From the mass defect to the reaction Q-value

Problem

Consider the fusion reaction $^{2}_{1}\text{H} +\ ^{3}_{1}\text{H} \to\ ^{4}_{2}\text{He} + n$. The atomic masses are: $^{2}\text{H} = 2{,}01410\ \text{u}$, $^{3}\text{H} = 3{,}01605\ \text{u}$, $^{4}\text{He} = 4{,}00260\ \text{u}$, neutron $= 1{,}00867\ \text{u}$. Calculate the mass defect $\Delta m$ and the energy released $Q$ in MeV, knowing that $1\ \text{u} \cdot c^2 = 931{,}5\ \text{MeV}$.

Solution

Mass defect. It is the difference between the total mass of the reactants and that of the products:

$\Delta m = (m_{^2\text{H}} + m_{^3\text{H}}) - (m_{^4\text{He}} + m_n)$

Substituting:

$\Delta m = (2{,}01410 + 3{,}01605) - (4{,}00260 + 1{,}00867) = 5{,}03015 - 5{,}01127 = 0{,}01888\ \text{u}$

Energy released. Convert the mass defect into energy with the factor $931{,}5\ \text{MeV/u}$:

$Q = \Delta m \cdot 931{,}5 = 0{,}01888 \cdot 931{,}5 \approx 17{,}6\ \text{MeV}$

$\ev{Q \approx 17{,}6\ \text{MeV}}$

This is the deuterium-tritium reaction, the most promising for controlled fusion: it releases about $17{,}6\ \text{MeV}$ from extremely light nuclei, an enormous amount compared with chemical reactions.

Connections

Topics: Nuclear physics Concepts: Nuclear fusion · Mass-energy equivalence