# Atomic structure

> Physics · CIE A-Level AS
> Source: https://www.owlsprep.com/study/cie-9702-u11-atomic-structure/

This sub-topic covers the fundamental structure of the atom, the development of the nuclear model from Rutherford scattering, key nuclear properties and nuclide notation. It forms the foundation for all subsequent AS particle physics topics.

**Prerequisites:** Basic atomic structure (IGCSE level); Alpha particle properties and charge

## Learning objectives

- Describe the structure of the atom and key properties of the nucleus
- Explain the evidence for the nuclear model from Rutherford alpha scattering
- Use standard nuclide notation to calculate proton, neutron and electron counts
- Apply the nuclear radius formula and understand constant nuclear density

## 1. Rutherford Scattering and the Nuclear Model

Before Rutherford's famous experiment, the accepted atomic model was J.J. Thomson's *plum pudding model*, which described negative electrons embedded evenly in a uniform sphere of positive charge.

**Plum Pudding Model** — Early atomic model where negative electrons are distributed evenly through a uniform sphere of positive charge

Rutherford, Geiger and Marsden tested this model by firing positively charged alpha particles at thin gold foil. Most passed through undeflected, but a tiny fraction deflected through angles >90°, which could not be explained by the plum pudding model.

**Worked example:** State three conclusions from Rutherford's experiment, linking each to the corresponding observation.

1. Observation 1: Most alpha particles pass straight through the foil. Conclusion 1:
2. Most of the atom is empty space, with mass concentrated in a small central region.
3. Observation 2: Some alpha particles are deflected through small angles. Conclusion 2:
4. The central nucleus is positively charged, and repels positively charged alpha particles.
5. Observation 3: Very few alpha particles deflect through angles >90°. Conclusion 3:
6. Almost all the mass of the atom is concentrated in a very small, dense central nucleus.

**Exam command terms**

- **Explain** — Link observation directly to the conclusion, do not just list results *(You must explain why large deflection leads to a dense nucleus, not just state the conclusion)*

> **Exam tip:** Never mix up observations and conclusions — examiners always test this distinction.

## 2. Nuclear Notation and Key Terms

All atomic nuclei are made of nucleons (protons and neutrons). We use standard notation to describe any nuclide, with two key numbers: proton number (Z) and nucleon number (A).

**Nuclide** — A specific type of atom defined by its proton and nucleon count

*Notation:* ^A_Z \text{X}

*Example:* Carbon-12 is written $^{12}_{\phantom{1}6} \text{C}$

- **Proton number (Z):** Number of protons in the nucleus, equal to the positive charge of the nucleus
- **Nucleon number (A):** Total number of protons + neutrons in the nucleus
- **Neutron number:** $N = A - Z$, number of neutrons in the nucleus
- **Isotope:** Nuclides of the same element (same Z) with different N/A

**Worked example:** How many protons, neutrons and electrons are in a neutral $^{238}_{\phantom{1}92} \text{U}$ atom?

1. Proton number $Z = 92$, so there are 92 protons.
2. For a neutral atom, electron count equals proton count, so 92 electrons.
3. Calculate neutron number:
4. $$N = A - Z = 238 - 92 = 146$$
5. Final answer: 92 protons, 92 electrons, 146 neutrons.

## 3. Nuclear Size and Density

Experiments confirm that nuclear radius follows a simple proportional relationship with nucleon number:

$$R = R_0 A^{1/3}$$

Where $R_0 \approx 1.2 \times 10^{-15}\ \text{m} = 1.2\ \text{fm}$. This relationship leads to a very useful property: all nuclei have approximately the same density, regardless of size.

**Worked example:** Calculate the radius of an oxygen-16 nucleus, given $R_0 = 1.2$ fm.

1. Oxygen-16 has nucleon number $A = 16$.
2. Substitute into the formula:
3. $$R = 1.2 \times (16)^{1/3}$$
4. Calculate $(16)^{1/3} \approx 2.5198$
5. Final result: $R \approx 1.2 \times 2.52 = 3.02$ fm, or $3.02 \times 10^{-15}$ m.

> **info**
>
> Since volume $V \propto R^3$, and $R^3 = R_0^3 A$, so $V \propto A$. Mass is proportional to $A$, so density $\rho = \frac{m}{V}$ is constant for all nuclei.

## 4. Isotope Properties

Isotopes of the same element have identical chemical properties, but different physical and nuclear properties. This is because chemical reactions depend on the number and arrangement of electrons, which is determined by proton number, which is the same for all isotopes of an element.

Differences in neutron number change the mass and nuclear stability of isotopes, leading to different physical properties (e.g. density, boiling point) and nuclear behaviour (e.g. radioactivity).

**Worked example:** State one similarity and one difference between $^{16}_8 \text{O}$ and $^{18}_8 \text{O}$.

1. Similarity: Both have 8 protons and 8 electrons when neutral, so they have identical chemical properties.
2. Difference: $^{16}_8 \text{O}$ has 8 neutrons, $^{18}_8 \text{O}$ has 10 neutrons. They have different masses and different nuclear stabilities.

## Common pitfalls

- **Wrong:** Confusing Rutherford scattering observations with conclusions
  - Why it fails: Students often list what was seen instead of what was deduced from the results
  - Correct: Always link each observation directly to its corresponding conclusion in exam answers
- **Wrong:** Calculating neutron number as $Z - A$ instead of $A - Z$
  - Why it fails: Mixing up the order of proton and nucleon number in the subtraction
  - Correct: Remember: neutrons = total nucleons minus protons = $A - Z$
- **Wrong:** Claiming isotopes have different chemical properties
  - Why it fails: Confusing nuclear/physical properties with chemical properties that depend on electrons
  - Correct: Isotopes of the same element have identical chemical properties, only differing in mass and nuclear stability
- **Wrong:** Forgetting electron count equals Z for neutral atoms
  - Why it fails: Assuming all atoms are ions without checking the question
  - Correct: Only adjust electron count for charged ions; for neutral atoms, electrons = Z

## Cheatsheet

| Term | Notation/Formula | Meaning |
| --- | --- | --- |
| Proton Number | $Z$ | Number of protons in nucleus |
| Nucleon Number | $A$ | Total number of protons + neutrons |
| Neutron Number | $N = A-Z$ | Number of neutrons in nucleus |
| Nuclide Notation | $^A_Z X$ | Standard notation for nuclides |
| Nuclear Radius | $R = R_0 A^{1/3}$ | $R_0 = 1.2$ fm |
| Isotopes | Same $Z$, different $A$ | Same element, different neutron count |

## What's next

Understanding atomic structure and nuclear properties is the core foundation for all further topics in CIE A-Level AS particle and nuclear physics. Now that you have mastered the basic structure, nuclide notation and key experimental evidence from Rutherford scattering, you are ready to explore the composition of nucleons and the classification of fundamental particles. This knowledge will also be critical for later topics including radioactivity, nuclear fission and fusion, where you will apply your understanding of nuclide notation to balance decay equations and calculate key nuclear properties.

- [Isotopes](https://www.owlsprep.com/study/cie-9702-u11-isotopes/)
- [Alpha, beta and gamma radiation](https://www.owlsprep.com/study/cie-9702-u11-alpha-beta-and-gamma-radiation/)
- [Radioactive decay](https://www.owlsprep.com/study/cie-9702-u11-radioactive-decay/)

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