Nuclear Binding Energy
Energy required to disassemble a nucleus
Formula
E = Δm × c²
Δm = (Z mp + N mn - Mnucleus)
- E = binding energy (J)
- Δm = mass defect (kg)
- c = speed of light (3.0 × 10⁸ m/s)
- Z = number of protons
- N = number of neutrons
- mp = proton mass (1.007276 u)
- mn = neutron mass (1.008665 u)
Example: Helium-4
Given: Z = 2, N = 2, Mnucleus = 4.0015 u.
Δm = 2(1.007276) + 2(1.008665) - 4.0015 = 0.0304 u
Convert to kg: 0.0304 u × (1.66054 × 10⁻²⁷ kg/u) = 5.05 × 10⁻²⁹ kg
E = (5.05 × 10⁻²⁹) × (3.0 × 10⁸)² ≈ 4.5 × 10⁻¹² J
In MeV: 4.5 × 10⁻¹² / (1.602 × 10⁻¹³) ≈ 28.3 MeV
Answer: E ≈ 28.3 MeV (or ≈ 7.1 MeV/nucleon)
Notes
- Higher binding energy per nucleon = more stable nucleus.
- Fe-56 has maximum binding energy per nucleon (≈8.8 MeV/nucleon).
- Fusion of light nuclei and fission of heavy nuclei both release energy.