Which of the following nuclear reaction occurs in nature of the formation of tritium? — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Conservation Check for Each Option We verify both atomic number (Z) and mass number (A) conservation for each option, and check which occurs naturally in the atmosphere. Step 2 - Option A: $\ce{^{6}_{3}Li + ^{1}_{0}n -> ^{4}_{2}He + ^{3}_{1}H}$ $$Z: 3+0 = 2+1 = 3\ \checkmark \quad A: 6+1 = 4+3 = 7\ \checkmark$$ This reaction is balanced but occurs in nuclear reactors (controlled neutron bombardment of Li-6), not naturally in nature. Step 3 - Option B: $\ce{^{10}_{5}B + ^{1}_{0}n -> 2^{4}_{2}He + ^{3}_{1}H}$ $$Z: 5+0 = 4+1 = 5\ \checkmark \quad A: 10+1 = 8+3 = 11\ \checkmark$$ This also occurs in reactors/laboratories using boron moderators, not naturally. Step 4 - Option C: $\ce{^{14}_{7}N + ^{1}_{0}n -> ^{12}_{6}C + ^{3}_{1}H}$ $$Z: 7+0 = 6+1 = 7\ \checkmark \quad A: 14+1 = 12+3 = 15\ \checkmark$$ This reaction occurs **naturally** in the upper atmosphere. Cosmic rays produce high-energy neutrons, which collide with abundant nitrogen-14 in the atmosphere to produce tritium ($\ce{^{3}H}$) and Carbon-12. **This is the natural source of atmospheric tritium. Correct.** Step 5 - Option D: $\ce{^{9}_{4}Be + ^{2}_{1}H -> 2^{4}_{2}He + ^{3}_{1}H}$ $$Z: 4+1 = 4+1 = 5\ \checkmark \quad A: 9+2 = 8+3 = 11\ \checkmark$$ Requires deuteron bombardment — not a natural process. $$\boxed{\text{Answer: C — } \ce{^{14}_{7}N + ^{1}_{0}n -> ^{12}_{6}C + ^{3}_{1}H}}$$