# B.4 Mass, energy and matter structure

> IB Physics HL · IB Physics Higher Level 2025+
> Source: https://www.owlsprep.com/study/ib-physics-hl-u2-b-4-mass-energy-and/

This module covers Einstein's mass-energy equivalence, mass defect, nuclear binding energy, and mass-energy conversion in nuclear processes and particle interactions. You will learn to calculate energy changes from mass differences for nuclear and particle reactions.

**Prerequisites:** [Atomic structure and nuclear composition](https://www.owlsprep.com/study/ib-physics-hl-u2-b-3-the-atom/); Basic energy conservation for closed systems

## Learning objectives

- Distinguish between rest mass and total energy of a particle
- Calculate mass defect and binding energy for atomic nuclei
- Apply mass-energy equivalence to find energy changes in nuclear reactions
- Explain pair production and annihilation in terms of mass-energy conversion

## Mass-Energy Equivalence

**Rest Mass** — The invariant mass of a particle measured in its rest frame; it is constant and independent of the particle's motion relative to an observer.

*Notation:* $m_0$

*Example:* A proton always has a rest mass of ~1.00728 u, regardless of its speed.

Einstein's mass-energy equivalence is the foundational relationship that connects mass and energy, proving mass can be converted to energy and vice versa. The relationship for rest energy (the energy a particle has when at rest) is:

$$E_0 = m_0 c^2$$

Where $c$ is the speed of light in vacuum (~$3.00 \times 10^8$ m/s). Total energy of a moving particle is the sum of its rest energy and kinetic energy.

**Worked example:** Calculate the rest energy of an electron, given its rest mass is $9.11 \times 10^{-31}$ kg. Express the result in joules and mega-electron volts (MeV).

1. Start with the rest energy formula:
2. $$E_0 = m_0 c^2$$
3. Substitute the given values:
4. $$E_0 = (9.11 \times 10^{-31}) \times (3.00 \times 10^8)^2$$
5. Calculate the result in joules:
6. $$E_0 = 8.20 \times 10^{-14} \ \text{J}$$
7. Convert to MeV (1 MeV = $1.60 \times 10^{-13}$ J):
8. $$E_0 = \frac{8.20 \times 10^{-14}}{1.60 \times 10^{-13}} = 0.512 \ \text{MeV}$$

> **tip**
>
> Almost all IB nuclear questions ask for energy in MeV, not joules. Always check the required units!

> **Exam tip:** Remember that the value of 1 atomic mass unit (u) in energy is 931.5 MeV, which is given in the data booklet.

## Mass Defect and Nuclear Binding Energy

**Mass Defect** — The difference between the total mass of individual separated nucleons (protons and neutrons) and the measured mass of the intact nucleus.

*Notation:* $\Delta m$

*Example:* For helium-4, total mass of 2 protons + 2 neutrons = ~4.032 u, nucleus mass = ~4.0015 u, so Δm = 0.0305 u.

The mass defect is converted into binding energy, the energy that holds the nucleus together. By mass-energy equivalence:

$$E_b = \Delta m c^2$$

Binding energy per nucleon ($\frac{E_b}{A}$, where $A = Z+N$ is total nucleon number) is the key indicator of nuclear stability: higher binding energy per nucleon means a more stable nucleus.

**Worked example:** Calculate the mass defect and total binding energy for carbon-12. Given: $m_p = 1.00728$ u, $m_n = 1.00866$ u, $M_{C-12} = 12.00000$ u, $1 \text{ u} c^2 = 931.5$ MeV.

1. Carbon-12 has 6 protons and 6 neutrons, so Z=6, N=6:
2. Calculate total mass of separated nucleons:
3. $$Z m_p + N m_n = 6(1.00728) + 6(1.00866)$$
4. $$= 6.04368 + 6.05196 = 12.09564 \ \text{u}$$
5. Calculate mass defect:
6. $$\Delta m = 12.09564 - 12.00000 = 0.09564 \ \text{u}$$
7. Convert to binding energy:
8. $$E_b = \Delta m c^2 = 0.09564 \times 931.5 = 89.1 \ \text{MeV}$$

## Energy Changes in Nuclear Reactions

In any nuclear reaction (fission, fusion, decay), the total rest mass of products differs from the total rest mass of reactants. This mass difference gives the energy released or absorbed in the reaction, called the Q-value:

$$Q = (m_{\text{reactants}} - m_{\text{products}}) c^2$$

If $Q > 0$, total mass of products is less than reactants, so energy is released (exoergic reaction). If $Q < 0$, energy must be absorbed to make the reaction proceed (endoergic reaction).

**Worked example:** A uranium-235 fission reaction has total reactant mass of 236.0526 u and total product mass of 235.8673 u. Calculate the energy released.

1. Calculate the mass difference:
2. $$\Delta m = m_{\text{reactants}} - m_{\text{products}} = 236.0526 - 235.8673 = 0.1853 \ \text{u}$$
3. Convert to energy:
4. $$Q = 0.1853 \times 931.5 \approx 172.6 \ \text{MeV}$$
5. Q is positive, so ~173 MeV of energy is released in this reaction.

**Check your understanding**

Test your understanding:

1. A fusion reaction has a mass defect of 0.0189 u. What is the energy released?

   - 1.76 MeV
   - 17.6 MeV
   - 176 MeV
   - 0.0189 MeV

   *Why:* Correct: $0.0189 \times 931.5 \approx 17.6$ MeV.

## Pair Production and Annihilation

Mass-energy equivalence applies to particle creation and annihilation. Pair production is when a high-energy photon interacts with a nucleus to produce a particle and its matching antiparticle (e.g. electron + positron).

The photon must have at least enough energy to supply the total rest energy of the two particles. Annihilation is the reverse: a particle and antiparticle collide, annihilate, and produce photons carrying away the total energy.

**Worked example:** Calculate the minimum photon energy required for electron-positron pair production.

1. An electron and a positron each have a rest energy of 0.511 MeV:
2. Minimum energy equals the sum of the two rest energies:
3. $$E_{\text{min}} = 2 \times 0.511 = 1.022 \ \text{MeV}$$
4. Any energy above this minimum becomes kinetic energy of the produced particles.

## Common pitfalls

- **Wrong:** Calculating mass defect as $M_{nucleus} - (Z m_p + N m_n)$ leading to negative binding energy
  - Why it fails: Mass defect is defined as the difference between separated nucleons and the intact nucleus, so the order of subtraction is critical
  - Correct: Always calculate $
Delta m = (Z m_p + N m_n) - M_{nucleus}$ to get a positive mass defect for stable nuclei
- **Wrong:** Confusing total binding energy with binding energy per nucleon when comparing nuclear stability
  - Why it fails: Larger nuclei always have higher total binding energy just because they have more nucleons, not because they are more stable
  - Correct: Always use binding energy per nucleon to compare stability between different nuclei
- **Wrong:** Getting the sign of Q wrong for energy released in reactions
  - Why it fails: Energy is released when products have less mass than reactants
  - Correct: Use the formula $Q = (m_{reactants} - m_{products})c^2$, so positive Q always means energy released
- **Wrong:** Forgetting to cancel electron masses when using atomic masses for nuclear calculations
  - Why it fails: Atomic masses include electron mass, which will add an error if the number of electrons is unbalanced
  - Correct: Check that the number of electrons on reactant and product sides are equal, so electron masses cancel out automatically
- **Wrong:** Forgetting that pair production requires a nucleus to conserve momentum
  - Why it fails: A photon cannot produce a pair in empty vacuum, momentum cannot be conserved
  - Correct: Always note that pair production only occurs near a massive nucleus that absorbs the recoil momentum

## Cheatsheet

| Concept | Formula | Key IB Fact |
| --- | --- | --- |
| Rest Energy | $E_0 = m_0 c^2$ | $m_0$ = invariant rest mass |
| Mass Defect | $\Delta m = Z m_p + N m_n - M_{nucleus}$ | Positive for all stable nuclei |
| Binding Energy | $E_b = \Delta m c^2$ | 1 u $c^2$ = 931.5 MeV (data booklet) |
| Nuclear Stability | $\frac{E_b}{A}$ | Higher = more stable |
| Reaction Q Value | $Q = (m_r - m_p)c^2$ | $Q>0$ = energy released |
| e⁻e⁺ Pair Production | $E_{min} = 1.022$ MeV | Minimum photon energy required |

## What's next

Mass-energy equivalence and binding energy are the foundation of all nuclear physics topics you will study next. These concepts explain why energy is released in nuclear fission and fusion, the processes that power nuclear reactors and stars, and why some nuclei are unstable and undergo radioactive decay. Understanding the relationship between binding energy per nucleon and nucleon number also lets you predict which reactions will release energy. You can now apply these core ideas to practical nuclear processes and more advanced particle interactions.

- [B.5 Current and electric circuits (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u2-b-5-current-and-electric/)
- [B.6 Heating effect of current and electric cells (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u2-b-6-heating-effect-of/)
- [Theme C: Wave behaviour](https://www.owlsprep.com/study/ib-physics-hl-u3-overview/)

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