Nuclear Fission Energy

Nuclear Fission Energy

When a heavy nucleus like uranium-235 absorbs a slow neutron, it splits into two smaller fragments and releases an enormous amount of energy. This is nuclear fission.

The Q-value

The energy released per reaction is given by the Q-value, computed from the mass difference between reactants and products:

$Q = (\sum m_{\text{reactants}} - \sum m_{\text{products}}) \times 931.494 \text{ MeV/u}$

The conversion factor $931.494$ MeV/u comes from Einstein's $E = mc^2$ applied to atomic mass units.

U-235 Fission Example

A typical reaction:

$^{235}\text{U} + n \rightarrow ^{92}\text{Kr} + ^{141}\text{Ba} + 3n$

Mass of reactants: $235.043930 + 1.008665 = 236.052595$ u
Mass of products: $91.926156 + 140.914411 + 3 \times 1.008665 = 235.857897$ u
$Q \approx 173.3$ MeV per fission event.

Energy Density

One kilogram of U-235 contains $N = N_A / M$ atoms (where $M = 235 \times 10^{-3}$ kg/mol). Multiplying by $Q$ (converted to joules via $1 \text{ MeV} = 1.602 \times 10^{-13}$ J) gives the total energy per kilogram — roughly $10^{13}$ J/kg, millions of times more than chemical explosives (~$4.6 \times 10^{6}$ J/kg for TNT).

Your Task

Implement the three functions below. All constants must be defined inside each function.

• fission_q_value(m_fuel_u, m_products_total_u) — Q-value in MeV
• fission_energy_per_kg(Q_MeV, molar_mass_kg_per_mol) — energy released per kg of fuel in J/kg ($N_A = 6.022 \times 10^{23}$, $1$ MeV $= 1.602 \times 10^{-13}$ J)
• fission_vs_chemical(Q_MeV, molar_mass_kg_per_mol) — ratio of fission energy to TNT ($4.6 \times 10^6$ J/kg)