Lesson 4: Binding Energy | Modern Physics

Lesson at a Glance | He Tirohanga Whakamua

Do NowWhy a Helium nucleus weighs less than 2 free protons + 2 free neutrons10 min

Mass Defect & E=Δmc²Calculating

Δ m

in atomic mass units (

u

) and Joules/MeV15 min

Binding Energy CurvePeak stability at Iron-56 (

^{56} Fe

) and why fusion vs fission release energy15 min

Portfolio EntryWrite Section 4: Mass Defect & Nuclear Binding Energy Map10 min

Exit DrillCalculate binding energy per nucleon for Helium-4 (

^{4} He

)5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

Mass Defect ( $Δ m$ ): The mass of a bound nucleus is always less than the sum of its constituent free nucleons ( $Δ m = Σ m_{f r e e} - m_{b o u n d}$ ).
Nuclear Binding Energy ( $E_{b} = Δ m c^{2}$ ): The energy released when free nucleons bind together under the strong nuclear force ( $1 u = 1.6605 \times 10^{- 27} kg \leftrightarrow 931.5 MeV$ ).
Binding Energy Per Nucleon Curve ( $\frac{E_{b}}{A}$ ): Peaks at Iron-56 ( $^{56} Fe, \approx 8.8 MeV/nucleon$ ). Fusing light nuclei or splitting heavy nuclei increases binding energy per nucleon, releasing net energy.

Students will demonstrate

By calculating mass defect in kg, stored binding energy in MeV, and binding energy per nucleon for Helium-4.
By completing Section 4 of their Level 3 Modern Physics Mastery Portfolio.

Do Now | Tīmatanga Whakaaro (10 min)

Missing Mass Paradox Prompt:

"If you weigh 2 separate protons ( $2 \times 1.007276 u$ ) and 2 separate neutrons ( $2 \times 1.008665 u$ ), their combined mass is 4.031882 u. But an assembled Helium-4 nucleus ( $^{4} He$ ) weighs only 4.001506 u. Where did the missing 0.030376 u of mass go?"

Unpack: The missing mass ( $Δ m$ ) was converted directly into energy ( $E = Δ m c^{2}$ ) when the strong force bound the nucleons together! That $28.3 MeV$ of binding energy must be supplied back to break the nucleus apart into free nucleons.

Binding Energy Per Nucleon Curve (15 min)

1. Mass Defect Equation

$Δ m = [Z m_{p} + (A - Z) m_{n}] - m_{n u c l e u s}$
$E_{b} = Δ m c^{2} = Δ m (in u) \times 931.5 MeV$ .

2. Stability Peak ( $^{56} Fe$ )

Iron-56 sits at the top of the $\frac{E_{b}}{A}$ curve. Fusion ( $A & l t; 56$ ) and Fission ( $A & g t; 56$ ) both move towards Iron-56, increasing stability and releasing energy.

📁 Physics Modern Portfolio — Section 4: Mass Defect & Binding Energy

Students open their Level 3 Physics Portfolio and complete Section 4:

Section 4 Requirements:

1. Binding Energy per Nucleon Curve: Plot $\frac{E_{b}}{A}$ vs Mass Number $A$ , marking $^{4} He$ , $^{56} Fe$ , and $^{235} U$ , shading the Fusion and Fission energy release zones.

2. Mass Defect Calculation: Calculate $Δ m$ in kg and total $E_{b}$ in MeV for Carbon-12 ( $^{12} C$ , $m = 12.000000 u$ ).

3. Excellence Strong Nuclear Force Rationale: 1-paragraph explanation of how short-range strong nuclear attractive forces overcome electrostatic proton repulsion in stable nuclei.

Exit Verification | Ka Mutu Hoki (5 min)

Exit Check:

"My Section 4 calculates binding energy Eb = 28.3 MeV for Helium-4, giving 7.07 MeV per nucleon, and explains why fusing light nuclei releases net energy."

Teacher Planning & NCEA Alignment

NCEA Level 3 Physics Alignment (3 Credits Internal):

Mass Defect & Binding Energy: Demonstrate understanding of nuclear mass defect ( $Δ m$ ), $E = Δ m c^{2}$ , atomic mass units, and binding energy per nucleon curves.
Nuclear Stability: Relate strong force attraction and electrostatic repulsion to nuclear stability.

Vocabulary: Nucleon ( $A$ ), proton ( $Z$ ), neutron ( $N$ ), mass defect ( $Δ m$ ), binding energy ( $E_{b}$ ), $E = m c^{2}$ , atomic mass unit ( $u$ ), Iron-56 peak.

Other teaching approach: Guided Viewing & Problem Practice →

Lesson 4: Nuclear Structure, Mass Defect & Binding Energy