Match List-I with List-II.
$$\begin{array}{|l|l|l|l|}\hline & \text{List-I} & & \text{List-II} \\ \hline \text{A.} & {}^1_0n + {}^{235}_{92}U \to {}^{140}_{54}Xe + {}^{94}_{38}Sr + 2\,{}^1_0n & \text{I.} & \text{Chemical reaction} \\ \hline \text{B.} & 2H_2 + O_2 \to 2H_2O & \text{II.} & \text{Fusion with +ve Q value} \\ \hline \text{C.} & {}^2_1H + {}^2_1H \to {}^3_2He + {}^1_0n & \text{III.} & \text{Fission} \\ \hline \text{D.} & {}^1_1H + {}^3_1H \to {}^2_1H + {}^2_1H & \text{IV.} & \text{Fusion with -ve Q value} \\ \hline \end{array}$$
Choose the correct answer from the options given below:
Answer: (B) A-III, B-I, C-II, D-IV
- A. A heavy nucleus splits after absorbing a neutron: fission → III
- B. Only electrons rearrange; nuclei are unchanged: chemical reaction → I
- C. Two deuterons fuse into helium-3 and release energy (D–D fusion, $Q \approx +3.3\ \text{MeV}$) → II
- D. Light nuclei combine, but the products (two deuterons, binding energy $2\times2.22 = 4.45\ \text{MeV}$) are less tightly bound than the reactants (tritium, $8.48\ \text{MeV}$), so energy must be supplied: $Q < 0$ → IV
Solution by Sreeraj P, M.Sc Physics