7N}^{14} + ------$

${^{}_1H}^1, {^{}_2He}^4$

The correct answer is Option (2) → ${^{}_1H}^1, {^{}_2He}^4$

For nuclear reactions, use conservation of mass number (A) and atomic number (Z).

(1) $^7_14\text{N} + ^2_4\text{He} \rightarrow ^8_{17}\text{O} + X$

Mass numbers: $14 + 4 = 17 + A_X \Rightarrow A_X = 1$

Atomic numbers: $7 + 2 = 8 + Z_X \Rightarrow Z_X = 1$

∴ $X = ^1_1\text{H}$ (proton)

(2) $^8_{16}\text{O} + ^1_2\text{H} \rightarrow ^7_{14}\text{N} + Y$

Mass numbers: $16 + 2 = 14 + A_Y \Rightarrow A_Y = 4$

Atomic numbers: $8 + 1 = 7 + Z_Y \Rightarrow Z_Y = 2$

∴ $Y = ^2_4\text{He}$ (alpha particle)

∴ The two possible nuclei are: $^1_1\text{H}$ and $^2_4\text{He}$