Lesson 6: Fission & Fusion Energy | Modern Physics

Lesson at a Glance | He Tirohanga Whakamua

Do NowWhy fusion powers stars while nuclear reactors currently use fission10 min

Fission EnergeticsU-235 neutron capture, chain reactions, and

Δ m

energy yield15 min

Fusion EnergeticsDeuterium-Tritium D-T fusion, Coulomb barrier, Tokamaks15 min

Portfolio EntryWrite Section 6: Fission vs Fusion Reaction Energetics Matrix10 min

Exit DrillCalculate energy release for D-T fusion in MeV (

Δ m = 0.01888 u

)5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

Nuclear Fission: Heavy nucleus ( $^{235} U$ ) absorbs a thermal neutron, splitting into lighter fragments ( $^{141} Ba +^{92} Kr + 3 n$ ), releasing energy because products have higher binding energy per nucleon.
Nuclear Fusion: Light nuclei ( $^{2} H +^{3} H \to^{4} He + n$ ) fuse at extreme temperatures ( $T & g t; 10^{7} K$ ) overcoming Coulomb repulsion, releasing ~4x more energy per kg than fission.
How to calculate reaction energy yield: $Δ E = [Σ m_{r e a c t a n t s} - Σ m_{p r o d u c t s}] \times c^{2}$ .

Students will demonstrate

By calculating exact energy release in Joules and MeV for a complete $^{235} U$ fission reaction.
By completing Section 6 of their Level 3 Modern Physics Mastery Portfolio.

Do Now | Tīmatanga Whakaaro (10 min)

Fission vs Fusion Power Prompt:

"Both splitting a heavy Uranium atom and fusing two light Hydrogen atoms release massive amounts of energy. Why do both processes release energy if one splits and the other joins?"

Unpack: Because both reactions move towards Iron-56 ( $^{56} Fe$ ) at the peak of the binding energy per nucleon curve! In both cases, the products have higher binding energy per nucleon than the reactants. The excess mass difference ( $Δ m$ ) is released as kinetic energy and gamma radiation ( $E = Δ m c^{2}$ ).

Mass-Energy Balance Equations (15 min)

1. Fission Reaction Example

$_{92}^{235} U +_{0}^{1} n \to_{56}^{141} Ba +_{36}^{92} Kr + 3_{0}^{1} n + Δ E$
$Δ m = 0.215 u \to Δ E \approx 200 MeV$ .

2. Fusion Reaction Example

$_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He +_{0}^{1} n + Δ E$
$Δ m = 0.01888 u \to Δ E \approx 17.6 MeV$ .

📁 Physics Modern Portfolio — Section 6: Fission vs Fusion Energetics

Students open their Level 3 Physics Portfolio and complete Section 6:

Section 6 Requirements:

1. Reaction Energy Calculation: Perform step-by-step mass balance calculation in kg and MeV for Deuterium-Tritium fusion ( $m (^{2} H) = 2.014102 u, m (^{3} H) = 3.016049 u, m (^{4} He) = 4.001506 u, m_{n} = 1.008665 u$ ).

2. Fission vs Fusion Comparison Table: Compare Fuel Abundance, Energy Density ( ${J kg}^{- 1}$ ), Radioactive Waste, and Coulomb Barrier Temperatures.

3. Excellence Tokamak Confinement Rationale: 1-paragraph explanation of why magnetic confinement (ITER Tokamak) is necessary to contain $10^{8} K$ fusion plasma away from reactor walls.

Exit Verification | Ka Mutu Hoki (5 min)

Exit Check:

"My Section 6 calculates mass defect delta m = 0.01898 u for D-T fusion, yielding 17.6 MeV (2.82 x 10^-12 J) energy release."

Teacher Planning & NCEA Alignment

NCEA Level 3 Physics Alignment (3 Credits Internal):

Nuclear Fission & Fusion: Demonstrate understanding of nuclear equations, mass defect ( $Δ m$ ), $E = Δ m c^{2}$ , chain reactions, and fusion plasma dynamics.
Energy Yield Calculations: Convert between mass defect in atomic mass units ( $u$ ) and energy in Joules / MeV.

Vocabulary: Fission, fusion, mass defect ( $Δ m$ ), $E = m c^{2}$ , chain reaction, critical mass, Coulomb barrier, Deuterium, Tritium, Tokamak.

Other teaching approach: Guided Viewing & Problem Practice →

Lesson 6: Nuclear Fission & Fusion Energy