# Mass defect and binding energy

> Physics · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9702-u27-mass-defect-and-binding-energy/

This module explains mass defect, the origin of binding energy in nuclei, and how to calculate these quantities. You will learn how binding energy per nucleon determines nuclear stability, a core concept for nuclear fission and fusion.

**Prerequisites:** [Einstein's mass-energy equivalence](https://www.owlsprep.com/study/cie-9702-u26-mass-energy-equivalence/); [Nuclear notation (proton and nucleon numbers)](https://www.owlsprep.com/study/cie-9702-u27-nuclear-structure/)

## Learning objectives

- Define mass defect and binding energy correctly
- Calculate mass defect and binding energy from nuclear data
- Relate binding energy per nucleon to nuclear stability
- Convert between atomic mass units and energy in MeV

## Mass Defect: Definition and Origin

**Mass defect** — The difference between the total mass of the individual, separate nucleons that make up a nucleus and the actual measured mass of the intact nucleus. Mass defect arises because some mass is converted to energy when nucleons bind together.

*Notation:* \Delta m

*Example:* Calculated for any nuclide using the formula below

$$\Delta m = (Z m_p + (A-Z) m_n) - M_{\text{nucleus}}$$

Where $Z$ = proton number, $A$ = nucleon number, $m_p$ = proton mass, $m_n$ = neutron mass, $M_{\text{nucleus}}$ = mass of the intact nucleus. When using atomic data, electron masses cancel out automatically, so you can use atomic masses directly without correction.

**Worked example:** Calculate the mass defect of a nitrogen-14 nucleus, given: mass of nitrogen-14 atom = 14.003074 u, mass of ¹H atom = 1.007825 u, mass of neutron = 1.008665 u. Nitrogen-14 has $Z=7$, $A=14$.

1. Step 1: Calculate total mass of 7 hydrogen atoms (protons + electrons) and 7 neutrons:
2. $$7 \times 1.007825 + 7 \times 1.008665 = 14.115430 \text{ u}$$
3. Step 2: Subtract the atomic mass of nitrogen-14 to get mass defect (electron masses cancel):
4. $$\Delta m = 14.115430 - 14.003074 = 0.112356 \text{ u}$$

> **Exam tip:** Always double-check the number of protons and neutrons matches the nuclide's proton and nucleon numbers given in the question.

## Binding Energy: Definition and Calculation

**Binding Energy** — The minimum energy required to completely separate a nucleus into its individual free nucleons. It is equal to the energy released when the nucleus is formed from separate nucleons, and is calculated from mass defect using Einstein's mass-energy relation.

*Notation:* E_b

*Example:* A higher binding energy means the nucleus is more tightly bound

For CIE exams, the standard conversion factor is $1 \text{ u} = 931.5 \text{ MeV}$, so binding energy can be calculated directly from mass defect in u by simple multiplication.

**Worked example:** Calculate the total binding energy of the nitrogen-14 nucleus from the previous example, where $\Delta m = 0.112356$ u.

1. Use the conversion factor 1 u = 931.5 MeV to convert mass defect to binding energy:
2. $$E_b = \Delta m \times 931.5 = 0.112356 \times 931.5 \approx 104.6 \text{ MeV}$$

**Check your understanding**

Check your understanding:

1. What is the binding energy of a nuclide with mass defect 0.250 u?

   - 233 MeV
   - 3726 MeV
   - 0.250 MeV
   - 931.5 MeV

   *Why:* Correct: $0.250 \times 931.5 = 232.875 \approx 233$ MeV. Other options use incorrect conversion factors.

## Binding Energy per Nucleon and Nuclear Stability

**Binding Energy per Nucleon** — The total binding energy of a nucleus divided by the number of nucleons ($A$) in the nucleus. It is the standard measure of how tightly bound a nucleus is, and therefore how stable it is.

*Notation:* \frac{E_b}{A}

*Example:* Higher binding energy per nucleon = more stable nuclide

A graph of binding energy per nucleon against nucleon number $A$ has a characteristic shape: it rises steeply for small $A$, peaks at around $A = 56$ (iron), then gradually decreases for large $A$. This shape explains why energy is released in both fission (large nuclei splitting) and fusion (small nuclei joining): both processes produce more tightly bound nuclides with higher binding energy per nucleon, converting mass to energy.

**Worked example:** Calculate the binding energy per nucleon for nitrogen-14, given total binding energy is 104.6 MeV and $A = 14$.

1. Divide total binding energy by the number of nucleons $A$:
2. $$\frac{E_b}{A} = \frac{104.6}{14} \approx 7.47 \text{ MeV per nucleon}$$

> **info**
>
> You will be expected to draw and interpret this curve in CIE exams, and use it to explain energy release from fission and fusion.

## Common pitfalls

- **Wrong:** Subtracting electron mass from atomic masses to get nuclear mass
  - Why it fails: This introduces an unnecessary error, because electron masses cancel out when using atomic data
  - Correct: Use atomic masses directly, the electron masses will cancel automatically in mass defect calculations
- **Wrong:** Calculating mass defect as $\Delta m = M_{\text{nucleus}} - \text{total nucleon mass}$, giving a negative value
  - Why it fails: This reverses the definition, leading to negative binding energy which is impossible
  - Correct: Mass defect is always positive: $\Delta m = \text{total mass of separate nucleons} - \text{nuclear mass}$
- **Wrong:** Using total binding energy instead of binding energy per nucleon to compare stability
  - Why it fails: Large nuclei always have higher total binding energy just because they have more nucleons, not because they are more stable
  - Correct: Always compare binding energy *per nucleon* when assessing relative nuclear stability
- **Wrong:** Using 1 u = 931.5 J instead of 931.5 MeV for conversion
  - Why it fails: This gives a binding energy 10⁶ times too large, leading to wrong final answers
  - Correct: Always check units: CIE expects 1 u = 931.5 MeV for all binding energy calculations

## Cheatsheet

| Quantity | Formula/Rule | Key Note |
| --- | --- | --- |
| Mass defect | $\Delta m = Zm_p + (A-Z)m_n - M$ | Electron masses cancel with atomic data |
| Total binding energy | $E_b = \Delta m \times 931.5$ MeV | 1 u = 931.5 MeV |
| Binding energy per nucleon | $\frac{E_b}{A} = \frac{E_b}{A}$ | Higher value = more stable |
| Peak of binding energy curve | A ≈ 56 (Iron) | Most stable common nuclide |

## What's next

Mass defect and binding energy are the foundation for understanding all nuclear energy processes, which are heavily tested in CIE A-Level Physics papers 1 and 2. The shape of the binding energy per nucleon curve directly explains why energy is released in nuclear fission (used in nuclear power) and nuclear fusion (the energy source of stars), the next core topics in nuclear physics. Mastery of mass defect calculations is also required for questions on radioactive decay energy, so a strong grasp of this sub-topic will help you with all subsequent nuclear physics questions.

- [Radioactive decay law](https://www.owlsprep.com/study/cie-9702-u27-radioactive-decay-law/)
- [Half-life](https://www.owlsprep.com/study/cie-9702-u27-half-life/)
- [Nuclear fission and fusion](https://www.owlsprep.com/study/cie-9702-u27-nuclear-fission-and-fusion/)

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