For the following fusion reaction, $4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \beta^{+}+2 \nu+\gamma$ the Q-value (energy of the reaction) in MeV (rounded off to one decimal place) is ____ . (Given: Mass of ${ }^{1} \mathrm{H}$ nucleus is 1.007825 u and mass of ${ }^{4} \mathrm{He}$ nucleus is 4.002604 u )
Answer
SELF-PRACTICE — the source book printed no answer.
Nothing is invented here, so this question has no answer on record.
Explanation
Answer: 26.7 MeV — Δm = 4 × 1.007825 − 4.002604 = 0.028696 u (the given values are atomic masses, so positron annihilation is included); Q = 0.028696 × 931.5 = 26.7 MeV.