Match the LIST-I with LIST-II
| LIST-I Nuclear reaction | LIST-II Energy released (E) |
| A. 11H + 11H → 21H + e+ + ν + E | I. 5.49 MeV |
| B. e+ + e– → γ + γ + E | II. 12.86 MeV |
| C. 21H + 11H → 32He + γ + E | III. 0.42 MeV |
| D. 32He + 32He → 42He + 11H + 11H + E | IV. 1.02 MeV |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A - III, B - IV, C - I, D - II
The energy released is the mass defect converted to energy. The reactions are steps of the proton–proton cycle in the Sun, plus electron–positron annihilation.
(A) 11H + 11H → 21H + e+ + ν + E — this matches (III) 0.42 MeV. Two protons fuse into a deuteron, releasing the least energy.
(B) e+ + e– → γ + γ + E — this matches (IV) 1.02 MeV. In annihilation the rest mass of the pair becomes energy: \( 2 \times 0.511 \) MeV.
(C) 21H + 11H → 32He + γ + E — this matches (I) 5.49 MeV. A deuteron captures a proton to form helium-3.
(D) 32He + 32He → 42He + 11H + 11H + E — this matches (II) 12.86 MeV. Forming tightly bound helium-4 releases the most energy.
NCERT (2026-27):
Nuclear Fusion – Energy Generation in Stars – “… The proton-proton (p, p) cycle by which this occurs is represented by the following sets of reactions:
\( \displaystyle \begin{array}{l} {}^{1}_{1}\mathrm{H} + {}^{1}_{1}\mathrm{H} \rightarrow {}^{2}_{1}\mathrm{H} + e^{+} + \nu + 0.42\ \mathrm{MeV} \qquad \text{(i)} \\ e^{+} + e^{-} \rightarrow \gamma + \gamma + 1.02\ \mathrm{MeV} \qquad \text{(ii)} \\ {}^{2}_{1}\mathrm{H} + {}^{1}_{1}\mathrm{H} \rightarrow {}^{3}_{2}\mathrm{He} + \gamma + 5.49\ \mathrm{MeV} \qquad \text{(iii)} \\ {}^{3}_{2}\mathrm{He} + {}^{3}_{2}\mathrm{He} \rightarrow {}^{4}_{2}\mathrm{He} + {}^{1}_{1}\mathrm{H} + {}^{1}_{1}\mathrm{H} + 12.86\ \mathrm{MeV} \qquad \text{(iv)} \end{array} \qquad (13.14) \)”