The fusion reaction D + T $\to$ ^{4}He + n releases energy. Given atomic masses: m(^{2}H)=2.014102 u, m(^{3}H)=3.016049 u, m(^{4}He)=4.002603 u, m(n)=1.008665 u. (1 u = 931.5 MeV/$c^2$). (a) Calculate the total energy released per reaction. (b) If 1 g of deuterium (pure $D_2$ gas) reacts completely with tritium in stoichiometric proportion, find the total energy released in joules.

Step-by-step solution

1. Compute the total mass of reactants: $m_{react}$=2.014102 u + 3.016049 u = 5.030151 u.
2. Compute the total mass of products: $m_{prod}$=4.002603 u + 1.008665 u = 5.011268 u.
3. Find the mass defect: $\Delta$ m = $m_{react}-m_{prod}$=5.030151 u - 5.011268 u = 0.018883 u.
4. Convert the mass defect to energy: $E_{reaction}$=0.018883 u $\times$ 931.5 $\frac{MeV}{u}$=17.595 MeV $\approx$ 17.6 MeV.
5. Convert MeV to joules (1 MeV = 1.602$\times 10^{-13}$ J): $E_{reaction}$=17.595 $\times$ 1.602$\times 10^{-13}$ J = 2.818$\times 10^{-12}$ J per reaction.
6. Determine the number of deuterium nuclei in 1 g of D: atomic mass of D ≈2 g·$mo{l}^{-1}$, so 1 g corresponds to 0.5 mol = 0.5 $\times$ 6.022$\times 10^{23}$=3.011$\times 10^{23}$ nuclei.
7. Each reaction consumes one deuterium nucleus, thus the number of reactions equals 3.011$\times 10^{23}$.
8. Total energy released: $E_{total}$=3.011$\times 10^{23}$ $\times$ 2.818$\times 10^{-12}$ J = 8.48$\times 10^{11}$ J $\approx$ 8.5$\times 10^{11}$ J.

Concept tested: Energy from nuclear fusion reaction

Published 8 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked