The number of neutrons accompanying the formation of _54Xe^139 and _38Sr^94 from the absorption of s — Nuclear Chemistry and Radioactivity Chemistry Question
Question
The number of neutrons accompanying the formation of _54Xe^139 and _38Sr^94 from the absorption of slow neutron by _92U^235 by nuclear fission is
💡 Solution & Explanation
Step 1 - Write the Nuclear Fission Equation A slow neutron absorbed by $\ce{^{235}_{92}U}$ causes fission: $$\ce{^{235}_{92}U + ^{1}_{0}n -> ^{139}_{54}Xe + ^{94}_{38}Sr + k \cdot ^{1}_{0}n}$$ Step 2 - Conserve Mass Number (A) $$235 + 1 = 139 + 94 + k \times 1$$ $$236 = 233 + k$$ $$k = 3$$ Step 3 - Verify with Atomic Number (Z) Conservation $$92 + 0 = 54 + 38 + 0 \implies 92 = 92\ \checkmark$$ Step 4 - Evaluate Options - **(A) 0**: Violates mass number conservation ($236 \neq 233$). Incorrect. - **(B) 2**: $236 \neq 235$. Incorrect. - **(C) 1**: $236 \neq 234$. Incorrect. - **(D) 3**: $236 = 233 + 3$ ✓. **Correct.** $$\boxed{\text{Answer: D — 3 neutrons are produced}}$$