OxfordAQA International AS and A-level Physics: Atomic structure and elementary particles (9630)

This guide covers sections 3.3.1 (Constituents of the atom) and 3.3.2 (Elementary particles) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, which is written for exams from May/June 2020 onwards. Both sections are shared content, so they apply at International AS and at International A-level. Sections 3.6 to 3.13 are International A-level only; these two are not.

The specification says the atomic mass unit is not required at International AS (it appears in section 3.12.2, International A-level), so every mass below is in kilograms or in relative units. Work through this guide, then use the revision notes and the practice set. Course pages: physics hub and printable checklist. Not sure where you stand? Try the free 10-minute diagnostics.

What you must be able to do

Section	Outcome	Level
3.3.1	Describe the simple model of the atom (proton, neutron, electron)	AS and A-level
3.3.1	Describe Rutherford scattering as evidence for a nucleus	AS and A-level
3.3.1	Appreciate how understanding of nuclear structure has changed	AS and A-level
3.3.1	Give charge and mass of the three particles in SI and relative units	AS and A-level
3.3.1	Calculate specific charge of the proton, electron, nuclei and ions	AS and A-level
3.3.1	Use proton number Z, nucleon number A and nuclide notation	AS and A-level
3.3.2	Classify particles; know each particle has an antiparticle	AS and A-level
3.3.2	Know the four named antiparticle pairs	AS and A-level
3.3.2	Describe annihilation and pair production and the energies involved	AS and A-level
3.3.2	Compare mass, charge and rest energy (MeV) of particle and antiparticle	AS and A-level

3.3.1 The simple model of the atom

An atom has a small, dense, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons, collectively known as nucleons. In a neutral atom the number of electrons matches the number of protons, so the net charge is zero.

Most of the atom’s volume is empty space. An atom is roughly 10⁻¹⁰ m in radius and a nucleus roughly 10⁻¹⁵ m. A radius ratio of about 10⁵ means a volume ratio of about 10¹⁵. Nearly all the mass sits in that tiny nucleus, because a nucleon is about 1800 times as massive as an electron.

Evidence for a nucleus: Rutherford scattering

In the scattering experiment, a narrow beam of alpha particles from a radioactive source was directed at a very thin metal foil (gold). A detector that flashed when struck, a zinc sulfide screen viewed through a microscope, was moved around the foil to count alpha particles at different angles. The apparatus was in a vacuum, because air would absorb or scatter the alpha particles. The foil was thin so that each alpha particle was scattered by at most one nucleus.

Alpha particles are positively charged. The observations, and what each tells you:

Observation	Conclusion
Most alpha particles passed straight through or deflected only slightly	The atom is largely empty space
Some alpha particles were deflected by large angles	A small, concentrated region of positive charge repels the alpha particles
A very small fraction rebounded back towards the source	The charge is concentrated in a nucleus that also holds most of the mass; a light particle cannot turn back a heavy one

Say “repelled” or “electrostatic repulsion” when you explain a large deflection. Alpha particles carry positive charge, so the nucleus must be positive.

How understanding of the nucleus has changed

You need an appreciation of how the picture developed, not exact dates. A good short account runs in this order.

• Early model. Atoms were pictured with positive charge spread through the whole atom and electrons embedded in it (the “plum pudding” model). This predicts only small deflections of alpha particles, so it cannot explain the large-angle scattering.
• Nuclear model. The scattering results replaced it with a small positive nucleus and orbiting electrons.
• Neutron. The neutron was identified by Chadwick in 1932. It explained why nuclei are heavier than their proton count alone suggests.
• Later. Protons and neutrons are now known to have internal structure. Section 3.3 does not ask for that detail.

The point to make is that a model is kept until new evidence contradicts it, then replaced by one that explains more.

Charge and mass of the proton, neutron and electron

The values below are standard values to 3 significant figures. Use the numbers a question supplies if it gives any.

Particle	Relative charge	Relative mass	Rest energy (MeV)	Charge (C)	Mass (kg)
Proton	+1	1	938.3	+1.60 × 10⁻¹⁹	1.67 × 10⁻²⁷
Neutron	0	1	939.6	0	1.67 × 10⁻²⁷
Electron	−1	about 1/1840	0.511	−1.60 × 10⁻¹⁹	9.11 × 10⁻³¹

Relative charge is measured in units of the proton charge e. Relative mass is measured in units of the proton mass. The proton and neutron masses differ in the fourth significant figure, which is why both round to 1.67 × 10⁻²⁷ kg here.

Specific charge

Specific charge is charge per unit mass:

specific charge = charge / mass, measured in C kg⁻¹

For a nucleus or ion, the charge is the net charge and the mass is the mass of the particle. For a nucleus, use mass ≈ A × (mass of one nucleon). For an ion, the electrons’ mass changes the answer by far less than the precision of the data, so you can ignore it unless asked.

Worked example 1: proton and electron

Proton: 1.60 × 10⁻¹⁹ ÷ 1.67 × 10⁻²⁷ = 9.58 × 10⁷ C kg⁻¹.

Electron: 1.60 × 10⁻¹⁹ ÷ 9.11 × 10⁻³¹ = 1.76 × 10¹¹ C kg⁻¹.

The electron’s specific charge is about 1800 times greater. The charges are equal in size, so the difference is entirely down to mass.

Worked example 2: an ion

Find the specific charge of the iron ion with 26 protons, 30 neutrons and 23 electrons.

1. The nucleon number is 26 + 30 = 56.
2. The net charge is +3e = 3 × 1.60 × 10⁻¹⁹ = 4.80 × 10⁻¹⁹ C.
3. The mass is about 56 × 1.67 × 10⁻²⁷ = 9.35 × 10⁻²⁶ kg.
4. Specific charge = 4.80 × 10⁻¹⁹ ÷ 9.35 × 10⁻²⁶ = 5.13 × 10⁶ C kg⁻¹.

For the bare nucleus of the same atom, the charge is 26e = 4.16 × 10⁻¹⁸ C, so the specific charge is 4.16 × 10⁻¹⁸ ÷ 9.35 × 10⁻²⁶ = 4.45 × 10⁷ C kg⁻¹.

Worked example 3: an ion with extra electrons

A sulfide ion has 16 protons, 16 neutrons and 18 electrons.

1. Net charge = (16 − 18)e = −2e = −3.20 × 10⁻¹⁹ C.
2. Mass ≈ 32 × 1.67 × 10⁻²⁷ = 5.34 × 10⁻²⁶ kg (adding the eighteen electrons changes this by about 0.03%).
3. Specific charge = −5.99 × 10⁶ C kg⁻¹, with the negative sign carried from the charge.

Proton number, nucleon number and nuclide notation

• Proton number Z: number of protons. It sets which element the atom is.
• Nucleon number A: protons plus neutrons.
• Neutron number = A − Z.

You must be familiar with the notation with A as a superscript and Z as a subscript to the left of the symbol:

 56
   Fe        A = 56, Z = 26
 26

This nuclide has 26 protons and 56 − 26 = 30 neutrons. A neutral atom has 26 electrons. Isotopes of an element have the same Z and different A, so they differ in neutron number.

3.3.2 Classification and antiparticles

The specification asks you to classify particles and to know that for every type of particle there is a corresponding antiparticle. It does not list further families, so concentrate on the particles named in the section. The four pairs you must know:

Particle	Antiparticle	Charge change	Rest energy (MeV)
Electron	Positron	−1 to +1	0.511 for both
Proton	Antiproton	+1 to −1	938.3 for both
Neutron	Antineutron	stays 0	939.6 for both
Neutrino	Antineutrino	stays 0	very small for both

A particle and its antiparticle have the same mass and rest energy and opposite charge. A neutral particle’s antiparticle is still a different particle, so the antineutron is not the neutron. At this level, treat the neutrino rest energy as negligible.

The specification says that E = mc² is not required in calculations. Work directly with rest energies in MeV and convert with 1 MeV = 1.60 × 10⁻¹³ J.

Annihilation

If a particle meets its own antiparticle, the pair vanishes and the energy is carried away as photons (gamma rays). Charge is conserved: the total charge is zero before and after. At least two photons are produced, travelling in opposite directions, so that momentum is conserved.

Total energy released = rest energies of both + any kinetic energy they had.

Worked example 4: electron and positron

An electron and a positron, each with kinetic energy 0.80 MeV, collide head-on and annihilate into two photons.

1. Energy of each particle = 0.511 + 0.80 = 1.311 MeV.
2. Total energy = 2 × 1.311 = 2.622 MeV.
3. The photons travel in opposite directions with equal energy: 2.622 ÷ 2 = 1.31 MeV each, which is 1.31 × 1.60 × 10⁻¹³ = 2.10 × 10⁻¹³ J.

If both particles were at rest, each photon would carry 0.511 MeV.

Worked example 5: proton and antiproton

A proton and an antiproton at rest annihilate. Total energy = 2 × 938.3 = 1876.6 MeV (about 1.88 GeV), so each of two photons has 938.3 MeV.

Pair production

Pair production is the reverse process. A photon of sufficient energy turns into a particle and its antiparticle. Charge is conserved, so the pair has zero net charge. It occurs close to a nucleus, which takes up a little momentum. You can ignore that recoil in calculations.

The minimum photon energy is the sum of the two rest energies:

• electron-positron pair: 2 × 0.511 = 1.022 MeV
• proton-antiproton pair: 2 × 938.3 = 1876.6 MeV

Any photon energy above this appears as kinetic energy of the pair.

Worked example 6

A photon of energy 3.50 MeV produces an electron-positron pair.

1. Energy used to create the pair = 1.022 MeV.
2. Kinetic energy available = 3.50 − 1.022 = 2.478 MeV.
3. Shared equally: 1.24 MeV each.

A photon of 0.80 MeV cannot produce this pair, since 0.80 is less than 1.022.

Common errors

• Writing the positron’s charge as negative, or the antiproton’s as positive.
• Saying annihilation “destroys energy”. Mass-energy is conserved; the energy goes to photons.
• Taking one rest energy as the pair-production threshold. Both particles must be created, so you need two.
• Counting A as the number of neutrons. Neutrons = A − Z.
• Dropping the sign on the specific charge of a negative ion.
• Forgetting to convert MeV to J (× 1.60 × 10⁻¹³) when a question asks for joules.
• Blaming an electron for a rebound. An electron is far too light to turn an alpha particle round.
• Treating the neutron and antineutron as the same particle because both are neutral.

For significant figures and unit checking, see the measurements revision notes.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), sections 3.3.1 and 3.3.2, Version 4.4 (for exams May/June 2020 onwards), published by OxfordAQA.