Nuclear Fusion Energy

Nuclear Fusion Energy

Fusion combines light nuclei to form a heavier nucleus, releasing far more energy per unit mass than fission. It powers the Sun and is the goal of tokamak research reactors.

Deuterium–Tritium (D-T) Fusion

The most promising reaction for terrestrial reactors:

$^2\text{H} + ^3\text{H} \rightarrow ^4\text{He} + n$

$Q = (2.014102 + 3.016049 - 4.002602 - 1.008665) \times 931.494 \approx 17.59 \text{ MeV}$

Energy per Kilogram

The energy per kg of D-T reactant mixture is:

$E/\text{kg} = \frac{Q_{\text{J}}}{m_{\text{pair}}} = \frac{Q \times 1.602 \times 10^{-13}}{(m_D + m_T) \times 1.66054 \times 10^{-27}}$

This gives roughly $3.4 \times 10^{14}$ J/kg — about 7 million times more than gasoline ($4.6 \times 10^7$ J/kg).

Lawson Criterion

For a D-T plasma to sustain fusion, the product of ion density $n$ and energy confinement time $\tau_E$ must exceed:

$n \tau_E > \frac{10^{20}}{T_{\text{keV}}} \quad [\text{m}^{-3}\text{s}]$

At $T = 10$ keV, the threshold is $n\tau_E > 10^{19}$ m$^{-3}$s.

Your Task

All constants must be defined inside each function.

• dt_fusion_q() — D-T Q-value in MeV (no parameters)
• fusion_energy_per_kg(Q_MeV, m_reactants_u_per_reaction) — energy in J/kg ($1$ u $= 1.66054 \times 10^{-27}$ kg)
• lawson_criterion_DT(T_keV) — minimum $n\tau_E$ threshold in m$^{-3}$s