1. Identify each reaction in List-I

A.

${}_0^1n + {}_{92}^{235}U \rightarrow {}_{54}^{140}Xe + {}_{38}^{94}Sr + 2{}_0^1n$ A heavy nucleus splits into two medium nuclei with neutrons emitted.

• This is nuclear fission.
• So, $A \to III$

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B.

$2H_2 + O_2 \rightarrow 2H_2O$ This is an ordinary reaction involving rearrangement of electrons and atoms, not nuclei.

• This is a chemical reaction.
• So, $B \to I$

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C.

${}_1^2H + {}_1^2H \rightarrow {}_2^3He + {}_0^1n$ Two light nuclei combine to form a heavier nucleus.

• This is fusion.
• Now check whether $Q$ is positive or negative.

Using approximate atomic masses: $m({}^2H) \approx 2.0141\,u$ So initial mass: $2\times 2.0141 = 4.0282\,u$

Final masses: $m({}^3He) \approx 3.0160\,u, \quad m(n) \approx 1.0087\,u$ $3.0160 + 1.0087 = 4.0247\,u$

Mass defect: $\Delta m = 4.0282 - 4.0247 = 0.0035\,u > 0$ Hence, $Q = \Delta m c^2 > 0$ So this is fusion with positive $Q$ value.

Therefore, $C \to II$

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D.

${}_1^1H + {}_1^3H \rightarrow {}_1^2H + {}_1^2H$ Again, light nuclei are involved, so classify as fusion/transmutation-type nuclear process and check $Q$.

Initial mass: $m({}^1H) + m({}^3H) \approx 1.0078 + 3.0160 = 4.0238\,u$

Final mass: $2m({}^2H) \approx 2\times 2.0141 = 4.0282\,u$

Mass defect: $\Delta m = 4.0238 - 4.0282 = -0.0044\,u < 0$ Thus, $Q < 0$ So this corresponds to fusion with negative $Q$ value.

Therefore, $D \to IV$

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2. Final matching

Thus the correct match is: $A-III, \; B-I, \; C-II, \; D-IV$

3. Compare with options

This corresponds to Option B.

4. Comparison with stored answer

Stored correct answer = B.

Our derived answer also = B.

So, the stored answer is correct.