In the fusion reaction ${}^3\mathrm{H}{+}^3\mathrm{He}{\to }^6\mathrm{Li}+\gamma$, the atomic masses are: $m{(}^3\mathrm{H})=3.016049$ u, $m{(}^3\mathrm{He})=3.016029$ u, $m{(}^6\mathrm{Li})=6.015122$ u, and photon mass is zero. (a) Calculate the energy released per reaction in MeV. (b) If 0.5 g of tritium reacts completely with 0.5 g of helium‑3, find the total energy released in joules.

Step-by-step solution

1. 1. (a) Write total initial mass: $m_i=3.016049+3.016029=6.032078$ u.
2. 2. Final mass: $m_f=6.015122$ u (photon has zero mass).
3. 3. Mass defect $Deltam=m_i-m_f=6.032078-6.015122=0.016956$ u.
4. 4. Convert to energy: $Q=\Delta m\times 931.5=0.016956\times 931.5=15.80$ MeV. (This is too high, check).
5. 5. Re‑examine masses: Actual mass defect should be small. Use precise masses: $m{(}^6\mathrm{Li})=6.015122$ u is correct. Compute again: $Deltam=6.032078-6.015122=0.016956$ u. This yields $Q=15.80$ MeV which is realistic for this reaction.
6. 6. (b) Energy per reaction: $E_f=15.80\text{MeV}=15.80\times 1.602\times 10^{-13}=2.53\times 10^{-12}$ J.
7. 7. Number of tritium nuclei in 0.5 g: $n_t=0.5\times 10^{-3},\text{kg}/(3\times 1.6605\times 10^{-27})=0.5\times 10^{-3}/(4.9815\times 10^{-27})=1.00\times 10^{23}$ nuclei.
8. 8. Number of helium‑3 nuclei in 0.5 g: $n_{He}=0.5\times 10^{-3},\text{kg}/(3\times 1.6605\times 10^{-27})=1.00\times 10^{23}$ nuclei (same).
9. 9. Each reaction consumes one tritium and one helium‑3, so total reactions = $1.00\times 10^{23}$.
10. 10. Total energy released: $E_{total}=1.00\times 10^{23}\times 2.53\times 10^{-12}=2.53\times 10^{11}$ J.
11. 11. Final answers: (a) 15.80 MeV, (b) $2.53\times 10^{11}$ J.

Concept tested: Fusion reaction energy calculation

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked