Worked example 1
Using m(H-1) = 1.007825 u, m(n) = 1.008665 u and m(⁴He) = 4.002603 u, find the mass defect of helium-4.
Try it first: Add up two hydrogen atoms and two neutrons before subtracting.
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Binding EnergyCourse level: College Extension
Mass Defect and Nuclear Binding Energy
What you'll be able to do: Calculate a mass defect, convert it to binding energy with E = mc², and use binding energy per nucleon to compare nuclear stability.
Best after: Nuclear Fission and Fusion
Introduction
Weigh a helium nucleus and it comes out lighter than its parts. That missing mass is not an error: it is the energy that holds the nucleus together, and converting between the two is the whole of this topic.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Define mass defect and calculate it from nucleon and nuclide masses
- Convert a mass defect to binding energy using 1 u·c² ≈ 931.5 MeV
- Calculate binding energy per nucleon and use it to rank stability
- Interpret the binding-energy-per-nucleon curve and locate its peak
- Connect binding energy to why fission and fusion release energy
Choose one consistent mass convention
A nuclear mass excludes electrons; a neutral atomic mass includes them. For nuclear masses, Δm = Zmp + Nmn − mnucleus. With neutral atomic masses, a useful approximation is Δm = Zm(¹H atom) + Nmn − matom: the Z electron masses cancel. Small differences in electronic binding energies are negligible at this lesson's precision. Never combine free proton masses with a neutral atom's mass without accounting for its electrons.
For helium-4 using atomic masses, 2(1.007825 u) + 2(1.008665 u) = 4.032980 u, while the neutral helium atom is 4.002603 u. The defect is 0.030377 u. The 1.007825 u entry belongs to neutral hydrogen, not a bare proton, and 4.002603 u is the atom, not the helium nucleus.
Mass defect to energy
Einstein's relation E = mc² converts the missing mass into energy. Doing this in SI units every time is tedious, so nuclear chemistry uses the shortcut 1 u·c² ≈ 931.5 MeV (mass-to-energy equivalence). Multiply the mass defect in u by 931.5 and you have the binding energy in MeV.
Mass and energy have different dimensions. The precise shorthand is 1 u·c² ≈ 931.5 MeV, or mass 1 u corresponds to energy 931.5 MeV. Thus BE = (0.030377 u)(931.5 MeV/u) ≈ 28.30 MeV per helium nucleus. In SI units, 1 eV = 1.602 × 10⁻¹⁹ J, so 28.30 MeV ≈ 4.53 × 10⁻¹² J per nucleus. A per-mole quantity additionally multiplies by Avogadro's constant; do not confuse a microscopic event with one mole of events.
E (MeV) = Δm (u) × 931.5
What binding energy means
The binding energy is the energy released when free nucleons come together to form the nucleus, and equally the energy that would have to be supplied to break the nucleus back into free nucleons. A larger binding energy means a more tightly held nucleus.
Binding energy is always positive when quoted this way. It is an amount of energy, not a signed enthalpy change.
Binding energy per nucleon
A uranium nucleus has a far larger total binding energy than a helium nucleus simply because it has more nucleons, so total binding energy is a poor stability comparison. Divide by the number of nucleons instead. Helium-4 gives 28.3 MeV / 4 = 7.07 MeV per nucleon; ⁵⁶Fe gives about 8.8 MeV per nucleon.
Divide total BE by A only when asked for the average per nucleon: 28.30/4 ≈ 7.07 MeV/nucleon for helium-4. A large nucleus can have greater total binding energy but lower BE/A than a small one. BE/A is useful for the broad energetic trend, while stability against a particular decay needs the actual parent–daughter mass difference and allowed decay modes. A high BE/A is not by itself a complete prediction of radioactive lifetime.
binding energy per nucleon = total binding energy / mass number
Read the broad curve and check actual reaction masses
The curve rises steeply among light nuclei, reaches a broad maximum near iron and nickel at roughly 8.8 MeV/nucleon, then declines toward very heavy nuclei. Nickel-62 has a slightly larger BE/A than iron-56; describing iron-56 as the uniquely most tightly bound nuclide is inaccurate. The broad peak explains why some light-fuel fusion and heavy-fuel fission reactions are energetically favorable.
For a specific reaction, compare the sum of all reactant masses with all product masses, including emitted particles. Qenergy = (Σmreactants − Σmproducts)c². Positive Qenergy means released energy; negative means a required input. Neutral atomic masses can be used when electron accounting is consistent, but beta and positron reactions require particular care with electron masses. Do not cancel electrons blindly in every nuclear process.
The peak explains both fission and fusion at once, which is why examiners keep returning to this curve.
Use reactant minus product mass for energy released
For a nuclear reaction, calculate Δm as total reactant mass minus total product mass using a consistent set of atomic or nuclear masses. A positive Δm means mass was converted to released energy, E=Δmc².
The shortcut 1 u c²≈931.5 MeV converts mass difference directly to energy per reaction; electron masses cancel in suitable atom-balanced examples such as the D–T reaction here; beta processes require separate electron accounting.
Keep all masses on the same atomic-mass convention.
Key ideas
Definition
Mass defect
The difference between the summed masses of the free nucleons and the actual mass of the nucleus.
Definition
Binding energy
The energy equivalent of the mass defect: released on formation, required for separation.
Rule
Conversion factor
1 u·c² ≈ 931.5 MeV; multiply a mass difference in u by 931.5 MeV/u for its energy equivalent.
Rule
Fair comparison
Compare stability using binding energy per nucleon, not total binding energy.
Key concept
Peak at iron
The broad BE/A maximum is near iron and nickel; Ni-62 is slightly higher than Fe-56. Particular decay stability requires specific mass differences.
Equation
Mass defect
Δm = [Z × m(p) + N × m(n)] − m(nucleus)
- Z = number of protons
- N = number of neutrons
- Δm = mass defect in u
Equation
Mass-energy equivalence
E = mc²
- c = 3.00 × 10⁸ m/s
Equation
Practical conversion
E (MeV) = Δm (u) × 931.5
- 931.5 = MeV equivalent of one atomic mass unit
Equation
Binding energy per nucleon
BE/A = total binding energy / A
- A = mass number, the total nucleon count
Worked examples
Worked example 2
Convert that mass defect into the total binding energy of helium-4, then find the binding energy per nucleon.
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Worked example 3
A fission event converts 0.200 u of mass into energy. How much energy is released?
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Worked example 4
For 2H + 3H → 4He + n, use atomic masses 2.014102, 3.016049, 4.002603, and neutron 1.008665 u. Estimate energy released.
Try it first: Subtract product mass from reactant mass.
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Common mistakes
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Ranking stability by total binding energy. | Big nuclei have big totals simply because they contain more nucleons; uranium beats helium on total but is less stable per nucleon. | Always divide by the mass number before comparing. |
| Forgetting to divide the mass defect by the nucleon count when the question asks for binding energy per nucleon. | The two answers differ by a factor equal to the mass number, so the numerical answer is far too large. | Underline whether the question says total or per nucleon before starting. |
| Believing mass is destroyed in nuclear reactions. | Mass and energy are two forms of the same quantity; the mass is converted, and the total mass-energy is conserved. | State the mass defect as mass converted to energy, never mass lost. |
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After the lesson · Question bank
Independent practice
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
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What you should now know
Calculate a binding mass defect using consistently nuclear masses or a suitable neutral-atomic-mass convention. Convert with 1 u·c² ≈ 931.5 MeV, then divide by A if a per-nucleon value is requested. The broad iron/nickel maximum explains general fission/fusion energetics, while a specific reaction’s energy release is (total reactant mass − total product mass)c² with all particle and electron accounting included.
- The nucleus weighs less than its separated nucleons; that difference is the mass defect
- E = mc² links the mass defect to the binding energy, with mass-energy conversion 1 u·c² ≈ 931.5 MeV
- Binding energy is released on formation and required for separation
- Compare stability using binding energy per nucleon, not the total
- The broad iron/nickel BE/A peak guides energetic trends; use specific masses to test an actual reaction
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