Which one of the following particles is used to bombard _13Al^27 to give _15P^30 and a neutron? — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Set Up the Nuclear Reaction Let the unknown bombarding particle be $\ce{^{a}_{z}X}$: $$\ce{^{27}_{13}Al + ^{a}_{z}X -> ^{30}_{15}P + ^{1}_{0}n}$$ Step 2 - Conserve Atomic Number (Z) $$13 + z = 15 + 0 \implies z = 2$$ Step 3 - Conserve Mass Number (A) $$27 + a = 30 + 1 \implies a = 4$$ Step 4 - Identify the Particle $z = 2$, $a = 4$ → This is the **alpha particle** $\ce{^{4}_{2}He}$. The complete reaction is: $$\ce{^{27}_{13}Al + ^{4}_{2}He -> ^{30}_{15}P + ^{1}_{0}n}$$ Step 5 - Evaluate Options - **(A) $\ce{_1H^2}$ (deuteron)**: $z=1 \neq 2$. Would not conserve charge. Incorrect. - **(B) γ (gamma photon)**: $z=0, a=0$; gives $P^{27}$, not $P^{30}$. Incorrect. - **(C) α particle**: $z=2, a=4$ ✓. Conserves both charge and mass. **Correct.** - **(D) β**: $z=-1$; would give a product with $Z=16 \neq 15$. Incorrect. $$\boxed{\text{Answer: C — Alpha particle (}\ \ce{^{4}_{2}He}\ \text{)}}$$