Step 1: Write down the given nuclear reaction. The reaction is: $$^{14}_{7}N + ^{1}_{0}n \rightarrow ^{14}_{6}C + X$$ Step 2: Apply the conservation of mass number (A). The sum of the mass numbers (superscripts) on the reactant side must equal the sum of the mass numbers on the product side. For the reactants: $14 + 1 = 15$ For the products: $14 + A_X$ So, $15 = 14 + A_X$ $A_X = 15 - 14 = 1$ Step 3: Apply the conservation of atomic number (Z). The sum of the atomic numbers (subscripts) on the reactant side must equal the sum of the atomic numbers on the product side. For the reactants: $7 + 0 = 7$ For the products: $6 + Z_X$ So, $7 = 6 + Z_X$ $Z_X = 7 - 6 = 1$ Step 4: Identify particle X based on its mass number and atomic number. Particle X has a mass number $A_X = 1$ and an atomic number $Z_X = 1$. This corresponds to a proton, which is represented as $^{1}_{1}p$ or $^{1}_{1}H$. Step 5: Compare with the given options. A: an electron ($^{0}_{-1}e$) B: a proton ($^{1}_{1}p$) C: a neutron ($^{1}_{0}n$) D: $\gamma$-radiation ($^{0}_{0}\gamma$) The particle with $A=1$ and $Z=1$ is a proton. The final answer is $\boxed{\text{B: a proton}}$.