# E.2 Nuclear structure and reactions

> IB Physics HL · Theme E: Nuclear and quantum physics
> Source: https://www.owlsprep.com/study/ib-physics-hl-u5-e-2-nuclear-structure-and/

This subtopic covers nuclear composition, mass defect, binding energy, and balancing common nuclear reactions, forming the core foundation for all further nuclear physics topics tested in IB Physics HL.

**Prerequisites:** [Fundamental atomic structure concepts](https://www.owlsprep.com/study/ib-physics-hl-u5-e-1-atomic-structure/)

## Learning objectives

- Describe nuclear composition and interpret nuclide notation
- Calculate mass defect and binding energy for any nuclide
- Explain the significance of the binding energy per nucleon curve
- Balance nuclear reaction equations for all common reaction types

## Nuclear Composition and Nuclide Notation

**Nuclide** — A distinct nuclear species with a specific number of protons (Z) and neutrons (N). Protons have charge +e, neutrons are neutral, both have approximately 1 atomic mass unit (u) of mass.

*Notation:* ^A_ZX

*Example:* Carbon-12 is written as $^{12}_{\,\;6}C$

Isotopes are nuclides of the same element, so they share the same atomic number $Z$ but have different mass numbers $A$, meaning different numbers of neutrons. The relationship between the values is always: $A = Z + N$, where $N$ is neutron number.

**Worked example:** How many protons, neutrons and electrons are in a neutral atom of $^{235}_{92}U$?

1. The atomic number $Z = 92$, so number of protons = $Z = 92$.
2. The mass number $A = 235$, so neutron number $N = A - Z = 235 - 92 = 143$.
3. For a neutral atom, number of electrons equals number of protons, so electrons = $92$.

## Mass Defect and Binding Energy

**Mass Defect** — The difference between the total mass of separate individual nucleons and the measured mass of the intact nucleus. The missing mass is converted into nuclear binding energy.

*Notation:* \Delta m = Zm_p + Nm_n - m_{\text{nucleus}}

Mass-energy equivalence, given by Einstein's famous relation, connects mass defect to the binding energy that holds the nucleus together:

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

In nuclear physics, 1 atomic mass unit (u) is equivalent to approximately 931.5 MeV of energy, so this conversion factor is used to simplify calculations.

**Worked example:** Calculate the total binding energy of helium-4 ($^{4}_{2}He$). Given: $m_p = 1.00728$ u, $m_n = 1.00866$ u, $m_{He} = 4.00151$ u.

1. Identify values: $Z = 2$, $N = 4 - 2 = 2$
2. Calculate mass defect:
3. $$\Delta m = (2 \times 1.00728 + 2 \times 1.00866) - 4.00151$$
4. $\Delta m = 4.03188 - 4.00151 = 0.03037$ u
5. Convert to binding energy using $1$ u $c^2 = 931.5$ MeV:
6. $$E_b = 0.03037 \times 931.5 \approx 28.3 \text{ MeV}$$

## Binding Energy per Nucleon Curve

Binding energy per nucleon ($E_b/A$) is the average energy required to remove one nucleon from a nucleus. When plotted against mass number $A$, the curve peaks at $A \approx 56$ (iron-56), the most stable nuclide. Lighter and heavier nuclides have lower binding energy per nucleon.

This curve explains why fusion of light nuclides and fission of heavy nuclides both release energy: the products of the reaction have higher average binding energy per nucleon than the reactants, so the difference in total binding energy is released as kinetic energy and radiation.

**Worked example:** Uranium-235 ($A=235$) has a binding energy per nucleon of ~7.6 MeV. When it fissions, it splits into two equal product nuclides ($A \approx 117$ each) with binding energy per nucleon ~8.5 MeV. Calculate the total energy released per fission.

1. Total number of nucleons is conserved, so total $A = 235$ for reactants and products.
2. Initial total binding energy: $235 \times 7.6 = 1786$ MeV
3. Final total binding energy: $(117 + 117) \times 8.5 = 1989$ MeV
4. Energy released equals the increase in total binding energy: $1989 - 1786 = 203$ MeV, which matches typical experimental values for U-235 fission.

## Balancing Nuclear Reactions

All nuclear reactions must obey two core conservation rules for balancing: conservation of total charge (atomic number $Z$) and conservation of total mass number ($A$). The sum of all $Z$ values on the left side of the reaction must equal the sum on the right, and the same rule applies to $A$.

This rule applies to all reaction types: radioactive alpha/beta decay, fusion, fission, and induced transmutation.

**Worked example:** Complete the following fission reaction: $^{235}_{92}U + ^1_0n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + X \cdot ^1_0n$. Find $X$.

1. First balance total mass number $A$: Left total $A = 235 + 1 = 236$
2. Right total $A$ from known particles: $141 + 92 + (X \times 1) = 233 + X$
3. Equate both sides: $236 = 233 + X \rightarrow X = 3$
4. Verify charge balance: Left total $Z = 92 + 0 = 92$, right total $Z = 56 + 36 + 0 = 92$, which balances. So 3 neutrons are produced.

## Common pitfalls

- **Wrong:** Trying to manually correct for electron mass when using atomic mass for mass defect
  - Why it fails: Tabulated neutral atomic masses include electrons, but the number of electrons cancels out automatically in most calculations, so extra corrections lead to wrong Δm
  - Correct: Use tabulated neutral atomic masses directly for mass defect calculations, no extra correction is needed for electron mass
- **Wrong:** Confusing total binding energy with energy released in a reaction
  - Why it fails: Binding energy is the energy required to break a nucleus apart, not directly the energy released by a fission/fusion reaction
  - Correct: Energy released in a reaction equals the difference between the total final binding energy of products and total initial binding energy of reactants
- **Wrong:** Claiming iron has the highest total binding energy
  - Why it fails: Heavy nuclei like uranium have much higher total binding energy than iron, only the binding energy *per nucleon* peaks at iron
  - Correct: Always specify that binding energy per nucleon peaks at iron, which makes it the most stable nuclide
- **Wrong:** Only balancing mass number and forgetting to balance atomic number when completing nuclear reactions
  - Why it fails: Some missing particles have the same mass number but different charge, so only balancing A gives the wrong answer
  - Correct: Always check and balance *both* total A and total Z separately when finding unknown particles in a reaction

## Cheatsheet

| Concept | Rule/Formula | Key Value |
| --- | --- | --- |
| Nuclide Notation | $^A_Z X$, $A = Z + N$ | $Z=$protons, $N=$neutrons |
| Mass Defect | $\Delta m = Zm_p + Nm_n - m_{nucleus}$ |  |
| Binding Energy | $E_b = \Delta m c^2$ | 1 u = 931.5 MeV/c² |
| Reaction Balance | Total $A$ left = Total $A$ right, Total $Z$ left = Total $Z$ right |  |
| Energy Released | $\Delta E = E_{b, products} - E_{b, reactants}$ | Positive ΔE = exothermic |

## What's next

This subtopic forms the foundational knowledge for all further nuclear physics topics in IB Physics HL. Mastery of mass defect calculations, binding energy interpretation, and nuclear reaction balancing is required for almost every exam question on nuclear energy, radioactive decay, and nuclear interactions. Questions testing these concepts appear regularly in both Paper 1 multiple choice and Paper 2 extended response sections, so it is critical to be comfortable with routine calculations here. Next, you will move on to learning about the kinetics of radioactive decay, the properties of nuclear radiation, and then the practical applications of fission and fusion for energy generation.

- [E.3 Quantum physics: photons and matter waves](https://www.owlsprep.com/study/ib-physics-hl-u5-e-3-quantum-physics-photons/)
- [E.4 The nuclear atom](https://www.owlsprep.com/study/ib-physics-hl-u5-e-4-the-nuclear-atom/)

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