B.4 Mass, energy and matter structure

IB Physics HL· Theme B: The particulate nature of matter· 15 min read

1. Mass-Energy Equivalence★★☆☆☆⏱ 4 min

📘 Definition

Rest Mass

$m_{0}$

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.

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 (~ $3.00 \times 1 0^{8}$ 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 $9.11 \times 1 0^{- 31}$ 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 = $1.60 \times 1 0^{- 13}$ J):
8
$E_0 = \frac{8.20 \times 10^{-14}}{1.60 \times 10^{-13}} = 0.512 \ \text{MeV}$

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.

2. Mass Defect and Nuclear Binding Energy★★★☆☆⏱ 5 min

📘 Definition

Mass Defect

$Δ m$

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

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, $1 u c^{2} = 931.5$ 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
$E_b = \Delta m c^2 = 0.09564 \times 931.5 = 89.1 \ \text{MeV}$

3. Energy Changes in Nuclear Reactions★★★☆☆⏱ 4 min

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.

✓ Quick check

Test your understanding:

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
Reveal answer
17.6 MeV —
Correct: $0.0189 \times 931.5 \approx 17.6$ MeV.

4. Pair Production and Annihilation★★★★☆HL only⏱ 2 min

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.

5. Common Pitfalls

Wrong move:

Calculating mass defect as $M_{n u c l e u s} - (Z m_{p} + N m_{n})$ leading to negative binding energy

Why:

Mass defect is defined as the difference between separated nucleons and the intact nucleus, so the order of subtraction is critical

Correct move:

Always calculate $ Delta m = (Z m_p + N m_n) - M_{nucleus}$ to get a positive mass defect for stable nuclei

Wrong move:

Confusing total binding energy with binding energy per nucleon when comparing nuclear stability

Why:

Larger nuclei always have higher total binding energy just because they have more nucleons, not because they are more stable

Correct move:

Always use binding energy per nucleon to compare stability between different nuclei

Wrong move:

Getting the sign of Q wrong for energy released in reactions

Why:

Energy is released when products have less mass than reactants

Correct move:

Use the formula $Q = (m_{r e a c t an t s} - m_{p r o d u c t s}) c^{2}$ , so positive Q always means energy released

Wrong move:

Forgetting to cancel electron masses when using atomic masses for nuclear calculations

Why:

Atomic masses include electron mass, which will add an error if the number of electrons is unbalanced

Correct move:

Check that the number of electrons on reactant and product sides are equal, so electron masses cancel out automatically

Wrong move:

Forgetting that pair production requires a nucleus to conserve momentum

Why:

A photon cannot produce a pair in empty vacuum, momentum cannot be conserved

Correct move:

Always note that pair production only occurs near a massive nucleus that absorbs the recoil momentum

6. Quick Reference Cheatsheet

Concept	Formula	Key IB Fact
Rest Energy	$E_{0} = m_{0} c^{2}$	$m_{0}$ = invariant rest mass
Mass Defect	$Δ m = Z m_{p} + N m_{n} - M_{n u c l e u s}$	Positive for all stable nuclei
Binding Energy	$E_{b} = Δ m c^{2}$	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

7. Frequently Asked

Is relativistic mass used in IB Physics 2025?

No, IB uses the modern convention where mass always refers to invariant rest mass. Relativistic energy changes are described via total energy, not changing mass.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 1
Mass defect calculation
2022 · Paper 2
Binding energy per nucleon
2021 · Paper 1
Pair production minimum energy

Going deeper

referenceIB Physics 2025 Data BookletContains 1 u = 931.5 MeV/c² conversion

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.