The proton-proton cycle in the sun is made up of the following reactions:
(i) \({}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}\rightarrow{}^{2}_{1}\mathrm{H}+e^{+}+\nu+0.42\ \mathrm{MeV}\)
(ii) \(e^{+}+e^{-}\rightarrow\gamma+\gamma+1.02\ \mathrm{MeV}\)
(iii) \({}^{2}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}\rightarrow{}^{3}_{2}\mathrm{He}+\gamma+5.49\ \mathrm{MeV}\)
(iv) \({}^{3}_{2}\mathrm{He}+{}^{3}_{2}\mathrm{He}\rightarrow{}^{4}_{2}\mathrm{He}+{}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}+Q\)
The combination 2(i) + 2(ii) + 2(iii) + (iv) releases a total energy of 26.7 MeV. The value of \(Q\) is about
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 12.8 MeV
Reaction (iv) needs two \( {}^{3}_{2}\mathrm{He} \) nuclei. So reactions (i), (ii) and (iii) must each take place twice for one reaction (iv). The total 26.7 MeV is the sum of all these energies.
\( \begin{aligned} E_{\mathrm{(i)-(iii)}} &= 2\times(0.42+1.02+5.49) \\ &= 2\times6.93 \\ &= 13.86\ \mathrm{MeV} \end{aligned} \)
\( \begin{aligned} Q &= 26.7-13.86 \\ &= 12.84\ \mathrm{MeV} \\ &\approx 12.8\ \mathrm{MeV} \end{aligned} \)
Therefore, reaction (iv) releases about 12.8 MeV. The more exact value is 12.86 MeV. The small difference comes from rounding the total to 26.7 MeV.
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:”
\( {}^{3}_{2}\mathrm{He}+{}^{3}_{2}\mathrm{He}\rightarrow{}^{4}_{2}\mathrm{He}+{}^{1}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}+ \)\( 12.86\ \mathrm{MeV} \)\( \qquad \text{(iv)} \)
“For the fourth reaction to occur, the first three reactions must occur twice, in which case two light helium nuclei unite to form ordinary helium nucleus. …”