JEE Main 2025 · 3 April · morning · Single correct · +4 / −1
Match the LIST-I with LIST-II
Choose the correct answer from the options given below:
| List - I | List - II | ||
|---|---|---|---|
| A. | ₀¹ n+ ₉₂²³⁵ U → ₅₄¹⁴⁰ Xe+ ₃₈⁹⁴ Sr+2 ₀¹ n | I. | Chemical reaction |
| B. | 2 H₂+O₂ → 2 H₂ O | II. | Fusion with +ve Q value |
| C. | ₁² H+ ₁² H → ₂³ He+ ₀¹ n | III. | Fission |
| D. | ₁¹ H+ ₁³ H → ₁² H+ ₁² H | IV. | Fusion with -ve Q value |
- (a) A-II, B-I, C-III, D-IV
- (b) A-III, B-I, C-II, D-IV
- (c) A-III, B-I, C-IV, D-II
- (d) A-II, B-I, C-IV, D-III
Answer and solution
Answer (b) A-III, B-I, C-II, D-IV
Solution
Physics sir · 3 steps
Split a heavy nucleus or join light ones, then let binding energy decide the sign of Q.
- Sort by type. In A a neutron hits ²³⁵U and it breaks into two medium nuclei, so that is fission. In B only electrons rearrange between atoms and the nuclei stay unchanged, so that is a chemical reaction. C and D both combine light nuclei, so they are the two fusion entries.
- Sign of Q. Q is the total binding energy of the products minus that of the reactants. Use B(²H) ≈ 2.22 MeV, B(³H) ≈ 8.48 MeV and B(³He) ≈ 7.72 MeV. A free proton or neutron has zero binding energy.
Q = Σ B_(products) - Σ B_(reactants)
Reaction What happens Q Match A ²³⁵U splits into Xe and Sr large, positive III (fission) B 2H₂ + O₂ → 2H₂O, nuclei unchanged a few eV only I (chemical) C ²H + ²H → ³He + n 7.72 - 4.45 = +3.27 MeV II (fusion, +ve Q) D ¹H + ³H → 2 ²H 4.45 - 8.48 = -4.03 MeV IV (fusion, −ve Q) - So. The matching is A-III, B-I, C-II, D-IV, which is option (b). D has a negative Q because the tightly bound triton turns into two loosely bound deuterons.
The idea to keep. To find the sign of Q in a nuclear reaction, compare total binding energies. If the products are bound more tightly than the reactants, energy is released and Q is positive.
Answer from the NTA final answer key. Solution by PrepWiser.