Playground
Drag the mass defect slider to see nucleons bind and release energy. Protons (red) and neutrons (blue) assemble into a glowing nucleus.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Binding energyoutput Energy required to separate the nucleus into free nucleons | J | M·L²·T⁻² | 0 – 1e-10 | |
| Mass defect Difference between nucleon mass sum and nuclear mass | kg | M | 0 – 1e-27 | |
| Speed of light Speed of light in vacuum | m/s | L·T⁻¹ | 299792458 – 299792458 |
Deep dive
Derivation
Consider Z protons and N = A − Z neutrons combining to form a nucleus of mass M(A,Z). The mass defect is Δm = Z·m_p + N·m_n − M(A,Z). By Einstein's mass-energy equivalence, the energy released in forming the nucleus (equal to the energy needed to disassemble it) is B = Δm·c². This binding energy accounts for the strong nuclear force overcoming Coulomb repulsion at short range.
Experimental verification
Aston's mass spectrograph (1920s) first revealed nuclear mass defects. Modern Penning trap mass spectrometry measures atomic masses to parts in 10¹¹, confirming binding energies. The binding energy curve (B/A vs A, peaking near iron-56) is reproduced across thousands of nuclides.
Common misconceptions
- Binding energy is not energy stored in the nucleus — it is energy you must ADD to break the nucleus apart. A nucleus with high binding energy is MORE stable, not less.
- The mass defect is not 'lost' mass — it is the mass equivalent of the energy radiated away when the nucleus formed.
- Binding energy per nucleon, not total binding energy, determines nuclear stability. Iron-56 has maximum B/A, not maximum B.
Real-world applications
- Nuclear power generation (fission of uranium releases ~0.9 MeV/nucleon of binding energy difference)
- Stellar nucleosynthesis (fusion of hydrogen to helium releases 6.7 MeV/nucleon)
- Medical isotope production (understanding which nuclei are stable enough to be useful)
- Nuclear weapons design (maximizing energy release from binding energy differences)
Worked examples
Binding energy of Helium-4
Given:
- Δm_u:
- 0.0304
- c:
- 299792458
Find: B
Solution
B = 0.0304 u × 931.494 MeV/u = 28.3 MeV
Binding energy of Deuterium
Given:
- Δm_u:
- 0.002388
- c:
- 299792458
Find: B
Solution
B = 0.002388 u × 931.494 MeV/u = 2.224 MeV
Scenarios
What if…
- scenario:
- What if the strong force were 2% weaker?
- answer:
- Deuterium would be unbound (B < 0), hydrogen fusion in stars would fail, and the universe would contain almost no elements heavier than hydrogen.
- scenario:
- What if we could measure mass defect to infinite precision?
- answer:
- We could predict nuclear reaction energies exactly. In practice, Penning trap measurements at 10⁻¹¹ precision already constrain binding energies to sub-keV accuracy.
- scenario:
- What if binding energy per nucleon kept increasing with A?
- answer:
- Fusion would be energetically favorable for all elements, not just light ones. Stars would fuse all the way to the heaviest elements instead of stopping at iron.
Limiting cases
- condition:
- Δm → 0
- result:
- B → 0
- explanation:
- Zero mass defect means the nucleus weighs exactly as much as its free nucleons — no binding energy and the nucleus is unbound.
- condition:
- Δm large (e.g., iron-56)
- result:
- B ≈ 492 MeV
- explanation:
- Iron-56 has one of the largest binding energies per nucleon (~8.8 MeV/nucleon), making it the most tightly bound common nucleus.
- condition:
- Δm < 0
- result:
- B < 0 (unphysical for stable nuclei)
- explanation:
- A negative mass defect would mean the nucleus is heavier than its parts — it would spontaneously disassemble.
Context
Francis Aston · 1920
Aston's mass spectrograph revealed that nuclei weigh less than the sum of their nucleons, confirming Einstein's mass-energy equivalence on nuclear scales.
Hook
Why does a helium atom weigh less than its parts?
A helium-4 nucleus has a mass defect of 0.0304 u. What is its binding energy in MeV?
Dimensions:
- lhs:
- B → [M·L²·T⁻²]
- rhs:
- Δm·c² → [M]·[L·T⁻¹]² = [M·L²·T⁻²]
- check:
- Both sides are [M·L²·T⁻²] = Joules. ✓
Validity: Valid for any bound nuclear system. The mass defect must be computed using precise atomic or nuclear masses. Applies to all nuclei from deuterium (A=2) to the heaviest known elements.